M= \(\left(\frac{a}{a^2-4}+\frac{2}{2-a}+\frac{1}{a+2}\right):\left(\frac{10-a^2}{a+2}-2+a\right)\)
a, Rút gọn M
b, Tìm x thuộc z để M thuộc Z
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A= \(\frac{1,11+0,19-1,3.2}{2,06+0,54}-\left(\frac{1}{2}+\frac{1}{3}\right):2=\frac{-\frac{131}{100}}{\frac{13}{5}}-\frac{5}{6}:2\)
=\(-\frac{131}{260}-\frac{5}{12}=-\frac{359}{390}\)
B= \(\left(5\frac{7}{8}-2\frac{1}{4}-0,5\right):2\frac{23}{26}=\left(\frac{47}{8}-\frac{9}{4}-\frac{1}{2}\right):\frac{75}{26}=\frac{25}{8}.\frac{26}{75}=\frac{13}{12}\)
b) ta có : A=\(-\frac{359}{390}\approx-0,9\)
B= \(\frac{13}{12}\approx1,08\)
=> A<x<B mà x nguyên => x=0 hoặc x=1
a: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
b: \(M=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right):\dfrac{x^2-4+10-x^2}{x+2}\)
\(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{6}\)
\(=\dfrac{-1}{x-2}\)
d: Để M nguyên thì \(x-2\in\left\{1;-1\right\}\)
hay \(x\in\left\{3;1\right\}\)
a: ĐKXĐ: \(a\notin\left\{0;1;-1\right\}\)
\(A=\dfrac{a^2}{\left(a-1\right)\left(a+1\right)}-\dfrac{a^2}{a^2+1}\cdot\dfrac{a^2+1}{a\left(a+1\right)}\)
\(=\dfrac{a^2}{\left(a-1\right)\left(a+1\right)}-\dfrac{a}{a+1}\)
\(=\dfrac{a^2-a^2+a}{\left(a-1\right)\left(a+1\right)}=\dfrac{a}{\left(a-1\right)\left(a+1\right)}=\dfrac{a}{a^2-1}\)
b: Để A=3 thì \(3a^2-3=a\)
\(\Leftrightarrow2a^2=3\)
hay \(a\in\left\{\dfrac{\sqrt{6}}{2};-\dfrac{\sqrt{6}}{2}\right\}\)
\(\left(\frac{1}{x+1}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3}{x^2-x+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
\(\left(\frac{x^2-x+1}{x^3+1}-\frac{3}{x^3+1}+\frac{3\left(x+1\right)}{x^3+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
\(\left(\frac{x^2-x+1-3+3x+3}{x^3+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
tới đây bạn biến đổi tiếp, gõ = cái này lâu quá, gõ mathtype nhanh hơn