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21 tháng 9 2018

(=) \(2\sqrt{2}+\sqrt{x.\left(x+1\right)}=\sqrt{\left(x+1\right)\left(x+9\right)}\)(nhân cả 2 vế cho \(\sqrt{x+1}\) ).

(=) \(8+4\sqrt{2x\left(x+1\right)}+x\left(x+1\right)=\left(x+1\right)\left(x+9\right)\) \(\Leftrightarrow4\sqrt{2x^2+2x}=x^2+10x+9-x^2-x-8\)     

(=) \(4\sqrt{2x^2+2x}=9x+1\) (=) \(16\left(2x^2+2x\right)=81x^2+18x+1\)(=) \(0=49x^2-14x+1\)

(=)\(\left(7x-1\right)^2=0\) (=) \(x=\frac{1}{7}\)

19 tháng 6 2016

\(\left(\frac{\sqrt{x}-4x}{1-4x}-1\right):\left(\frac{1+2x}{1-4x}-\frac{2\sqrt{x}}{2\sqrt{x}-1}-1\right)\left(ĐK:0\le x\ne\frac{1}{4}\right)\)

\(=\frac{\sqrt{x}-4x+4x-1}{1-4x}:\frac{\left(1+2x\right)+2\sqrt{x}\left(1+2\sqrt{x}\right)+4x-1}{1-4x}\)

\(=\frac{\sqrt{x}-1}{1-4x}.\frac{1-4x}{10x+2\sqrt{x}}=\frac{\sqrt{x}-1}{2\sqrt{x}\left(5\sqrt{x}+1\right)}\)

19 tháng 8 2021

Để căn thức có nghĩa\(\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{2}{x+1}\ge0\\x+1\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+1\le0\\x+1\ne0\end{matrix}\right.\)\(\Leftrightarrow x+1< 0\Leftrightarrow x< -1\)

Vậy...

ĐKXĐ: x<-1

10 tháng 7 2019

\(3,\)Áp dụng bđt Mincopski \(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\)hai lần có

\(VT\ge\sqrt{\left(\sqrt{x}+\sqrt{y}\right)^2+\left(\sqrt{yz}+\sqrt{zx}\right)^2}+\sqrt{z+xy}\)

       \(\ge\sqrt{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2+\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}\)

       \(=\sqrt{x+y+z+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)+\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}\)

       \(=\sqrt{1+2t+t^2}\left(t=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)
        \(=\sqrt{\left(t+1\right)^2}=t+1=VP\left(Đpcm\right)\)

10 tháng 7 2019

\(2,\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\frac{2\sqrt{ab}}{2\sqrt{\sqrt{a}.\sqrt{b}}}=\sqrt{\sqrt{ab}}\left(đpcm\right)\)

27 tháng 7 2018

\(\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\right)^2\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)\left(DKXD:x>0;x\ne1\right)\)

\(\Leftrightarrow\left(\frac{\sqrt{x}.\sqrt{x}-1}{2\sqrt{x}}\right)^2\left(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

\(\Leftrightarrow\frac{\left(x-1\right)^2}{\left(2\sqrt{x}\right)^2}\left(\frac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{x-1}\right)\)

\(\Leftrightarrow\frac{\left(x-1\right)^2}{4x}.\frac{\left(\sqrt{x}-1-\sqrt{x}-1\right)\left(\sqrt{x}-1+\sqrt{x}-1\right)}{x-1}\)

\(\Leftrightarrow\frac{\left(x-1\right)^2}{4x}.\frac{-2.2\sqrt{x}}{x-1}\)

\(\Leftrightarrow\frac{\left(x-1\right)^2.-4\sqrt{x}}{4x.\left(x-1\right)}\)

\(\Leftrightarrow\frac{x-1}{-\sqrt{x}}\Leftrightarrow\frac{1+x}{\sqrt{x}}\Leftrightarrow\frac{\left(1+x\right).\sqrt{x}}{\sqrt{x}.\sqrt{x}}\Leftrightarrow\frac{\sqrt{x}+x\sqrt{x}}{x}\)

27 tháng 7 2018

\(VT=\left(\frac{x\sqrt{x}+3\sqrt{3}}{x-\sqrt{3x}+3}-2\sqrt{x}\right)\left(\frac{\sqrt{x}+\sqrt{3}}{3-x}\right)\)

\(=\left(\frac{\left(\sqrt{x}+\sqrt{3}\right)\left(x-\sqrt{3x}+3\right)}{x-\sqrt{3x}+3}-2\sqrt{x}\right)\left(\frac{\sqrt{x}+\sqrt{3}}{\left(\sqrt{3}-\sqrt{x}\right)\left(\sqrt{3}+\sqrt{x}\right)}\right)\)

\(=\left(\sqrt{x}+\sqrt{3}-2\sqrt{x}\right)\left(\frac{1}{\sqrt{3}-\sqrt{x}}\right)\)

\(=\frac{\sqrt{3}-\sqrt{x}}{\sqrt{3}-\sqrt{x}}=1=VP\left(dpcm\right)\)