cho p=(x^2/2x-4+x-2/2x+4+-8/x^2-4):4/x-2
a, tìm đkxđ
b,rút gọn p
c, tính gtri của bt p khi x=-1/1/3
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21 tháng 8 2021
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne2\\x\ne4\\x\ge0\end{matrix}\right.\)
24 tháng 6 2021
`a)P=(x/(x+2)-(x^3-8)/(x^3+8)*(x^2-2x+4)/(x^2-4)):4/(x+2)`
`đk:x ne 0,x ne -2`
`P=(x/(x+2)-((x-2)(x^2+2x+4))/((x+2)(x^2-2x+4))*(x^2-2x+4)/((x-2)(x+2)))*(x+2)/4`
`=(x/(x+2)-(x^2+2x+4)/(x+2)^2)*(x+2)/4`
`=(x^2+2x-x^2-2x-4)/(x+2)^2*(x+2)/4`
`=-4/(x+2)^2*(x+2)/4`
`=-1/(x+2)`
`b)P<0`
`<=>-1/(x+2)<0`
Vì `-1<0`
`<=>x+2>0`
`<=>x> -2`
`c)P=1/x+1(x ne 0)`
`<=>-1/(x+2)=1/x+1`
`<=>1/x+1+1/(x+2)=0``
`<=>x+2+x(x+2)+x=0`
`<=>x^2+4x+2=0`
`<=>` \(\left[ \begin{array}{l}x=\sqrt2-2\\x=-\sqrt2-2\end{array} \right.\)
`d)|2x-1|=3`
`<=>` \(\left[ \begin{array}{l}2x=4\\2x=-2\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2(l)\\x=-1(tm)\end{array} \right.\)
`x=-1=>P=-1/(-1+2)=-1`
`e)P=-1/(x+2)` thì nhỏ nhất cái gì nhỉ?
24 tháng 6 2021
a) đk: \(x\ne-2;2\)
\(P=\left[\dfrac{x}{x+2}-\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}.\dfrac{x^2-2x+4}{\left(x-2\right)\left(x+2\right)}\right]:\dfrac{4}{x+2}\)
= \(\left[\dfrac{x}{x+2}-\dfrac{x^2+2x+4}{\left(x+2\right)^2}\right].\dfrac{x+2}{4}\)
= \(\dfrac{x^2+2x-x^2-2x-4}{\left(x+2\right)^2}.\dfrac{x+2}{4}\) = \(\dfrac{-4}{4\left(x+2\right)}=\dfrac{-1}{x+2}\)
b) Để P < 0
<=> \(\dfrac{-1}{x+2}< 0\)
<=> x +2 > 0
<=> x > -2 ( x khác 2)
c) Để P= \(\dfrac{1}{x}+1\)
<=> \(\dfrac{-1}{x+2}=\dfrac{1}{x}+1\)
<=> \(\dfrac{1}{x}+\dfrac{1}{x+2}+1=0\)
<=> \(\dfrac{x+2+x+x\left(x+2\right)}{x\left(x+2\right)}=0\)
<=> x2 + 4x + 2 = 0
<=> (x+2)2 = 2
<=> \(\left[{}\begin{matrix}x=\sqrt{2}-2\left(c\right)\\x=-\sqrt{2}-2\left(c\right)\end{matrix}\right.\)
d) Để \(\left|2x-1\right|=3\)
<=> \(\left[{}\begin{matrix}2x-1=3< =>x=2\left(l\right)\\2x-1=-3< =>x=-1\left(c\right)\end{matrix}\right.\)
Thay x = -1, ta có:
P = \(\dfrac{-1}{-1+2}=-1\)
5 tháng 8 2021
a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)
b) Ta có: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\)
\(=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
d) Để A>0 thì \(\sqrt{x}-2>0\)
hay x>4
23 tháng 3 2020
\(A=\frac{2x^2+4x}{x^3-4x}+\frac{x^2-4}{x^2+2x}+\frac{2}{2-x}\left(x\ne0;x\ne\pm2\right)\)
\(A=\frac{2x^2+4x}{x\left(x^2-4\right)}+\frac{\left(x-2\right)\left(x+2\right)}{x\left(x+2\right)}-\frac{2}{x-2}\)
\(A=\frac{2x^2+4x}{x\left(x-2\right)\left(x+2\right)}+\frac{\left(x-2\right)^2\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}-\frac{2x\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{2x^2+4x}{x\left(x-2\right)\left(x+2\right)}+\frac{x^3-2x^2-4x+8}{x\left(x-2\right)\left(x+2\right)}-\frac{2x^2+4x}{x\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{2x^2+4x+x^3-2x^2-4x+8-2x^2-4x}{x\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{-2x^2-4x+8}{x\left(x-2\right)\left(x+2\right)}=\frac{-2x\left(x+2\right)+8}{x\left(x-2\right)\left(x+2\right)}=\frac{-2x+8}{x\left(x-2\right)}\)
Vậy \(A=\frac{-2x+8}{x\left(x-2\right)}\left(x\ne0;x\ne\pm2\right)\)
b) \(A=\frac{-2x+8}{x\left(x-2\right)}\left(x\ne0;x\ne\pm2\right)\)
Ta có: x=4 (tmđk) thay vào A ta có:
\(A=\frac{-2\cdot4+8}{4\left(4-2\right)}=\frac{-8+8}{4\cdot2}=\frac{0}{8}=0\)
Vậy A=0 với x=4
AL
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