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Đáp án:
a.$A=\dfrac{x^2+2}{x+2}$
b.$A=\dfrac{33}{28}$
c.$x\in\{-1,-3,0,-4,1,-5,4,-8\}$
Giải thích các bước giải:
a.$A=\dfrac{1-x}{2+x}-\dfrac{x-1}{x-2}+\dfrac{4-x^3}{4-x^2}$
$\rightarrow A=\dfrac{1-x}{2+x}-\dfrac{1-x}{2-x}+\dfrac{4-x^3}{4-x^2}$
$\rightarrow A=(1-x)(\dfrac{1}{2+x}-\dfrac{1}{2-x})+\dfrac{4-x^3}{4-x^2}$
$\rightarrow A=(1-x)(\dfrac{2-x}{(2+x)(2-x)}-\dfrac{2+x}{(2+x)(2-x)})+\dfrac{4-x^3}{(2+x)(2-x)}$
$\rightarrow A=(1-x)\dfrac{2-x-(2+x)}{(2+x)(2-x)}+\dfrac{4-x^3}{(2+x)(2-x)}$
$\rightarrow A=(1-x)\dfrac{-2x}{(2+x)(2-x)}+\dfrac{4-x^3}{(2+x)(2-x)}$
$\rightarrow A=\dfrac{-2x(1-x)}{(2+x)(2-x)}+\dfrac{4-x^3}{(2+x)(2-x)}$
$\rightarrow A=\dfrac{2x^2-2x}{(2+x)(2-x)}+\dfrac{4-x^3}{(2+x)(2-x)}$
$\rightarrow A=\dfrac{2x^2-2x+4-x^3}{(2+x)(2-x)}$
$\rightarrow A=\dfrac{x^3-2x^2+2x-4}{(2+x)(x-2)}$
$\rightarrow A=\dfrac{(x-2)x^2+2(x-2)}{(2+x)(x-2)}$
$\rightarrow A=\dfrac{(x-2)(x^2+2)}{(2+x)(x-2)}$
$\rightarrow A=\dfrac{x^2+2}{x+2}$
b.$x=-0,25\rightarrow A=\dfrac{(-0,25)^2+2}{-0,25+2}=\dfrac{33}{28}$
c.$A=\dfrac{x^2+2}{x+2}=\dfrac{x^2-4+6}{x+2}=\dfrac{(x-2)(x+2)+6}{x+2}=x-2+\dfrac{6}{x+2}$
$\begin{split}\rightarrow A\in Z&\leftrightarrow x-2+\dfrac{6}{x+2}\in Z\\&\leftrightarrow \dfrac{6}{x+2}\in Z\\&\leftrightarrow x+2\in U(6)=\{1,-1,2,-2,3,-3,6,-6\}\\&\leftrightarrow x\in\{-1,-3,0,-4,1,-5,4,-8\}\end{split}$
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