TOP 16 câu Trắc nghiệm Nhân, chia phân thức (Chân trời sáng tạo 2024) có đáp án - Toán 8

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: D

$A=\frac{x^2+y^2+xy}{x^2-y^2}:\frac{x^3-y^3}{x^2+y^2-2xy}$

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

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

Với x + y = 5 ta có $\text{A}=\frac{1}{5}$

$B=\frac{x^2-y^2}{x^2+y^2}:\frac{x^2-2xy+y^2}{x^4-y^4}$

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

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

Với x + y = 5 ta có $\text{B}=5^2=25$

$\text{A}=\frac{5^2-1}{3^2-1}:\frac{9^2-1}{7^2-1}:\frac{{13}^2-1}{{11}^2-1}:\mathrm{...}:\frac{{55}^2-1}{{53}^2-1}$

$=\frac{5^2-1}{3^2-1}\cdot \frac{7^2-1}{9^2-1}\cdot \frac{{11}^2-1}{{13}^2-1}\mathrm{...}\frac{{53}^2-1}{{55}^2-1}$

$=\frac{4.6}{2.4}\cdot \frac{6.8}{8.10}\cdot \frac{10.12}{12.14}\mathrm{...}\frac{52.54}{54.56}$

$=\frac{6}{2}\cdot \frac{6}{10}\cdot \frac{10}{14}\mathrm{...}\frac{52}{56}$

$=3\cdot \frac{6}{56}=\frac{9}{28}$

Đáp án đúng là: A

Đáp án đúng là: A

$A:B=\frac{x^3-x^2-x+11}{x-2}:\frac{x+2}{x-2}=\frac{x^3-x^2-x+11}{x-2}\cdot \frac{x-2}{x+2}$

$=\frac{x^3-x^2-x+11}{x+2}=\frac{x^3+2x^2-3x^2-6x+5x+10+1}{x+2}$

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

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

Để phân thức A chia hết cho phân thức B thì $\frac{\text{A}}{\text{B}}\in \mathbb{Z}$.

Suy ra $\left(x^2-3x+5+\frac{1}{x+2}\right)\in \mathbb{Z}$

Mà $\left(x^2-3x+5\right)\in \mathbb{Z}\forall x\in \mathbb{Z}$hay

Đáp án đúng là: A

Do $a+b+c=0$nên $a=-\left(b+c\right)$.

$=2bc-b^2-c^2=-{\left(b-c\right)}^2$

$=a^2+bc-ab-ac$$=\left(a^2-ab\right)-\left(ac-bc\right)$

$=a\left(a-b\right)-c\left(a-b\right)=\left(a-c\right)\left(a-b\right)$

Khi đó $\frac{4bc-a^2}{bc+2a^2}=\frac{-{\left(b-c\right)}^2}{\left(a-c\right)\left(a-b\right)}$.

Tương tự, ta có: $\frac{4ca-b^2}{ca+2b^2}=\frac{-{\left(c-a\right)}^2}{\left(b-a\right)\left(b-c\right)};$

$\frac{4ab-c^2}{ab+2c^2}=\frac{-{\left(a-b\right)}^2}{\left(c-a\right)\left(c-b\right)}$

$A=\frac{4bc-a^2}{bc+2a^2}\cdot \frac{4ca-b^2}{ca+2b^2}\cdot \frac{4ab-c^2}{ab+2c^2}$

$=\frac{-{\left(b-c\right)}^2}{\left(a-c\right)\left(a-b\right)}\cdot \frac{-{\left(c-a\right)}^2}{\left(b-a\right)\left(b-c\right)}\cdot \frac{-{\left(a-b\right)}^2}{\left(c-a\right)\left(c-b\right)}=1$.

Đáp án đúng là: A

$\text{A}=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\mathrm{...}\left(1-\frac{1}{n^2}\right)$

$=\frac{2^2-1}{2^2}\cdot \frac{3^2-1}{3^2}\cdot \frac{4^2-1}{4^2}\cdot \frac{5^2-1}{5^2}\cdots \frac{n^2-1}{n^2}$

$=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}.\frac{4.6}{5^2}\mathrm{...}\frac{\left(n-1\right)\left(n+1\right)}{n^2}$

$=\frac{\mathrm{1.2.3.4...}\left(n-1\right)}{\mathrm{2.3.4.5...}n}.\frac{\mathrm{3.4.5.6...}\left(n+1\right)}{\mathrm{2.3.4.5...}n}$

$=\frac{1}{n}.\frac{n+1}{2}=\frac{n+1}{2n}$

Đáp án đúng là: C

$A=\left(4x^2-16\right).\frac{7x-2}{3x+6}=\frac{\left(4x^2-16\right)\left(7x-2\right)}{3x+6}$

$=\frac{4\left(x-2\right)\left(x+2\right)7x-2}{3\left(x+2\right)}$$=\frac{4\left(x-2\right)7x-2}{3}$

$=\frac{4}{3}\left(7x^2-2x-14x+4\right)=\frac{4}{3}\left(7x^2-16x+4\right)$

Ta có: ${\left(\sqrt{7}x-\frac{8}{\sqrt{7}}\right)}^2\ge 0\forall x\Rightarrow {\left(\sqrt{7}x-\frac{8}{\sqrt{7}}\right)}^2-\frac{36}{7}\ge -\frac{36}{7}\forall x$

hay $A\ge -\frac{48}{7}$

Dấu “=” xảy ra khi và chỉ khi ${\left(\sqrt{7}x-\frac{8}{\sqrt{7}}\right)}^2=0$ hay $x=\frac{8}{7}$.

Vậy giá trị nhỏ nhất của biểu thức A là$-\frac{48}{7}$ khi $x=\frac{8}{7}$.

Đáp án đúng là: C

$\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)\left(1-\frac{1}{5^2}\right)\cdot \cdot \cdot \left(1-\frac{1}{n^2}\right)$

$=\frac{2^2-1}{2^2}\cdot \frac{3^2-1}{3^2}\cdot \frac{4^2-1}{4^2}\cdot \frac{5^2-1}{5^2}\cdots \frac{n^2-1}{n^2}$

$=\frac{1.3}{2^2}\cdot \frac{2.4}{3^2}\cdot \frac{3.5}{4^2}\cdot \frac{4.6}{5^2}\cdots \frac{\left(n-1\right)\left(n+1\right)}{n^2}$

$=\frac{\mathrm{1.2.3.4...}\left(n-1\right)}{\mathrm{2.3.4.5...}n}.\frac{\mathrm{3.4.5.6...}\left(n+1\right)}{\mathrm{2.3.4.5...}n}$

$=\frac{1}{n}.\frac{n+1}{2}=\frac{n+1}{2n}$

Áp dụng với n = 2010 , ta có:

$\text{A}=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdot \cdot \cdot \left(1-\frac{1}{{2010}^2}\right)=\frac{2010+1}{2.2010}=\frac{2011}{4020}$.

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: 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 $\text{A = }{\left(\text{x}-\text{2}\right)}^{\text{2}}\text{;\hspace{0.17em}\hspace{0.17em}B = 9}\left(\text{x + 2}\right)$.

Đáp án đúng là: B

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

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

Đáp án đúng là: D

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

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

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

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

Đáp án đúng là: D

$\frac{x+y}{x^3+x^{2}y+x{y}^2+y^3}:\frac{x^2+xy-2y^2}{x^4-y^4}$

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

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

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

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

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

Ta có $\frac{x+y}{x^3+x^{2}y+x{y}^2+y^3}:\frac{x^2+xy-2y^2}{x^4-y^4}=2$

$\frac{x+y}{x+2y}=2$

$x+y=2x+4y$

$x=-3y$