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chịu thua vô điều kiện xin lỗi nha : v
muốn biết câu trả lời lo mà sệt trên google ấy đừng có mà dis:v
a) \(ĐKXĐ:x>0\)
\(Y=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-1-\frac{2x+\sqrt{x}}{\sqrt{x}}\)
\(\Leftrightarrow Y=\frac{\sqrt{x}\left(x\sqrt{x}+1\right)}{\left(x-\sqrt{x}+1\right)}-1-2\sqrt{x}-1\)
\(\Leftrightarrow Y=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\left(x-\sqrt{x}+1\right)}-2\sqrt{x}-2\)
\(\Leftrightarrow Y=x+\sqrt{x}-2\sqrt{x}-2\)
\(\Leftrightarrow Y=x-\sqrt{x}-2\)
b) Ta có \(Y=x-\sqrt{x}-2=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{9}{4}\ge-\frac{9}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{1}{4}\)
Vậy \(Min_Y=-\frac{9}{4}\Leftrightarrow x=\frac{1}{4}\)
c) Để \(Y-\left|Y\right|=0\)
\(\Leftrightarrow Y=\left|Y\right|\)
\(\Leftrightarrow Y\ge0\)
\(\Leftrightarrow x-\sqrt{x}-2\ge0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)\ge0\)
\(\Leftrightarrow\sqrt{x}-2\ge0\) (Vì \(\sqrt{x}+1\ge0\))
\(\Leftrightarrow\sqrt{x}\ge2\)
\(\Leftrightarrow x\ge4\) (ĐPCM)
a)\(B=\frac{1}{\sqrt{x}+\sqrt{y}}=\frac{1}{\sqrt{0}+\sqrt{4}}=\frac{1}{2}\)
b)\(M=A+B=\frac{2\sqrt{y}}{x-y}+\frac{1}{\sqrt{x}-\sqrt{y}}+\frac{1}{\sqrt{x}+\sqrt{y}}\)\(=\frac{2\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}+\frac{1}{\sqrt{x}-\sqrt{y}}+\frac{1}{\sqrt{x}+\sqrt{y}}\)
\(=\frac{2\sqrt{y}+\sqrt{x}+\sqrt{y}+\sqrt{x}-\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)\(=\frac{2\sqrt{y}+2\sqrt{x}}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)\(=\frac{2\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)\(=\frac{2}{\sqrt{x}-\sqrt{y}}\)
c)\(M=\frac{2}{\sqrt{x}-\sqrt{y}}\)<=>\(1=\frac{2}{\sqrt{4y}-\sqrt{y}}\)<=>\(1=\frac{2}{2\sqrt{y}-\sqrt{y}}\)<=>\(1=\frac{2}{\sqrt{y}}\)<=> \(\sqrt{y}=2\)
<=> \(\left(\sqrt{y}\right)^2=2^2\)<=> \(y=4\)
=>\(x=4y=4\cdot4=16\)
a) \(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-\left(x-2\sqrt{xy}+y\right)\)
\(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
c) \(4x-\sqrt{8}+\frac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}=4x-\sqrt{8}+\frac{\sqrt{x^2\left(x+2\right)}}{x+2}=4x-\sqrt{8}+x=5x-\sqrt{8}\)
a/ ĐKXĐ: \(x>0\)
\(y=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\frac{2x+\sqrt{x}}{\sqrt{x}}\)
\(=\frac{\sqrt{x}\left[\left(\sqrt{x}\right)^3+1\right]}{x-\sqrt{x}+1}+1-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1-2\sqrt{x}-1\)
\(=\sqrt{x}\left(\sqrt{x}+1\right)+1-2\sqrt{x}-1=x-\sqrt{x}\)
Ta có: \(y=2\Leftrightarrow x-\sqrt{x}=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loai\right)\\x=1\end{matrix}\right.\)
Vậy ...............
b/ Ta có: \(x>1\Rightarrow\sqrt{x}>1\Rightarrow\sqrt{x}-1>0\) và \(\sqrt{x}>0\)
nên \(y=\sqrt{x}\left(\sqrt{x}-1\right)>0\). Khi đó \(\left|y\right|=y\)
\(\Rightarrow y-\left|y\right|=y-y=0\) (ĐPCM)
c/ \(y=x-\sqrt{x}=\left(\sqrt{x}\right)^2-2\sqrt{x}.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\)
\(=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
Dấu "=" xảy ra khi \(\sqrt{x}-\frac{1}{2}=0\Leftrightarrow\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\)
Vậy \(Min_y=-\frac{1}{4}\Leftrightarrow x=\frac{1}{4}\)