15 câu Trắc nghiệm Toán 8 Cánh diều Bài 6: Phép cộng, phép trừ phân thức đại số có đáp án

Đáp án đúng là: B

Ta có $\frac{\text{A}}{\text{B}}\text{+}\frac{\text{C}}{\text{B}}\text{=}\frac{\text{A + C}}{\text{B}}$.

Đáp án đúng là: D

$\frac{{\text{x}}^{2}}{\text{x}+2}-\frac{4}{\text{x}+2}=\frac{{\text{x}}^{2}-4}{\text{x}+2}=\frac{(\text{x}-2)(\text{x}+2)}{\text{x}+2}$

$=\frac{(\text{x}-2)(\text{x}+2):(\text{x}+2)}{(\text{x}+2):(\text{x}+2)}=\frac{\text{x}-2}{1}=\text{x}-2$.

Đáp án đúng là: B

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

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

$=\frac{-x-1-(x-1)}{{\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(x-2)}$$=\frac{-2}{x-2}=\frac{2}{2-x}$.

Đáp án đúng là: D

$\text{A}=\frac{1}{1\ .\ 2}+\frac{1}{2\ .\ 3}+\frac{1}{3\ .\ 4}+...+\frac{1}{99\ .\ 100}$

$=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+...+(\frac{1}{99}-\frac{1}{100})$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}$

$=1-\frac{1}{100}=\frac{99}{100}$.

Đáp án đúng là: D

Ta có 3y – x = 6 suy ra x = 3y – 63

Thay x = 3y – 6 vào $\text{A =}\frac{\text{x}}{\text{y}-\text{2}}\text{+}\frac{\text{2x}-\text{3y}}{\text{x}-\text{6}}$, ta được:

$\text{A}=\frac{3\text{y}-6}{\text{y}-2}+\frac{2(3\text{y}-6)-3\text{y}}{3\text{y}-6-6}$

$=\frac{3(\text{y}-2)}{\text{y}-2}+\frac{6\text{y}-12-3\text{y}}{3\text{y}-12}$

$=3+\frac{3\text{y}-12}{3\text{y}-12}=3+1=4$.

Đáp án đúng là: A

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

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

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

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

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

$=\frac{1}{2x(x-1)}$

Đáp án đúng là: C

Thay 2023 = abc vào biểu thức A ta được:

$\frac{\text{2023a}}{\text{ab + 2023a + 2023}}\text{+}\frac{\text{b}}{\text{bc + b + 2023}}\text{+}\frac{\text{c}}{\text{ac + 1 + c}}$

$=\frac{{a}^{2}bc}{ab+{a}^{2}bc+abc}+\frac{b}{bc+b+abc}+\frac{c}{ac+1+c}$

$=\frac{{a}^{2}bc}{ab(1+ac+c)}+\frac{b}{b(c+1+ac)}+\frac{c}{ac+1+c}$

$=\frac{ac}{1+ac+c}+\frac{1}{c+1+ac}+\frac{c}{ac+1+c}=1$.