13 Bài tập Tính chất cơ bản của phân thức đại số (có đáp án) - Kết nối tri thức Trắc nghiệm Toán 8

Theo tính chất cơ bản của phân thức đại số, ta có:

$\frac{\text{A}}{\text{B}}\text{=}\frac{\text{A.M}}{\text{B.M}}$ (với M khác đa thức 0) $\Rightarrow \frac{\text{A}}{\text{B}}\text{=}\frac{\text{A}\left(-\text{1}\right)}{\text{B}\left(-\text{1}\right)}\text{=}\frac{-\text{A}}{-\text{B}}$

$\frac{\text{A}}{\text{B}}\text{=}\frac{\text{A:N}}{\text{B:N}}$ (với N là một nhân tử chung, N khác đa thức 0)

Mệnh đề $\frac{\text{A}}{\text{B}}\text{=}\frac{\text{A + M}}{\text{B + M}}$ sai. Ví dụ: $\frac{2}{3}\ne \frac{3}{4}=\frac{2+1}{3+1}$

Đáp án cần chọn là: D

$\frac{{\text{x}}^{\text{2}}-\text{7x + 12}}{{\text{x}}^{\text{2}}-\text{6x + 9}}\text{=}\frac{{\text{x}}^{\text{2}}-\text{4x}-\text{3x + 12}}{{\left(\text{x}-\text{3}\right)}^{\text{2}}}$$\text{=}\frac{\text{x}\left(\text{x}-\text{4}\right)-\text{3}\left(\text{x}-\text{4}\right)}{{\left(\text{x}-\text{3}\right)}^{\text{2}}}\text{=}\frac{\left(\text{x}-\text{4}\right)\left(\text{x}-\text{3}\right)}{{\left(\text{x}-\text{3}\right)}^{\text{2}}}\text{=}\frac{\text{x}-\text{4}}{\text{x}-\text{3}}$

Đáp án cần chọn là: C

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

B. $\frac{9{\text{x}}^2+12\text{x}+4}{4-9{\text{x}}^2}=\frac{{\left(\text{3x + 2}\right)}^{\text{2}}}{\left(\text{2}-\text{3x}\right)\left(\text{2 + 3x}\right)}$ $\text{=}\frac{{\left(\text{3x + 2}\right)}^{\text{2}}\text{:}\left(\text{2 + 3x}\right)}{\left(\text{2}-\text{3x}\right)\left(\text{2 + 3x}\right)\text{:}\left(\text{2 + 3x}\right)}$ $\text{=}\frac{\text{2 + 3x}}{\text{2}-\text{3x}}\ne \frac{\text{2}-\text{3x}}{\text{2 + 3x}}$

C. $\frac{\text{4}-{\text{9x}}^{\text{2}}}{{\left(\text{2 + 3x}\right)}^{\text{2}}}\text{=}\frac{\left(\text{2}-\text{3x}\right)\left(\text{2 + 3x}\right)}{{\left(\text{2 + 3x}\right)}^{\text{2}}}$ $\text{=}\frac{\left(\text{2}-\text{3x}\right)\left(\text{2 + 3x}\right)\text{:}\left(\text{2 + 3x}\right)}{{\left(\text{2 + 3x}\right)}^{\text{2}}\text{:}\left(\text{2 + 3x}\right)}\text{=}\frac{\text{2}-\text{3x}}{\text{2 + 3x}}$

D. $\frac{\text{4}-\text{6x}}{\text{4 + 6x}}\text{=}\frac{\text{2}\left(\text{2}-\text{3x}\right)}{\text{2}\left(\text{2 + 3x}\right)}$$\text{=}\frac{\text{2}\left(\text{2}-\text{3x}\right)\text{:2}}{\text{2}\left(\text{2 + 3x}\right)\text{:2}}\text{=}\frac{\text{2}-\text{3x}}{\text{2 + 3x}}$

Đáp án cần chọn là: B

Mẫu thức của hai phân thức $\frac{5}{2\left(\text{x}-3\right)},\frac{7}{{\left(\text{x}-3\right)}^3}$ là 2(x – 3) và (x – 3)³ nên mẫu thức chung có phần hệ số là 2, phần biến số là (x - 3)³

$\Rightarrow$Mẫu thức chung là 2(x - 3)³

Đáp án cần chọn là: D

Mẫu chung của các phân thức là xyz

Nhân tử phụ của $\frac{\text{1}}{\text{x}}$ là $\text{yz}\Rightarrow \frac{\text{1}}{\text{x}}\text{=}\frac{\text{yz}}{\text{xyz}}$

Nhân tử phụ của $\frac{2}{y}$ là $\text{xz}\Rightarrow \frac{\text{2}}{\text{y}}\text{=}\frac{\text{2xz}}{\text{xyz}}$

Nhân tử phụ của $\frac{\text{3}}{\text{z}}$ là $\text{xy}\Rightarrow \frac{\text{3}}{\text{z}}\text{=}\frac{\text{3xy}}{\text{xyz}}$
Vậy quy đồng mẫu số các phân thức $\frac{\text{1}}{\text{x}}\text{,\hspace{0.17em}}\frac{\text{2}}{\text{y}}\text{,\hspace{0.17em}}\frac{\text{3}}{\text{z}}$ ta được $\frac{\text{1}}{\text{x}}\text{=}\frac{\text{yz}}{\text{xyz}}\text{,\hspace{0.17em}}$$\frac{\text{2}}{\text{y}}\text{=}\frac{\text{2xz}}{\text{xyz}}\text{,\hspace{0.17em}}$$\frac{\text{3}}{\text{z}}\text{=}\frac{\text{3xy}}{\text{xyz}}$

Đáp án cần chọn là: A

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

Đáp án cần chọn là: D

$\frac{{\left(\text{a + b}\right)}^{\text{2}}-{\text{c}}^{\text{2}}}{\text{a + b + c}}\text{=}\frac{\text{[}\left(\text{a + b}\right)-\text{c][}\left(\text{a + b}\right)\text{ + c]}}{\text{a + b + c}}$$\text{=}\frac{\left(\text{a + b}-\text{c}\right)\left(\text{a + b + c}\right)}{\text{a + b + c}}\text{=}\frac{\text{a + b}-\text{c}}{\text{1}}$

Vậy khi rút gọn phân thức $\frac{{\left(\text{a + b}\right)}^{\text{2}}-{\text{c}}^{\text{2}}}{\text{a + b + c}}$ ta được phân thức có tử là: a + b – c

Đáp án cần chọn là: D

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

Vậy mẫu thức của phân thức $\frac{{\text{x}}^{\text{2}}-\text{xy}-\text{x + y}}{{\text{x}}^{\text{2}}\text{+ xy}-\text{x}-\text{y}}$  sau khi thu gọn là x + y

Đáp án cần chọn là: C

Ta có các phân thức $\frac{\text{1}}{\text{2}-\text{x}}\text{,\hspace{0.17em}}\frac{\text{2x + 1}}{{\left(\text{x}-\text{2}\right)}^{\text{2}}}\text{,\hspace{0.17em}}\frac{{\text{3x}}^{\text{2}}-\text{1}}{{\text{x}}^{\text{2}}\text{+ 4x + 4}}$ có mẫu thức lần lượt là: $2-\text{x},{\left(\text{x}-2\right)}^2$ và ${\text{x}}^2+4\text{x}+4={\left(\text{x}+2\right)}^2$ nên mẫu thức chung là ${\left(\text{x}-2\right)}^2{\left(\text{x}+2\right)}^2$

Đáp án cần chọn là: C

Ta có: ${\text{x}}^3+1=\left(\text{x}+1\right)\left({\text{x}}^2-\text{x}+1\right);$ $3\text{x}+3=3\left(\text{x}+1\right);$ $2{\text{x}}^2-2\text{x}+2=2\left({\text{x}}^2-\text{x}+1\right)$ và BCNN(2;3) = 6 nên mẫu thức chung của các phân thức $\frac{\text{1}}{{\text{x}}^{\text{3}}\text{+ 1}}\text{,\hspace{0.17em}}\frac{\text{2}}{\text{3x + 3}}\text{,\hspace{0.17em}}\frac{\text{x}}{{\text{2x}}^{\text{2}}-\text{2x + 2}}$ là $6\left(\text{x}+1\right)\left({\text{x}}^2-\text{x}+1\right)=6\left({\text{x}}^3+1\right)$

Nhân tử phụ của $\frac{1}{{\text{x}}^3+1}$ là 6. $\Rightarrow \frac{\text{1}}{{\text{x}}^{\text{3}}\text{+ 1}}\text{=}\frac{\text{6}}{\text{6}\left({\text{x}}^{\text{3}}\text{+ 1}\right)}$

Nhân tử phụ của $\frac{\text{2}}{\text{3x + 3}}$ là $2\left({\text{x}}^2-\text{x}+1\right).$ $\Rightarrow \frac{\text{2}}{\text{3x + 3}}\text{=}\frac{\text{2.2}\left({\text{x}}^{\text{2}}-\text{x + 1}\right)}{\text{3}\left(\text{x + 1}\right)\text{2}\left({\text{x}}^{\text{2}}-\text{x + 1}\right)}$$\text{=}\frac{{\text{4x}}^{\text{2}}-\text{4x + 4}}{\text{6}\left({\text{x}}^{\text{3}}\text{+ 1}\right)}$

Nhân tử phụ của $\frac{\text{x}}{{\text{2x}}^{\text{2}}-\text{2x + 2}}$ là 3(x + 1).

$\Rightarrow \frac{\text{x}}{{\text{2x}}^{\text{2}}-\text{2x + 2}}$$\text{=}\frac{\text{x.3}\left(\text{x + 1}\right)}{\text{2}\left({\text{x}}^{\text{2}}-\text{x + 1}\right)\text{3}\left(\text{x + 1}\right)}\text{=}\frac{{\text{3x}}^{\text{2}}\text{+ 3x}}{\text{6}\left({\text{x}}^{\text{3}}\text{+ 1}\right)}$

Đáp án cần chọn là: C

${\text{a}}^{\text{2}}{\text{x + 2ax + 4 = a}}^{\text{2}}$

$\iff {\text{a}}^{\text{2}}{\text{x + 2ax = a}}^{\text{2}}-\text{4}$

$\iff \text{x}\left({\text{a}}^{\text{2}}\text{+ 2a}\right){\text{ = a}}^{\text{2}}-\text{4}$

$\iff \text{x = }\frac{{\text{a}}^{\text{2}}-\text{4}}{{\text{a}}^{\text{2}}\text{+ 2a}}$

$\iff \text{x = }\frac{\left(\text{a}-\text{2}\right)\left(\text{a + 2}\right)}{\text{a}\left(\text{a + 2}\right)}$

$\iff \text{x = }\frac{\text{a}-\text{2}}{\text{a}}$

Đáp án cần chọn là: B

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

Tại x = 1 ta có $\text{A}=\frac{1-2}{2.1}=\frac{-1}{2}$

Đáp án cần chọn là: D

$\text{A = }\frac{{\text{2a}}^{\text{2}}{\text{+ 8ab + 8b}}^{\text{3}}}{\text{a + 2b}}$$\text{=}\frac{\text{2}\left({\text{a}}^{\text{2}}{\text{+ 4ab + 4b}}^{\text{2}}\right)}{\text{a + 2b}}\text{=}\frac{\text{2}{\left(\text{a + 2b}\right)}^{\text{2}}}{\text{a + 2b}}$$\text{=2}\left(\text{a + 2b}\right)\text{ = 2.5 = 10}$

Đáp án cần chọn là: D