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a. ĐKXĐ: x\(\ne1\) x, \(\ne-1\)
b. \(\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{x^2-2x+1}{2}\)
=\(\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{2}\)
=\(\left(\dfrac{\left(\sqrt{x}-2\right).\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)^2}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{2}\)
\(\left(\dfrac{\left(\sqrt{x}-2\right).\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{2}=\left(\dfrac{\left(\sqrt{x}-2\right)-\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{2}=\left(\dfrac{\sqrt{x}-2-\sqrt{x}-2}{\sqrt{x}+1}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{2}=\dfrac{4}{\sqrt{x}-1}.\dfrac{\sqrt{x}-1^2}{2}=2\left(\sqrt{x}-1\right)=2\sqrt{x}-2\)
c. khi x=0,16 thì G=\(2\sqrt{x}-2=2\sqrt{0,16}-2=2.0,4-2=0,8-2=-1,2\)
Câu 2:
a: ĐKXĐ: x>=0; x<>1
b: \(=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=-\dfrac{2\sqrt{x}}{2}\cdot\left(\sqrt{x}-1\right)=-\sqrt{x}\left(\sqrt{x}-1\right)\)
c: Thay x=4/25 vào G, ta được:
\(G=-\dfrac{2}{5}\cdot\left(\dfrac{2}{5}-1\right)=\dfrac{-2}{5}\cdot\dfrac{-3}{5}=\dfrac{6}{25}\)
a: ĐKXĐ: x>=0; x<>1
b: \(G=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)^2}{2}\)
\(=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{2}\cdot\left(\sqrt{x}-1\right)\)
\(=-\sqrt{x}\left(\sqrt{x}-1\right)\)
c: Thay x=0,16 vào G, ta được:
\(H=-0,4\cdot\left(0,4-1\right)=-0,4\cdot0,3=-0,12\)
ĐKXĐ : x > 0 ; x ≠ 1 ; x ≠ 4
a) \(A=\left(1-\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{1}{\sqrt{x-1}}\right)\div\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\left(\frac{x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\div\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\left(\frac{x-1-4\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\div\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\times\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}-3}{\sqrt{x}-2}\)
b) Với x = \(11-6\sqrt{2}\)
\(A=\frac{\sqrt{11-6\sqrt{2}}-3}{\sqrt{11-6\sqrt{2}}-2}\)
\(=\frac{\sqrt{2-6\sqrt{2}+9}-3}{\sqrt{2-6\sqrt{2}+9}-2}\)
\(=\frac{\sqrt{\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot3+3^2}-3}{\sqrt{\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot3+3^2}-2}\)
\(=\frac{\sqrt{\left(\sqrt{2}-3\right)^2}-3}{\sqrt{\left(\sqrt{2}-3\right)^2}-2}\)
\(=\frac{\left|\sqrt{2}-3\right|-3}{\left|\sqrt{2}-3\right|-2}\)
\(=\frac{3-\sqrt{2}-3}{3-\sqrt{2}-2}=\frac{-\sqrt{2}}{1-\sqrt{2}}\)
c) Ta có : \(A=\frac{\sqrt{x}-3}{\sqrt{x}-2}=\frac{\sqrt{x}-2-1}{\sqrt{x}-2}=1-\frac{1}{\sqrt{x}-2}\)
Để A nguyên => \(\frac{1}{\sqrt{x}-2}\)nguyên
=> \(1⋮\sqrt{x}-2\)
=> \(\sqrt{x}-2\inƯ\left(1\right)=\left\{\pm1\right\}\)
=> \(\sqrt{x}\in\left\{3;1\right\}\)
=> \(x=9\)( không nhận x = 1 do ĐKXĐ )
d) Để A = -2
=> \(\frac{\sqrt{x}-3}{\sqrt{x}-2}=-2\)( x > 0 ; x ≠ 1 ; x ≠ 4 )
=> \(\sqrt{x}-3=-2\sqrt{x}+4\)
=> \(\sqrt{x}+2\sqrt{x}=4+3\)
=> \(3\sqrt{x}=7\)
=> \(9x=49\)( bình phương hai vế )
=> \(x=\frac{49}{9}\)( tm )
e) Để A có giá trị âm
=> \(\frac{\sqrt{x}-3}{\sqrt{x}-2}< 0\)
Xét hai trường hợp :
1.\(\hept{\begin{cases}\sqrt{x}-3>0\\\sqrt{x}-2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}>3\\\sqrt{x}< 2\end{cases}}\Leftrightarrow\hept{\begin{cases}x>9\\x< 4\end{cases}}\)( loại )
2. \(\hept{\begin{cases}\sqrt{x}-3< 0\\\sqrt{x}-2>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}< 3\\\sqrt{x}>2\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 9\\x>4\end{cases}}\Leftrightarrow4< x< 9\)
Vậy với 4 < x < 9 thì A có giá trị âm
f) Để A < -2
=> \(\frac{\sqrt{x}-3}{\sqrt{x}-2}< -2\)
=> \(\frac{\sqrt{x}-3}{\sqrt{x}-2}+2< 0\)
=> \(\frac{\sqrt{x}-3}{\sqrt{x}-2}+\frac{2\sqrt{x}-4}{\sqrt{x-2}}< 0\)
=> \(\frac{3\sqrt{x}-7}{\sqrt{x}-2}< 0\)
Xét hai trường hợp :
1. \(\hept{\begin{cases}3\sqrt{x}-7< 0\\\sqrt{x}-2>0\end{cases}}\Leftrightarrow\hept{\begin{cases}3\sqrt{x}< 7\\\sqrt{x}>2\end{cases}}\Leftrightarrow\hept{\begin{cases}9x< 49\\x>4\end{cases}}\Leftrightarrow\hept{\begin{cases}x< \frac{49}{9}\\x>4\end{cases}}\Leftrightarrow4< x< \frac{49}{9}\)
2. \(\hept{\begin{cases}3\sqrt{x}-7>0\\\sqrt{x}-2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}3\sqrt{x}>7\\\sqrt{x}< 2\end{cases}}\Leftrightarrow\hept{\begin{cases}9x>49\\x< 4\end{cases}}\Leftrightarrow\hept{\begin{cases}x>\frac{49}{9}\\x< 4\end{cases}}\)( loại )
Vậy với 4 < x < 49/9 thì A < -2
g) Để \(A>\sqrt{x}-1\)
=> \(\frac{\sqrt{x}-3}{\sqrt{x}-2}>\sqrt{x}-1\)
=> \(\frac{\sqrt{x}-3}{\sqrt{x}-2}-\left(\sqrt{x}-1\right)>0\)
=> \(\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-2}>0\)
=> \(\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{x-3\sqrt{x}+2}{\sqrt{x}-2}>0\)
=> \(\frac{-x+4\sqrt{x}-5}{\sqrt{x}-2}>0\)
Ta có : \(-x+4\sqrt{x}-5=-\left(x-4\sqrt{x}+4\right)-1=-\left(\sqrt{x}-2\right)^2-1\le-1< 0\left(\forall\ge0\right)\)
Nên để A > 0 thì ta chỉ cần xét \(\sqrt{x}-2< 0\)
\(\sqrt{x}-2< 0\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)
Kết hợp với ĐKXĐ => \(\hept{\begin{cases}0< x< 4\\x\ne1\end{cases}}\)thì tm
a) ĐKXĐ : x > 0 , x khác 1
b)Rút gọn
P = 6+ căn x trên căn x + 1
điều kiện xác định : \(x\ge0;x\ne1\)
a) ta có : \(G=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{x^2-2x+1}{2}\)
\(\Leftrightarrow G=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(x-1\right)^2}{2}\)
\(\Leftrightarrow G=\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\dfrac{\left(x-1\right)^2}{2}\) \(\Leftrightarrow G=\left(\dfrac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\dfrac{\left(x-1\right)^2}{2}=\sqrt{x}-x\)
b) thay \(x=0,16\) vào \(G\) ta có : \(G=\sqrt{0,16}-0,16=0,24\)
c) ta có : \(G=-\left(x-\sqrt{x}+\dfrac{1}{4}\right)-\dfrac{1}{4}=-\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge\dfrac{-1}{4}\)
\(\Rightarrow G_{max}=\dfrac{-1}{4}\) khi \(\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\)
d) ta có : \(G=\sqrt{x}-x\) \(\Rightarrow\) để \(G\in Z\) \(\Rightarrow x=a^2\ne1\)
e) ta có : \(G>0\Leftrightarrow\sqrt{x}-x>0\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}>0\\1-\sqrt{x}>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}< 0\\1-\sqrt{x}< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0< x< 1\\x\in\varnothing\end{matrix}\right.\) \(\Rightarrow\left(đpcm\right)\)
f) để \(G< 0\Leftrightarrow\sqrt{x}-x< 0\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}>0\\1-\sqrt{x}< 0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}< 0\\1-\sqrt{x}>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>1\\x\in\varnothing\end{matrix}\right.\) vậy \(x>1\)
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