13 Bài tập Phép cộng và phép trừ phân thức đại số (có đáp án) - Kết nối tri thức Trắc nghiệm Toán 8

$\frac{\text{A}}{\text{B}}\text{+}\frac{\text{C}}{\text{B}}\text{=}\frac{\text{A + C}}{\text{B}}$

Đáp án cần chọn là: B

Quy đồng mẫu thức $\frac{\text{A}}{\text{B}}$ và $\frac{\text{C}}{\text{D}}$:

$\frac{\text{A}}{\text{B}}\text{=}\frac{\text{AD}}{\text{BD}}\text{;\hspace{0.17em}}\frac{\text{C}}{\text{D}}\text{=}\frac{\text{BC}}{\text{BD}}$

Do đó $\frac{\text{A}}{\text{B}}-\frac{\text{C}}{\text{D}}\text{=}\frac{\text{AD}}{\text{BD}}-\frac{\text{BC}}{\text{BD}}$$\text{=}\frac{\text{AD}-\text{BC}}{\text{BD}}$

Đáp án cần chọn là: C

Phân thức đối của phân thức $\frac{2\text{x}-1}{\text{x}+1}$ là $-\frac{\text{2x}-\text{1}}{\text{x + 1}}\text{=}\frac{\text{1}-\text{2x}}{\text{x + 1}}$

Đáp án cần chọn là: B

$\frac{{\text{x}}^2}{\text{x}+2}-\frac{4}{\text{x}+2}=\frac{{\text{x}}^2-4}{\text{x}+2}$$=\frac{\left(\text{x}-2\right)\left(\text{x}+2\right)}{\text{x}+2}=\frac{\left(\text{x}-2\right)\left(\text{x}+2\right):\left(\text{x}+2\right)}{\left(\text{x}+2\right):\left(\text{x}+2\right)}$$=\frac{\text{x}-2}{1}=\text{x}-2$

Đáp án cần chọn là: D

$\frac{\text{x}+2}{3\text{x}+5}-\text{A}=\frac{\text{x}-1}{2}$

$\Rightarrow A=\frac{\text{x}+2}{3\text{x}+5}-\frac{\text{x}-1}{2}$

$=\frac{\left(x+2\right)2}{2\left(3x+5\right)}-\frac{\left(x-1\right)\left(3x+5\right)}{2\left(3x+5\right)}$

$=\frac{2x+4}{2\left(3x+5\right)}-\frac{3x^2-3x+5x-5}{2\left(3x+5\right)}$

$=\frac{\left(2x+4\right)-\left(3x^2-3x+5x-5\right)}{2\left(3x+5\right)}$

$=\frac{\left(2x+4\right)-\left(3x^2+2x-5\right)}{2\left(3x+5\right)}$

$=\frac{2x+4-3x^2-2x+5}{2\left(3x+5\right)}$

$=\frac{-3x^2+9}{2\left(3x+5\right)}$

Đáp án cần chọn là: C

A. $\frac{\text{x}-1}{\text{x}+1}-\frac{\text{x}+1}{\text{x}-1}=\frac{{\left(x-1\right)}^2-{\left(x+1\right)}^2}{\left(x+1\right)\left(x-1\right)}$

$=\frac{\left(x^2-2x+1\right)-\left(x^2+2x+1\right)}{x^2-1}$

$=\frac{x^2-2x+1-x^2-2x-1}{x^2-1}$

$=\frac{-4x}{x^2-1}\ne \frac{4x}{x^2-1}$

B. $\frac{\text{2x}-1}{\text{x}+1}-\frac{\text{2x}+1}{\text{x}-1}$$=\frac{\left(2x-1\right)\left(x-1\right)-\left(2x+1\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}$

$=\frac{\left(2x^2-x-2x+1\right)-\left(2x^2+x+2x+1\right)}{x^2-1}$

$=\frac{\left(2x^2-3x+1\right)-\left(2x^2+3x+1\right)}{x^2-1}$

$=\frac{2x^2-3x+1-2x^2-3x-1}{x^2-1}$

$=\frac{-6x}{x^2-1}\ne \frac{4x}{x^2-1}$

C. $\frac{\text{x}+1}{\text{x}-1}-\frac{\text{x}-1}{\text{x}+1}=\frac{{\left(x+1\right)}^2-{\left(x-1\right)}^2}{\left(x-1\right)\left(x+1\right)}$

$=\frac{\left(x^2+2x+1\right)-\left(x^2-2x+1\right)}{x^2-1}$

$=\frac{x^2+2x+1-x^2+2x-1}{x^2-1}$

$=\frac{4x}{x^2-1}$

D. $\frac{\text{2x}+1}{\text{x}-1}-\frac{\text{2x}-1}{\text{x}+1}$$=\frac{\left(2x+1\right)\left(x+1\right)-\left(2x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}$

$=\frac{\left(2x^2+x+2x+1\right)-\left(2x^2-x-2x+1\right)}{x^2-1}$

$=\frac{\left(2x^2+3x+1\right)-\left(2x^2-3x+1\right)}{x^2-1}$

$=\frac{2x^2+3x+1-2x^2+3x-1}{x^2-1}$

$=\frac{6x}{x^2-1}\ne \frac{4x}{x^2-1}$

Vậy phân thức $\frac{4x}{x^2-1}$ là kết quả của phép tính $\frac{\text{x}+1}{\text{x}-1}-\frac{\text{x}-1}{\text{x}+1}$

Đáp án cần chọn là: C

$\frac{3\text{x}+21}{{\text{x}}^2-9}+\frac{2}{\text{x}+3}-\frac{3}{\text{x}-3}$

$=\frac{3x+21}{\left(x-3\right)\left(x+3\right)}+\frac{2}{\text{x}+3}+\frac{-3}{\text{x}-3}$

$=\frac{3\text{x}+21}{\left(x-3\right)\left(x+3\right)}+\frac{2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}$$+\frac{-3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}$

$=\frac{3\text{x}+21+2\left(x-3\right)-3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}$

$=\frac{3x+21+2x-6-3x-9}{\left(x-3\right)\left(x+3\right)}$

$=\frac{2x+6}{\left(x-3\right)\left(x+3\right)}$

$=\frac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}$

$=\frac{2}{x-3}$

Đáp án cần chọn là: D

$\text{A = }\frac{\text{2x}-\text{1}}{{\text{6x}}^{\text{2}}-\text{6x}}-\frac{\text{3}}{{\text{4x}}^{\text{2}}-\text{4}}$

$=\frac{2x-1}{6x\left(x-1\right)}-\frac{3}{4\left(x^2-1\right)}$

$=\frac{2x-1}{6x\left(x-1\right)}-\frac{3}{4\left(x-1\right)\left(x+1\right)}$

$=\frac{2\left(2x-1\right)\left(x+1\right)-3.3x}{12x\left(x-1\right)\left(x+1\right)}$

$=\frac{2\left(2x^2-x+2x-1\right)-9x}{12x\left(x-1\right)\left(x+1\right)}$

$=\frac{2\left(2x^2+x-1\right)-9x}{12x\left(x-1\right)\left(x+1\right)}$

$=\frac{4x^2+2x-2-9x}{12x\left(x-1\right)\left(x+1\right)}$

$=\frac{4x^2-7x-2}{12x\left(x-1\right)\left(x+1\right)}$

Đáp án cần chọn là: C

A. $\frac{\text{x}}{\text{x}-\text{y}}\text{+}\frac{\text{y}}{\text{x + y}}\text{+}\frac{{\text{2y}}^{\text{2}}}{{\text{x}}^{\text{2}}-{\text{y}}^{\text{2}}}$$\text{=}\frac{x}{x-y}+\frac{y}{x+y}+\frac{2y^2}{\left(x-y\right)\left(x+y\right)}$

$=\frac{x\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}+\frac{y\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}$$+\frac{2y^2}{\left(x-y\right)\left(x+y\right)}$

$=\frac{x^2+xy+xy-y^2+2y^2}{\left(x-y\right)\left(x+y\right)}$

$=\frac{x^2+2xy+y^2}{\left(x-y\right)\left(x+y\right)}$

$=\frac{{\left(x+y\right)}^2}{\left(x-y\right)\left(x+y\right)}$

$=\frac{x+y}{x-y}\ne \frac{x-y}{x+y}$

B. $\frac{\text{1}}{\text{2x + 1}}-\frac{\text{1}}{\text{3x + 2}}$$\text{=}\frac{3x+2}{\left(\text{2x + 1}\right)\left(\text{3x + 2}\right)}-\frac{2x+1}{\left(\text{2x + 1}\right)\left(\text{3x + 2}\right)}$

$=\frac{\left(3x+2\right)-\left(2x+1\right)}{\left(\text{2x + 1}\right)\left(\text{3x + 2}\right)}$$=\frac{3x+2-2x-1}{\left(\text{2x + 1}\right)\left(\text{3x + 2}\right)}=\frac{\text{x + 1 }}{\left(\text{2x + 1}\right)\left(\text{3x + 2}\right)}$

C. $\frac{2\text{x}+3}{6}+\frac{\text{x}+1}{9}=\frac{3\left(2x+3\right)}{18}+\frac{2\left(x+1\right)}{18}$$=\frac{6x+9}{18}+\frac{2x+2}{18}$

$=\frac{6x+9+2x+2}{18}=\frac{8x+11}{18}\ne \frac{3\text{x}+4}{18}$

D. $\frac{\text{3}}{\text{x}-\text{1}}\text{+}\frac{\text{2x}}{{\text{x}}^{\text{2}}-\text{1}}=\frac{\text{3}}{\text{x}-\text{1}}\text{+}\frac{\text{2x}}{\left(x-1\right)\left(x+1\right)}$$=\frac{3\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{2x}{\left(x-1\right)\left(x+1\right)}$

$\text{=}\frac{3x+3}{\left(x-1\right)\left(x+1\right)}+\frac{2x}{\left(x-1\right)\left(x+1\right)}$$=\frac{3x+3+2x}{\left(x-1\right)\left(x+1\right)}=\frac{5x+3}{\left(x-1\right)\left(x+1\right)}\ne \frac{\text{3x + 5}}{{\text{x}}^{\text{2}}-\text{1}}$

Đáp án cần chọn là: B

A. $\frac{11\text{x}+13}{3\text{x}-3}+\frac{15\text{x}+17}{4-4\text{x}}$$=\frac{11x+13}{3\left(x-1\right)}+\frac{15x+17}{4\left(1-x\right)}$$=\frac{11x+13}{3\left(x-1\right)}-\frac{15x+17}{4\left(1-x\right)}$

$=\frac{4\left(11x+13\right)-3\left(15x+17\right)}{12\left(\text{x}-1\right)}$$=\frac{44x+52-45x-51}{12\left(\text{x}-1\right)}=\frac{-x+1}{12\left(\text{x}-1\right)}$$=\frac{-1}{12}\ne \frac{-\text{x}-1}{12\left(\text{x}-1\right)}$

B. $\frac{\text{xy}}{{\text{x}}^{\text{2}}-{\text{y}}^{\text{2}}}-\frac{{\text{x}}^{\text{2}}}{{\text{y}}^{\text{2}}-{\text{x}}^{\text{2}}}\text{=}\frac{xy}{{\text{x}}^{\text{2}}-{\text{y}}^{\text{2}}}+\frac{x^2}{{\text{x}}^{\text{2}}-{\text{y}}^{\text{2}}}$$=\frac{xy+x^2}{{\text{x}}^{\text{2}}-{\text{y}}^{\text{2}}}=\frac{x\left(y+x\right)}{\left(\text{x}-\text{y}\right)\left(x+y\right)}=\frac{\text{x}}{\text{x}-\text{y}}$

C. $\frac{1}{3\text{x}+4}-\frac{1}{3\text{x}+5}$$=\frac{3x+5}{\left(3\text{x}+4\right)\left(3\text{x}+5\right)}-\frac{3x+4}{\left(3\text{x}+4\right)\left(3\text{x}+5\right)}$

$=\frac{\left(3x+5\right)-\left(3x+4\right)}{\left(3\text{x}+4\right)\left(3\text{x}+5\right)}$$=\frac{3x+5-3x-4}{\left(3\text{x}+4\right)\left(3\text{x}+5\right)}=\frac{1}{\left(3\text{x}+4\right)\left(3\text{x}+5\right)}$

D. $\frac{\text{1}}{\text{x + 2}}-\frac{\text{1}}{\left(\text{x + 2}\right)\left(\text{4x + 7}\right)}$$=\frac{4x+7}{\left(\text{x + 2}\right)\left(\text{4x + 7}\right)}-\frac{1}{\left(\text{x + 2}\right)\left(\text{4x + 7}\right)}$

$\text{=}\frac{4x+7-1}{\left(\text{x + 2}\right)\left(\text{4x + 7}\right)}=\frac{4x+6}{\left(\text{x + 2}\right)\left(\text{4x + 7}\right)}$$=\frac{\text{2}\left(\text{2x + 3}\right)}{\left(\text{x + 2}\right)\left(\text{4x + 7}\right)}$

Đáp án cần chọn là: A

$\text{A = }\frac{{\text{2x}}^{\text{2}}\text{+ x}-\text{3}}{{\text{x}}^{\text{3}}-\text{1}}-\frac{\text{x}-\text{5}}{{\text{x}}^{\text{2}}\text{+ x + 1}}-\frac{\text{7}}{\text{x}-\text{1}}$

$=\frac{2x^2+x-3}{{\text{x}}^{\text{3}}-\text{1}}-\left(\frac{x-5}{x^2+x+1}+\frac{7}{x-1}\right)$

$=\frac{2x^2+x-3}{\left(x-1\right)\left(x^2+x+1\right)}$$-[\frac{\left(x-5\right)\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}+\frac{7\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x-1\right)}]$

$=\frac{2x^2+x-3}{\left(x-1\right)\left(x^2+x+1\right)}$$-[\frac{x^2-5x-x+5}{\left(x^2+x+1\right)\left(x-1\right)}+\frac{7x^2+7x+7}{\left(x^2+x+1\right)\left(x-1\right)}]$

$=\frac{2x^2+x-3}{\left(x-1\right)\left(x^2+x+1\right)}$$-\frac{x^2-5x-x+5+7x^2+7x+7}{\left(x-1\right)\left(x^2+x+1\right)}$

$=\frac{2x^2+x-3}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{8x^2+x+12}{\left(x-1\right)\left(x^2+x+1\right)}$

$=\frac{\left(2x^2+x-3\right)-\left(8x^2+x+12\right)}{\left(x-1\right)\left(x^2+x+1\right)}$

$=\frac{2x^2+x-3-8x^2-x-12}{\left(x-1\right)\left(x^2+x+1\right)}$

$=\frac{-6x^2-15}{\left(x-1\right)\left(x^2+x+1\right)}$

Đáp án cần chọn là: D

$\frac{\text{x}-\text{1}}{{\text{x}}^{\text{2}}-\text{2x}}\text{+ A = }\frac{-\text{x}-\text{1}}{{\text{x}}^{\text{2}}-\text{2x}}$

$\Rightarrow A=\frac{-x-1}{{\text{x}}^{\text{2}}-\text{2x}}-\frac{\text{x}-\text{1}}{{\text{x}}^{\text{2}}-\text{2x}}$

$=\frac{-x-1-\left(x-1\right)}{{\text{x}}^{\text{2}}-\text{2x}}$

$=\frac{-x-1-x+1}{{\text{x}}^{\text{2}}-\text{2x}}$

$=\frac{-2x}{{\text{x}}^{\text{2}}-\text{2x}}=\frac{-2x}{x\left(x-2\right)}$

$=\frac{-2}{x-2}=\frac{2}{2-x}$

Đáp án cần chọn là: B

$\text{A = }\frac{\text{5}}{\text{2x}}\text{+}\frac{\text{2x}-\text{3}}{\text{2x}-\text{1}}\text{+}\frac{{\text{4x}}^{\text{2}}\text{+ 3}}{{\text{8x}}^{\text{2}}-\text{4x}}$

$=\frac{\text{5}}{\text{2x}}\text{+}\frac{\text{2x}-\text{3}}{\text{2x}-\text{1}}\text{+}\frac{4x^2}{4x\left(2x-1\right)}$

$=\frac{\left(6x+7\right)\left(2x-1\right)}{4x\left(2x-1\right)}=\frac{6x+7}{4x}$

$=\frac{5.2\left(2x-1\right)}{4x\left(2x-1\right)}+\frac{4x\left(2x-3\right)}{4x\left(2x-1\right)}+\frac{4x^2+3}{4x\left(2x-1\right)}$

$=\frac{20x-10}{4x\left(2x-1\right)}+\frac{8x^2-12x}{4x\left(2x-1\right)}+\frac{4x^2+3}{4x\left(2x-1\right)}$

$=\frac{20x-10+8x^2-12x+4x^2+3}{4x\left(2x-1\right)}$

$=\frac{12x^2+8x-7}{4x\left(2x-1\right)}$

$=\frac{12x^2-6x+14x-7}{4x\left(2x-1\right)}$

$=\frac{6x\left(2x-1\right)+7\left(2x-1\right)}{4x\left(2x-1\right)}$

Với $\text{x}=\frac{1}{4}$ ta có: $\text{A}=\frac{6\cdot \frac{1}{4}+7}{4\cdot \frac{1}{4}}=\frac{\frac{3}{2}+7}{1}$$=\frac{3}{2}+7=\frac{3}{2}+\frac{14}{2}=\frac{17}{2}$

Đáp án cần chọn là: D