Cho biểu thức A = \(\frac{1}{x^2-3x+2}\).
Tìm gt của x để A đạt GTLN của biêu thức
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HN
19 tháng 2 2019
a) \(-ĐKXĐ:x\ne\pm2;1\)
Rút gọn : \(A=\left(\frac{1}{x+2}-\frac{2}{x-2}-\frac{x}{4-x^2}\right):\frac{6\left(x+2\right)}{\left(2-x\right)\left(x+1\right)}\)
\(=\left(\frac{1}{x+2}+\frac{-2}{x-2}+\frac{x}{x^2-4}\right).\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\left[\frac{x-2}{\left(x-2\right)\left(x+2\right)}+\frac{\left(-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x}{\left(x-2\right)\left(x+2\right)}\right]\)\(.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\left[\frac{x-2-2x-4+x}{\left(x-2\right)\left(x+2\right)}\right].\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)\(=\frac{x+1}{\left(x+2\right)^2}\)
b) \(A>0\Leftrightarrow\frac{x+1}{\left(x+2\right)^2}>0\Leftrightarrow\orbr{\begin{cases}x+1< 0;\left(x+2\right)^2< 0\left(voly\right)\\x+1>0;\left(x+2\right)^2>0\end{cases}}\)
\(\Leftrightarrow x>1;x>-2\Leftrightarrow x>1\)
Vậy với mọi x thỏa mãn x>1 thì A > 0
c) Ta có : \(x^2+3x+2=0\Leftrightarrow x^2+x+2x+2=0\)
\(\Leftrightarrow x\left(x+1\right)+2\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)
Vậy x = -1;-2
4 tháng 6 2019
a)Có A=\(\left(\frac{1}{x+2}-\frac{2}{x-2}-\frac{x}{4-x^2}\right):\frac{6\left(x+2\right)}{\left(2-x\right)\left(x+1\right)}\)(ĐKXĐ \(x\ne2,-2,-1\))
=\(\left(\frac{2-x}{\left(2-x\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(2-x\right)\left(x+2\right)}-\frac{x}{\left(2-x\right)\left(2+x\right)}\right):\frac{6\left(x+2\right)}{\left(2-x\right)\left(x+1\right)}\)
=\(\frac{2-x+2x+4-x}{\left(2-x\right)\left(x+2\right)}.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
=\(\frac{6\left(2-x\right)\left(x+1\right)}{6\left(2-x\right)\left(x+2\right)^2}\)
=\(\frac{x+1}{\left(x+2\right)^2}\)
b)Có A=\(\frac{x+1}{\left(x+2\right)^2}\)
Để A>0 <=> x+1>0 <=>x>-1
c) Có x2+3x+2=0
<=> x2+2x+x+2=0
<=> x(x+2)+(x+2)=0
<=>(x+1)(x+2)=0
<=> x=-1 hoặc x=-2
H
0
17 tháng 4 2022
B1: ĐXXĐ: \(x\ne\pm2;x\ne-1\)
\(=\left(\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}-\dfrac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{x}{\left(x+2\right)\left(x-2\right)}\right):\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)
\(=\left(\dfrac{x-2-2x-2+x}{\left(x+2\right)\left(x-2\right)}\right):\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)
\(=\dfrac{-4}{\left(x+2\right)\left(x-2\right)}:\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)
\(=\dfrac{-4}{\left(x+2\right)\left(x-2\right)}.\dfrac{\left(x-2\right)\left(x+1\right)}{-6\left(x+2\right)}=\dfrac{2\left(x+1\right)}{3\left(x+2\right)^2}\)
b, \(A=\dfrac{2\left(x+1\right)}{3\left(x+2\right)^2}>0\)
\(\Leftrightarrow2x+2>0\) (vì \(3\left(x+2\right)^2\ge0\forall x\))
\(\Leftrightarrow x>-1\).
-Vậy \(x\in\left\{x\in Rlx>-1;x\ne2\right\}\) thì \(A>0\).
22 tháng 3 2020
a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b, Ta có : \(A=\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\frac{\sqrt{x}-1}{2}\)
=> \(A=\left(\frac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{x\sqrt{x}-1}\right)\left(\frac{2}{\sqrt{x}-1}\right)\)
=> \(A=\left(\frac{x-2\sqrt{x}+1}{x\sqrt{x}-1}\right)\left(\frac{2}{\sqrt{x}-1}\right)\)
=> \(A=\frac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
=> \(A=\frac{2}{x+\sqrt{x}+1}\)
c, Ta có : \(A=\frac{2}{x+\sqrt{x}+1}=\frac{2}{\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}}\)
Ta thấy \(\frac{2}{\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}}>0\forall x\ne1\)
NV
0
Lớp 7 nhẩy lên lớp 9
\(\frac{1}{A}=\left(x-\frac{3}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\)
\(\hept{\begin{cases}A\le2\\A\left(\frac{3}{2}\right)=2\end{cases}}\) \(\Rightarrow x=\frac{3}{2}\)