cho bieur thức : \(A=\frac{x^5+x^2}{x^3-x^2+x}\)
a, rút gọn biểu thức A
b, tìm x để \(A-\left|A\right|=0\)
c, tìm x để A đạt GTNN
giải ra cụ thể mới đc tick !
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Những câu hỏi liên quan
20 tháng 7 2016
a) A=x^2+2
b) mình nghĩ x thuộc tập hợp R
c)GTNN của A=1/4 khi x=1/2
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 \(A>0\).
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
28 tháng 7 2023
a: \(A=\dfrac{x-2-2x-4+x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\dfrac{-6}{\left(x+2\right)}\cdot\dfrac{-\left(x+1\right)}{6\left(x+2\right)}=\dfrac{\left(x+1\right)}{\left(x+2\right)^2}\)
b: A>0
=>x+1>0
=>x>-1
c: x^2+3x+2=0
=>(x+1)(x+2)=0
=>x=-2(loại) hoặc x=-1(loại)
Do đó: Khi x^2+3x+2=0 thì A ko có giá trị
17 tháng 8 2019
\(a,x\ne2;x\ne-2;x\ne0\)
\(b,A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\frac{6}{x+2}\)
\(=\frac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)
\(=\frac{1}{2-x}\)
\(c,\)Để A > 0 thi \(\frac{1}{2-x}>0\Leftrightarrow2-x>0\Leftrightarrow x< 2\)
TN
12 tháng 11 2016
a)Ta có:
\(x^5+x^2=x^5-x^4+x^3+x^4-x^3+x^2\)
\(=x^2\left(x^3-x^2+x\right)+x\left(x^3-x^2+x\right)\)
\(=\left(x^2+x\right)\left(x^3-x^2+x\right)\)
Thay vào A ta có:\(A=\frac{x^5+x^2}{x^3-x^2+x}=\frac{\left(x^2+x\right)\left(x^3-x^2+x\right)}{x^3-x^2+x}=x^2+x\)
b)\(A-\left|A\right|=0\Leftrightarrow x^2+x-\left|x^2+x\right|=0\)
\(\left|x^2+x\right|=x^2+x\)\(\Leftrightarrow\orbr{\begin{cases}x^2+x=x^2+x\\x^2+x=-x^2-x\end{cases}}\)
giải tiếp chắc dễ
c)\(A=x^2+x\)\(=x^2-x+\frac{1}{4}-\frac{1}{4}\)
\(=\left(x+\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
Dấu = khi \(\left(x+\frac{1}{2}\right)^2=0\Leftrightarrow x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{2}\)
Vậy MinA=\(-\frac{1}{4}\Leftrightarrow x=-\frac{1}{2}\)
31 tháng 8 2021
a: Ta có: \(A=\left(1+\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}-x}\right)+\dfrac{5}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}:\dfrac{\sqrt{x}-1+1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{5}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}-1}{1}+\dfrac{5}{\sqrt{x}}\)
\(=\dfrac{x+4}{\sqrt{x}}\)
b: Để A=5 thì \(x+4=5\sqrt{x}\)
\(\Leftrightarrow x=16\)
NT
31 tháng 8 2021
a. \(A=\left(1+\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}-x}\right)+\dfrac{5}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}:\dfrac{1-\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}+\dfrac{5}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}.\dfrac{\sqrt{x}\left(1-\sqrt{x}\right)}{-\sqrt{x}}+\dfrac{5}{\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}}+\dfrac{5}{\sqrt{x}}=\dfrac{x-1+5}{\sqrt{x}}=\dfrac{x+4}{\sqrt{x}}\)
b. \(A=5\Leftrightarrow\dfrac{x+4}{\sqrt{x}}=5\Leftrightarrow x+4=5\sqrt{x}\Leftrightarrow x-5\sqrt{x}+4=0\)
\(\Leftrightarrow\left(\sqrt{x}-4\right)\left(\sqrt{x}-1\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=4\\\sqrt{x}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=16\\x=1\end{matrix}\right.\)
Vậy tất cả các x thỏa ycbt là x=1 hoặc x=16
c. \(A>4\Leftrightarrow\dfrac{x+4}{\sqrt{x}}>4\Leftrightarrow\dfrac{x+4}{\sqrt{x}}-4>0\Leftrightarrow\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}}>0\)
Vì \(\left(\sqrt{x}-2\right)^2\ge0\forall x\) nên \(\left\{{}\begin{matrix}\sqrt{x}-2\ne0\\\sqrt{x}>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne4\\x>0\end{matrix}\right.\)
Vậy tất cả các x thỏa mãn ycbt là x>0 và \(x\ne4\)
