\(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(\text{đ}k\text{x}\text{đ}:x\ge3\right)\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{2\sqrt{x}-9-\left(x-9\right)-\left(2x-4\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2\sqrt{x}-9-x+9-2x+4\sqrt{x}-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{5\sqrt{x}-3x+2}{x-5\sqrt{x}+6}\)

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Để \(M\in Z\) thì \(x-5\sqrt{x}+6\) thuộc ước của \(5\sqrt{x}-3x+2\)

\(\Rightarrow x-5\sqrt{x}+6=-5\sqrt{x}-3x+2\\ \Leftrightarrow x-5\sqrt{x}+6+5\sqrt{x}+3x-2=0\\ \Leftrightarrow4x-4=0\\ \Leftrightarrow4x=4\\ \Leftrightarrow x=1\)

dễ mà

Câu hỏi của Thư Nguyễn Nguyễn - Toán lớp 7 | Học trực tuyến

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\(\Leftrightarrow-x^3-x⋮x^2-2\)

\(\Leftrightarrow-x^3+2x-3x⋮x^2-2\)

\(\Leftrightarrow-3x^2⋮x^2-2\)

\(\Leftrightarrow x^2-2\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

hay \(x\in\left\{1;-1;2;-2\right\}\)

a) \(M=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)

\(\Leftrightarrow M=\left(\dfrac{-1}{x-1}+\dfrac{2}{x+1}+\dfrac{5-x}{x^2-1}\right):\dfrac{1-2x}{x^2-1}\)

\(\Leftrightarrow M=\left(\dfrac{-1}{x-1}+\dfrac{2}{x+1}+\dfrac{5-x}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1-2x}{x^2-1}\)

\(\Leftrightarrow M=\left(\dfrac{-\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{5-x}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1-2x}{x^2-1}\)

\(\Leftrightarrow M=\dfrac{-\left(x+1\right)+2\left(x-1\right)+\left(5-x\right)}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}\)

\(\Leftrightarrow M=\dfrac{-x-1+2x-2+5-x}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}\)

\(\Leftrightarrow M=\dfrac{2}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}\)

\(\Leftrightarrow M=\dfrac{2}{\left(x-1\right)\left(x+1\right)}.\dfrac{x^2-1}{1-2x}\)

\(\Leftrightarrow M=\dfrac{2\left(x^2-1\right)}{\left(x-1\right)\left(x+1\right)\left(1-2x\right)}\)

\(\Leftrightarrow M=\dfrac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(1-2x\right)}\)

\(\Leftrightarrow M=\dfrac{2}{1-2x}\)

b) \(M=\dfrac{2}{1-2x}=\dfrac{-2}{3}\)

\(\Rightarrow2.3=\left(1-2x\right).\left(-2\right)\)

\(\Rightarrow6=-2+4x\)

\(\Rightarrow4x=6-\left(-2\right)\)

\(\Rightarrow4x=6+2\)

\(\Rightarrow4x=8\)

\(\Rightarrow x=8:4\)

\(\Rightarrow x=2\)

Vậy \(M=\dfrac{-2}{3}\) thì \(x=2\)

c) Để \(M=\dfrac{2}{1-2x}\in Z\) \(\Leftrightarrow2⋮1-2x\)

\(\Rightarrow1-2x\in U\left(2\right)=\left\{-1;1;-2;2\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}1-2x=-1\Rightarrow x=1\\1-2x=1\Rightarrow x=0\\1-2x=-2\Rightarrow x=1,5\\1-2x=2\Rightarrow x=-0,5\end{matrix}\right.\)

Mà \(x\in Z\)

\(\Rightarrow x\in\left\{1;0\right\}\)

Vậy \(x=1\) hoặc \(x=0\) thì \(M\in Z\)

a) M = \(\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)

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

= \(\left(\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}-\dfrac{5-x}{\left(1-x\right)\left(1+x\right)}\right).\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)

= \(\dfrac{1+x+2-2x-5+x}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)\(=\dfrac{-2}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)

= \(\dfrac{2}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)

=\(\dfrac{2}{1-2x}\)

b) M = \(\dfrac{-2}{3}\Leftrightarrow\dfrac{2}{1-2x}=\dfrac{-2}{3}\)

=> 2 . 3 = -2 (1 - 2x) (tích chéo)

=> 6 = -2 + 4x

=> 6 + 2 - 4x = 0

=> 8 - 4x = 0

=> 4x = 8

=> x = 2 (thỏa mãn đkxđ)

Vậy để M = \(\dfrac{-2}{3}\) thì x = 2

giups mình với

TXĐ : \(x\ne\pm2\)

\(M=\left[\dfrac{1}{x+2}-\dfrac{2}{x-2}+\dfrac{x}{\left(x-2\right)\left(x+2\right)}\right]:\dfrac{10-x^2+\left(x-2\right)\left(x+2\right)}{x+2}\)

\(=\dfrac{x-2-2\left(x+2\right)+x}{\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{10-x^2+x^2-1}\)

\(=\dfrac{x-2-2x-4+x}{x-2}.\dfrac{1}{6}\)

\(=\dfrac{-6}{x-2}.\dfrac{1}{6}=\dfrac{1}{2-x}\)

1/2+1/3<x<=1+1/2+1/5

=>5/6<x<=1+7/10

=>5/6<x<17/10

mà x là số nguyên

nên x=1