TOP 15 câu Trắc nghiệm Phép cộng, phép trừ phân thức đại số (Cánh diều 2024) có đáp án - Toán 8

Đáp án đúng là: D

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

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

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

Mà $\frac{2}{\text{x}+3}+\frac{3}{{\text{x}}^2-9}=0$ nên $\frac{2x-3}{\left(x-3\right)\left(x+3\right)}=0$

2x - 3 = 0

2x = 3

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

Đáp án đúng là: D

A = $\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\mathrm{...}+\frac{1}{99.100}$

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

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

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

Đáp án đúng là: D

Ta có 3y – x = 6 suy ra x = 3y – 63

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

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

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

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

Đáp án đúng là: 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

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

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

Đáp án đúng là: C

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

Suy ra 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 đúng là: D

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

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

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

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

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

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

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

Đáp án đúng là: C

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 đúng là: A

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

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

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

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

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

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

Đáp án đúng là: A

A = $\frac{\text{ab}}{\left(\text{b}-\text{c}\right)\left(\text{c}-\text{a}\right)}\text{+}\frac{\text{bc}}{\left(\text{c}-\text{a}\right)\left(\text{a}-\text{b}\right)}\text{+}\frac{\text{ac}}{\left(\text{a}-\text{b}\right)\left(\text{b}-\text{c}\right)}$

= $\frac{ab\left(a-b\right)bc\left(b-c\right)+ac\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}$

= $\frac{ab\left(a-b\right)bc\left(b-c\right)+ac\left(c-b+b-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}$

= $\frac{\left(ab-ac\right)\left(a-b\right)+\left(bc-ac\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}$

= $\frac{a\left(b-c\right)\left(a-b\right)-c\left(a-b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}$

= $\frac{\left(a-c\right)\left(a-b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}$ = -1

Đáp án đúng là: A

Điều kiện: $\left\{\begin{array}{c}x-1\ne 0\\ x+1\ne 0\end{array}\right.\iff \left\{\begin{array}{c}x\ne 1\\ x\ne -1\end{array}\right.$

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

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

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

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

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

= x² + x + 1 - x + 1 = x² + 2

Ta có $x^2\ge 0\forall x\in ℝ$ nên $x^2+2\ge 2\forall x\in ℝ$ hay $\text{A}\ge 2$

Dấu “=” xảy ra khi và chỉ khi x² = 0 hay x = 0.

Vậy min A = 0 khi x = 0 .

Đáp án đúng là: C

Thay 2023 = abc vào biểu thức A ta được:

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

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

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

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

Đáp án đúng là: B

Ta có x + y + z = x + y + z + 0

= x + y + z + $\frac{{\text{x}}^{\text{2}}}{\text{y + z}}\text{+}\frac{{\text{y}}^{\text{2}}}{\text{x + z}}\text{+}\frac{{\text{z}}^{\text{2}}}{\text{x + y}}$

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

= $x\left(1+\frac{x}{\text{y + z}}\right)+y\left(1+\frac{y}{\text{x + z}}\right)+z\left(1+\frac{z}{\text{x + y}}\right)$

= $x\left(\frac{x+y+z}{\text{y + z}}\right)+y\left(\frac{x+y+z}{\text{x + z}}\right)+z\left(\frac{x+y+z}{\text{x + y}}\right)$

= (x + y + z)$\left(\frac{x}{\text{y + z}}+\frac{y}{\text{x + z}}+\frac{z}{\text{x + y}}\right)$

Khi đó x + y + z = (x + y + z)$\left(\frac{x}{\text{y + z}}+\frac{y}{\text{x + z}}+\frac{z}{\text{x + y}}\right)$.

Do đó $\frac{x}{\text{y + z}}+\frac{y}{\text{x + z}}+\frac{z}{\text{x + y}}=1$.

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

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

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

Đáp án đúng là: D

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:

A = $\frac{6\cdot \frac{1}{4}+7}{4\cdot \frac{1}{4}}=\frac{\frac{3}{2}+7}{1}=\frac{3}{2}+7=\frac{3}{2}+\frac{14}{2}=\frac{17}{2}$.