\(A=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\frac{\left(1-x\right)^2}{2},\)
a) Rút gọn A
b)tìm x để A dương
c)tìm giá trị lớn nhất của A
K
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1 tháng 8 2017
E mới 7 - 8 thui !!! nhưng e sẽ cố giúp
a) \(A=\frac{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(x-1\right)}{\left(x-1\right)\left(x+2\sqrt{x}+1\right)}.\frac{1-x^2}{2}\)
\(=\frac{x\sqrt{x}-3\sqrt{x}-2-x\sqrt{x}+\sqrt{x}-2x+2}{\left(x-1\right)\left(\sqrt{x}+1\right)^2}.\frac{1-x^2}{2}\)
\(=\frac{-2\sqrt{x}-2x}{\left(x-1\right)\left(\sqrt{x}+1\right)^2}.\frac{1-x^2}{2}\)
\(=\frac{-2\sqrt{x}\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(1-x\right)\left(x+1\right)}{2}\)
\(=\frac{2\left(\sqrt{x}+1\right)\left(x-1\right)\left(x+1\right)\sqrt{x}}{2\left(\sqrt{x}+1\right)\left(x-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(x+1\right)}{\sqrt{x}+1}\)
b )
ĐKXĐ : \(x\ge0\)
Vì \(\sqrt{x}+1>0\forall x\) Để \(A=\frac{\sqrt{x}\left(x+1\right)}{\sqrt{x}+1}>0\) \(\Leftrightarrow\sqrt{x}\left(x+1\right)>0\)
\(\Rightarrow\hept{\begin{cases}\sqrt{x}\ne0\\x+1>0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x>-1\end{cases}}}\) Mà theo đxxd thì \(x\ge0\) nên \(x>0\)
Vậy với \(x>0\) thì \(A>0\)
c ) Lớp 7 chưa bt làm :((
1 tháng 8 2017
E ghi rõ nèk
\(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+1\right)-\left(x-1\right)\left(\sqrt{x}+2\right)}{\left(x-1\right)\left(x+2\sqrt{x}+1\right)}\)
\(=\frac{\left(x\sqrt{x}+2x+\sqrt{x}-2x-4\sqrt{x}-2\right)-\left(x\sqrt{x}+2x-\sqrt{x}-2\right)}{\left(x-1\right)\left(x+2\sqrt{x}+1\right)}\)
\(=\frac{x\sqrt{x}-3\sqrt{x}-2-x\sqrt{x}-2x+\sqrt{x}-2}{\left(x-1\right)\left(x+2\sqrt{x}+1\right)}\)
MN
4 tháng 4 2020
Bài 1 :
a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)
\(A=\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{1}{\sqrt{x}-2}\)
\(\Leftrightarrow A=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
b) Để \(A< -1\)
\(\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< -1\)
\(\Leftrightarrow\sqrt{x}-2< -\sqrt{x}-1\)
\(\Leftrightarrow2\sqrt{x}< 1\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{2}\)
\(\Leftrightarrow x< \frac{1}{4}\)
Vậy để \(A< -1\Leftrightarrow x< \frac{1}{4}\)
ĐK \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
a, \(A=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\frac{\left(\sqrt{x}-1\right)^2.\left(\sqrt{x}+1\right)^2}{2}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2.\left(\sqrt{x}+1\right)^2}{2}\)
\(=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2.\left(\sqrt{x}+1\right)^2}{2}=-\sqrt{x}\left(\sqrt{x}-1\right)\)
b. \(A>0\Rightarrow-\sqrt{x}\left(\sqrt{x}-1\right)>0\Rightarrow\sqrt{x}-1< 0\Rightarrow0\le x< 1\)
c. \(A=-\left(x-\sqrt{x}\right)=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\Rightarrow A\le\frac{1}{4}\)
Vậy \(MaxA=\frac{1}{4}\Leftrightarrow\sqrt{x}-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}\)