Cho M=(1/x-2-1/x+2):2/x+2
a)tìm điều kiện xác định của M
b)rút gọn M
K
Khách
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Y
6 tháng 4 2023
\(M=\dfrac{4}{x+2}+\dfrac{3}{x-2}-\dfrac{5x+2}{x^2-4}\left(dkxd:x\ne\pm2\right)\)
\(=\dfrac{4}{x+2}+\dfrac{3}{x-2}-\dfrac{5x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4\left(x-2\right)+3\left(x+2\right)-\left(5x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4x-8+3x+6-5x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2}{x+2}\)
Để \(M=\dfrac{2}{5}\) thì \(\dfrac{2}{x+2}=\dfrac{2}{5}\)
Suy ra :
\(2.5=2\left(x+2\right)\)
\(\Leftrightarrow2x+4=10\)
\(\Leftrightarrow x=3\)
Vậy \(M=\dfrac{2}{5}\) thì x = 3
NH
23 tháng 12 2022
a)
\(ĐKXĐ:\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\\x^2-4\ne0\end{matrix}\right.< =>\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
b)
\(\dfrac{1}{x-2}-\dfrac{1}{x+2}+\dfrac{x^2+4x}{x^2-4}\)
\(=\dfrac{1}{x-2}-\dfrac{1}{x+2}+\dfrac{x\left(x+4\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x\left(x+4\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x+2-x+2+x^2+4x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2+4x+4}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x+2}{x-2}\)
c)
\(\dfrac{x+2}{x-2}=\dfrac{x-2+4}{x-2}=\dfrac{x-2}{x-2}+\dfrac{4}{x-2}=1+\dfrac{4}{x-2}\)
vậy M nhận giá trị nguyên thì 4⋮x-2
=> x-2 thuộc ước của 4
\(Ư\left(4\right)\in\left\{-1;1;-2;2;;4;-4\right\}\)
ta có bảng sau
| x-2 | -1 | 1 | -2 | 2 | 4 | -4 |
| x | 1(tm) | 3(tm) | 0(tm) | 4(tm) | 6(tm | -2(loại) |
12 tháng 11 2019
a) Biểu thức M xác định <=> \(\hept{\begin{cases}2-2x\ne0\\2-2x^2\ne0\end{cases}}\) <=> \(\hept{\begin{cases}2x\ne2\\2x^2\ne2\end{cases}}\) <=> \(\hept{\begin{cases}x\ne1\\x^2\ne1\end{cases}}\) <=> \(\hept{\begin{cases}x\ne1\\x\ne\pm1\end{cases}}\)
Vậy đk xác định biểu thức M <=> x \(\ne\)\(\pm\)1
b) Ta có:
M = \(\frac{x}{2-2x}-\frac{x^2+1}{2-2x^2}\)
M = \(\frac{x}{2\left(1-x\right)}-\frac{x^2+1}{2\left(1-x^2\right)}\)
M = \(\frac{x}{2\left(1-x\right)}-\frac{x^2+1}{2\left(1-x\right)\left(x+1\right)}\)
M = \(\frac{x\left(x+1\right)}{2\left(1-x\right)\left(x+1\right)}-\frac{x^2+1}{2\left(1-x\right)\left(x+1\right)}\)
M = \(\frac{x^2+x-x^2-1}{2\left(1-x\right)\left(x+1\right)}\)
M = \(\frac{x-1}{-2\left(x-1\right)\left(x+1\right)}\)
M = \(-\frac{1}{2\left(x+1\right)}\) (đk : x + 1 \(\ne\)0 => x \(\ne\)-1)
20 tháng 1 2019
a) \(M=\frac{x^4+2}{x^6+1}+\frac{x^2-1}{x^4-x^2+1}+\frac{x^2+3}{x^4+4x^2+3}\)
\(M=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{x^2+3}{x^4+3x^2+x^2+3}\)
\(M=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{x^2+3}{x^2\left(x^2+3\right)+x^2+3}\)
\(M=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{x^2+3}{\left(x^2+3\right)\left(x^2+1\right)}\)
\(M=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\)
\(M=\frac{x^4+2+x^4-1-x^4+x^2-1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\)
\(M=\frac{0+x^4+x^2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\)
\(M=\frac{x^2\left(x^2+1\right)}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\)
\(M=\frac{x^2}{x^4-x^2+1}\)
4 tháng 7 2023
a: ĐKXĐ: x<>0; x<>5; x<>5/2; x<>-5
b: \(M=\left(\dfrac{x}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{x\left(x+5\right)}\right):\dfrac{2x-5}{x\left(x+5\right)}\)
\(=\dfrac{x^2-x^2+10x-25}{x\left(x-5\right)\left(x+5\right)}\cdot\dfrac{x\left(x+5\right)}{2x-5}=\dfrac{1}{x-5}\)
10 tháng 11 2021
\(a,ĐK:x\ne\pm1;x\ne0\\ M=\dfrac{1-x+2x}{\left(1+x\right)\left(1-x\right)}:\dfrac{1-x}{x}\\ M=\dfrac{x+1}{\left(x+1\right)\left(1-x\right)}\cdot\dfrac{x}{1-x}=\dfrac{x}{\left(1-x\right)^2}\\ b,ĐK:x\ge0;x\ne4\\ N=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ N=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
Tất cả đều phải tìm điều kiện
AT
20 tháng 6 2021
a) ĐKXĐ: \(x,y\ge0\)
\(M=\dfrac{x\sqrt{y}-\sqrt{y}-y\sqrt{x}+\sqrt{x}}{1+\sqrt{xy}}=\dfrac{x\sqrt{y}-y\sqrt{x}+\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}\)
\(=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)+\left(\sqrt{x}-\sqrt{y}\right)}{1+\sqrt{xy}}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(1+\sqrt{xy}\right)}{1+\sqrt{xy}}=\sqrt{x}-\sqrt{y}\)
b) \(x=\left(1-\sqrt{3}\right)^2\Rightarrow\sqrt{x}=\sqrt{\left(1-\sqrt{3}\right)^2}=\left|1-\sqrt{3}\right|=\sqrt{3}-1\)
\(y=3-\sqrt{8}\Rightarrow\sqrt{y}=\sqrt{3-\sqrt{8}}=\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}=\left|\sqrt{2}-1\right|=\sqrt{2}-1\)
\(\Rightarrow M=\left(\sqrt{3}-1\right)-\left(\sqrt{2}-1\right)=\sqrt{3}-\sqrt{2}\)
a) Để \(M\) xác định thì \(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\\\dfrac{2}{x+2}\ne0\end{matrix}\right.\Rightarrow x\ne\pm2\)
Khi đó: \(M=\left(\dfrac{1}{x-2}-\dfrac{1}{x+2}\right):\dfrac{2}{x+2}\)
\(=\left[\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\cdot\dfrac{x+2}{2}\)
\(=\dfrac{x+2-x+2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{2}\)
\(=\dfrac{4}{2\left(x-2\right)}=\dfrac{2}{x-2}\)