30 câu Trắc nghiệm Phép nhân, phép chia phân thức đại số (có đáp án 2024) – Toán 8 Cánh diều

Đáp án đúng là: A

$\frac{\text{A}}{\text{B}}\cdot \frac{\text{C}}{\text{D}}\text{=}\frac{\text{A\hspace{0.17em}.\hspace{0.17em}C}}{\text{B\hspace{0.17em}.\hspace{0.17em}D}}$

Đáp án đúng là: B

Phân thức nghịch đảo của phân thức $\frac{2\text{x}+1}{\text{x}+2}$ là $\frac{\text{x}+2}{2\text{x}+1}$.

Đáp án đúng là: C

Muốn chia phân thức $\frac{\text{A}}{\text{B}}$ cho phân thức $\frac{\text{C}}{\text{D}}\left(\frac{\text{C}}{\text{D}}\ne \text{0}\right)$ ta nhân $\frac{\text{A}}{\text{B}}$ với phân thức nghịch đảo của $\frac{\text{C}}{\text{D}}$.

Đáp án đúng là: D

$\frac{3\text{x}+12}{4\text{x}-16}\cdot \frac{8-2\text{x}}{\text{x}+4}=\frac{3\left(\text{x}+4\right)}{4\left(\text{x}-4\right)}\cdot \frac{2\left(4-\text{x}\right)}{\text{x}+4}$

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

Đáp án đúng là: C

$\frac{4\text{x}+12}{{\left(\text{x}+4\right)}^2}:\frac{3\left(\text{x}+3\right)}{\text{x}+4}=\frac{4\left(\text{x}+3\right)}{{\left(\text{x}+4\right)}^2}:\frac{3\left(\text{x}+3\right)}{\text{x}+4}$

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

Đáp án đúng là: A

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

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

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

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

Vậy kết quả của phép chia $\frac{{\text{x}}^3+1}{{\text{x}}^2+2\text{x}+1}:\frac{3{\text{x}}^2-3\text{x}+3}{{\text{x}}^2-1}$ có tử thức là x − 1.

Đáp án đúng là: A

$\frac{\text{x + 3}}{{\text{x}}^{\text{2}}-\text{4}}\text{.}\frac{\text{8}-{\text{12x + 6x}}^{\text{2}}-{\text{x}}^{\text{3}}}{\text{9x + 27}}$

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

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

Vậy A = (x - 2)2; B = 9(x + 2)

Đáp án đúng là: B

$\frac{{\text{x}}^{\text{2}}-\text{25}}{\text{3x + 9}}\cdot \frac{\text{7}}{\text{x + 5}}$ = $\frac{\left(\text{x}-\text{5}\right)\left(\text{x + 5}\right)}{\text{3}\left(\text{x + 3}\right)}\cdot \frac{\text{7}}{\text{x + 5}}$ = $\frac{\text{7}\left(\text{x}-\text{5}\right)}{\text{3}\left(\text{x + 3}\right)}$

Do đó,  khi thực hiện phép nhân $\frac{{\text{x}}^2-25}{3\text{x}+9}\cdot \frac{3}{\text{x}+5}$ ta được phân thức có mẫu thức là 3(x+3).

Đáp án đúng là: D

$\frac{x+3y}{4x+8y}.A=\frac{x^2-9y^2}{x+2y}$

A = $\frac{x^2-9y^2}{x+2y}:\frac{x+3y}{4x+8y}$

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

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

Đáp án đúng là: B

A = $\frac{5x+10}{x-6}:\frac{x-2}{2x+12}\cdot \frac{2x-4}{x^2-36}$

= $\frac{5\left(x+2\right)}{x-6}:\frac{x-2}{x+6}\cdot \frac{2\left(x-2\right)}{\left(x-6\right)\left(x+6\right)}$

= $\frac{5\left(x+2\right)}{x-6}\cdot \frac{x+6}{x-2}\cdot \frac{2\left(x-2\right)}{\left(x-6\right)\left(x+6\right)}$ = $\frac{10\left(x+2\right)}{{\left(x-6\right)}^2}$

Đáp án đúng là: B

$\frac{3x+15}{x^2-4}:\frac{x+5}{x-2}=\frac{3\left(x+5\right)}{\left(x-2\right)\left(x+2\right)}:\frac{x+5}{x-2}$

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

Khi đó $\frac{3x+15}{x^2-4}:\frac{x+5}{x-2}=1$

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

x + 2 = 3

x = 1 (TM)

Đáp án đúng là: C

Điều kiện: $x\ne 6;x\ne 5$

$\frac{x^2+10x+25}{x+6}:\left(x+5\right)=\frac{{\left(x+5\right)}^2}{x+6}:\frac{x+5}{1}$

$=\frac{{\left(x+5\right)}^2}{x+6}.\frac{1}{x+5}=\frac{x+5}{x+6}=1-\frac{1}{x+6}$

Để $\frac{x^2+10x+25}{x+6}:\left(x+5\right)$ nguyên thì (x + 6) ∈ Ư(1) = {±1}.

Ta có bảng sau

Vậy để $\frac{x^2+10x+25}{x+6}:\left(x+5\right)$ thì x = −7.

Đáp án đúng là: C

A = $\frac{x-6}{x^2+1}\cdot \frac{3x^2-3x+3}{x^2-36}+\frac{x-6}{x^2+1}\cdot \frac{3x}{x^2-36}$

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

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

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

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

Khi x = 994, ta có A = $\frac{3}{994+6}=\frac{3}{1000}$.

Đáp án đúng là: B

A = $\frac{2x^3{y}^2}{x^2{y}^5{z}^2}:\frac{5x^{2}y}{4x^2{y}^5}:\frac{-8x^3{y}^2{z}^3}{15x^5{y}^2}$

= $\frac{2x^3{y}^2}{x^2{y}^5{z}^2}.\frac{4x^2{y}^5}{5x^{2}y}.\frac{15x^5{y}^2}{-8x^3{y}^2{z}^3}$

= $\frac{120x^{10}{y}^9}{-40x^7{y}^8{z}^5}=-\frac{3x^{3}y}{z^5}$.

Với x = 4, y = 1, z = −2 ta có: A = $\frac{\left(-3\right).4^3.1}{{\left(-2\right)}^5}$ = 6.

Đáp án đúng là: C

A = $\frac{27-x^3}{5x+5}:\frac{2x-6}{3x+3}=\frac{\left(3-x\right)\left(x^2+3x+9\right)}{5\left(x+1\right)}:\frac{2\left(x-3\right)}{3\left(x+1\right)}$

$=\frac{\left(3-x\right)\left(x^2+3x+9\right)}{5\left(x+1\right)}\cdot \frac{3\left(x+1\right)}{2\left(x-3\right)}=-\frac{3\left(x^2+3x+9\right)}{10}$

$=-\frac{3}{10}\left[\left(x^2+3x+\frac{9}{4}\right)+\frac{27}{4}\right]=-\frac{3}{10}\left[{\left(x+\frac{3}{2}\right)}^2+\frac{27}{4}\right]$

Ta có ${\left(x+\frac{3}{2}\right)}^2\ge 0\forall x\in ℝ$ nên ${\left(x+\frac{3}{2}\right)}^2+\frac{27}{4}\ge \frac{27}{4}\forall x\in ℝ$

Khi đó $\left(-\frac{3}{10}\right)\left[{\left(x+\frac{3}{2}\right)}^2+\frac{27}{4}\right]\le \left(-\frac{3}{10}\right)\frac{27}{4}=-\frac{81}{40}$ hay $A\le -\frac{81}{40}$

Dấu “=” xảy ra khi và chỉ khi ${\left(x+\frac{3}{2}\right)}^2=0$ nên $x+\frac{3}{2}=0$ hay $x=-\frac{3}{2}$.

Vậy giá trị lớn nhất của biểu thức A = $\frac{27-x^3}{5x+5}:\frac{2x-6}{3x+3}$ là $-\frac{81}{40}$ khi và chỉ khi x = $-\frac{3}{2}$.