Đáp án đúng là: B

$\text{B}=\frac{1}{\text{x}-23}-\frac{1}{\text{x}-3}=\frac{x-3}{\left(x-23\right)\left(x-3\right)}-\frac{x-23}{\left(x-23\right)\left(x-3\right)}$

$=\frac{\left(x-3\right)-\left(x-23\right)}{\left(x-23\right)\left(x-3\right)}=\frac{x-3-x+23}{\left(x-23\right)\left(x-3\right)}=\frac{20}{\left(x-23\right)\left(x-3\right)}$

Với x = 2023, ta có:

$\text{B}=\frac{20}{\left(2023-23\right)\left(2023-3\right)}=\frac{20}{2000.2020}$

$=\frac{20}{20.100.2020}=\frac{1}{100.2020}=\frac{1}{202000}$.

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 $\frac{2x-3}{\left(x-3\right)\left(x+3\right)}=0$

$2x-3=0$

$2x=3$

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

Vậy $x=\frac{3}{2}$.

Đáp án đúng là: D

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

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

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

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

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

$=\frac{2x^2+x-3-8x^2-x-12}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{-6x^2-15}{\left(x-1\right)\left(x^2+x+1\right)}$

Đáp án đúng là: D

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

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

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

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

$=\frac{20x-10+8x^2-12x+4x^2+3}{4x\left(2x-1\right)}=\frac{12x^2+8x-7}{4x\left(2x-1\right)}$

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

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

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 đúng là: D

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

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

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

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

Câu 6. Chọn khẳng định đúng.

A. $\frac{A}{B}-\frac{C}{D}=\frac{A-C}{B-D}$

B. $\frac{A}{B}-\frac{C}{D}=\frac{AD}{BC}$

C. $\frac{A}{B}-\frac{C}{D}=\frac{AD-BC}{BD}$

D. $\frac{A}{B}-\frac{C}{D}=\frac{A-C}{BD}$

Câu 7. Phân thức đối của phân thức $\frac{2\text{x}-1}{\text{x}+1}$là

A. $\frac{2\text{x}+1}{\text{x}+1}$

B. $\frac{1-2\text{x}}{\text{x}+1}$

C. $\frac{\text{x}+1}{2\text{x}-1}$

D. $\frac{\text{x}+1}{1-2\text{x}}$

Câu 8. Thực hiện phép tính sau:$\frac{{\text{x}}^2}{\text{x}+2}-\frac{4}{\text{x}+2}\left(\text{x}\ne -2\right)$

A. x + 2

B. 2x

C. x

D. x – 2

Câu 9. Tìm phân thức A thỏa mãn: $\frac{x-1}{x^2-2x}+A=\frac{-x-1}{x^2-2x}$.

A. $\frac{2}{\text{x}-2}$

B. $\frac{2}{2-\text{x}}$

C. $\frac{1}{\text{x}}$

D. $\frac{1}{\text{x}+2}$

Câu 10. Cho $A=\frac{2x-1}{6x^2-6x}-\frac{3}{4x^2-4}$. Phân thức thu gọn của A có tử thức là:

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

B. $4{\text{x}}^2-7\text{x}+2$

C. $4{\text{x}}^2-7\text{x}-2$

D. $\text{12x}\left(\text{x}-\text{1}\right)\left(\text{x + 1}\right)$

Đáp án đúng là: D

Ta có 3y – x = 6 nên 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:

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

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

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

Đáp án đúng là: A

$\text{A}=\frac{10}{\left(\text{x}+2\right)\left(3-\text{x}\right)}-\frac{12}{\left(3-\text{x}\right)\left(3+\text{x}\right)}-\frac{1}{\left(\text{x}+3\right)\left(\text{x}+2\right)}$

$=\frac{10}{\left(\text{x}+2\right)\left(3-\text{x}\right)}-\frac{11x+27}{\left(3-x\right)\left(x+3\right)\left(x+2\right)}$

$=\frac{10\left(x+3\right)}{\left(3-x\right)\left(x+2\right)\left(x+3\right)}-\frac{11x+27}{\left(3-x\right)\left(x+2\right)\left(x+3\right)}$

$=\frac{10\left(x+3\right)-\left(11x+27\right)}{\left(3-x\right)\left(x+2\right)\left(x+3\right)}=\frac{10x+30-11x-27}{\left(3-x\right)\left(x+2\right)\left(x+3\right)}$

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

Tại $\text{x}=-\frac{3}{4}$ta có $\text{A}=\frac{1}{\left(\frac{-3}{4}+2\right)\left(\frac{-3}{4}+3\right)}=\frac{1}{\frac{5}{4}\cdot \frac{9}{4}}=\frac{1}{\frac{45}{16}}=\frac{16}{45}$

Vậy 0 < A < 1.

Đáp án đúng là: D

$\text{A = }\frac{{\text{6x}}^{\text{2}}\text{+ 8x + 7}}{{\text{x}}^{\text{3}}-\text{1}}\text{+}\frac{\text{x}}{{\text{x}}^{\text{2}}\text{+ x + 1}}-\frac{\text{6}}{\text{x}-\text{1}}$

$=\frac{6x^2+8x+7}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x}{x^2+x+1}-\frac{6}{x-1}$

$=\frac{6x^2+8x+7+x\left(x-1\right)-6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}$

$=\frac{6x^2+8x+7+x^2-x-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}$

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

Để $\text{A}\in \mathbb{Z}$hay $\frac{1}{\text{x}-1}\in \mathbb{Z}$ thì x – 1 ∈ Ư(1) = {−1; 1}.

Ta có bảng sau:

Đáp án đúng là: C

Điều kiện:

$\text{A = }\frac{\text{3}}{\text{x}-\text{3}}-\frac{{\text{x}}^{\text{2}}}{\text{4}-{\text{x}}^{\text{2}}}-\frac{\text{4x}-\text{12}}{{\text{x}}^{\text{3}}-{\text{3x}}^{\text{2}}-\text{4x + 12}}$

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

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

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

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

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

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

Để 

Ta có bảng sau:

Vậy có 3 giá trị của x để biểu thức A có giá trị là một số nguyên.

Đáp án đúng là: A

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

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

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

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

Đáp án đúng là: A

$\frac{1}{1-\text{x}}+\frac{1}{1+\text{x}}+\frac{2}{1+{\text{x}}^2}+\frac{4}{1+{\text{x}}^4}+\frac{8}{1+{\text{x}}^8}$

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

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

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

$=\frac{4}{1-{\text{x}}^4}+\frac{4}{1+{\text{x}}^4}+\frac{8}{1+{\text{x}}^8}$

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

$=\frac{8}{1-{\text{x}}^8}+\frac{8}{1+{\text{x}}^8}=\frac{8\left(1+{\text{x}}^8\right)+8\left(1-{\text{x}}^8\right)}{\left(1-{\text{x}}^8\right)\left(1+{\text{x}}^8\right)}=\frac{16}{1-x^{16}}$.

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