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(=) \(2\sqrt{2}+\sqrt{x.\left(x+1\right)}=\sqrt{\left(x+1\right)\left(x+9\right)}\)(nhân cả 2 vế cho \(\sqrt{x+1}\) ).
(=) \(8+4\sqrt{2x\left(x+1\right)}+x\left(x+1\right)=\left(x+1\right)\left(x+9\right)\) \(\Leftrightarrow4\sqrt{2x^2+2x}=x^2+10x+9-x^2-x-8\)
(=) \(4\sqrt{2x^2+2x}=9x+1\) (=) \(16\left(2x^2+2x\right)=81x^2+18x+1\)(=) \(0=49x^2-14x+1\)
(=)\(\left(7x-1\right)^2=0\) (=) \(x=\frac{1}{7}\)
\(\left(\frac{\sqrt{x}-4x}{1-4x}-1\right):\left(\frac{1+2x}{1-4x}-\frac{2\sqrt{x}}{2\sqrt{x}-1}-1\right)\left(ĐK:0\le x\ne\frac{1}{4}\right)\)
\(=\frac{\sqrt{x}-4x+4x-1}{1-4x}:\frac{\left(1+2x\right)+2\sqrt{x}\left(1+2\sqrt{x}\right)+4x-1}{1-4x}\)
\(=\frac{\sqrt{x}-1}{1-4x}.\frac{1-4x}{10x+2\sqrt{x}}=\frac{\sqrt{x}-1}{2\sqrt{x}\left(5\sqrt{x}+1\right)}\)
\(3,\)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
\(VT\ge\sqrt{\left(\sqrt{x}+\sqrt{y}\right)^2+\left(\sqrt{yz}+\sqrt{zx}\right)^2}+\sqrt{z+xy}\)
\(\ge\sqrt{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2+\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}\)
\(=\sqrt{x+y+z+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)+\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}\)
\(=\sqrt{1+2t+t^2}\left(t=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)
\(=\sqrt{\left(t+1\right)^2}=t+1=VP\left(Đpcm\right)\)
\(2,\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\frac{2\sqrt{ab}}{2\sqrt{\sqrt{a}.\sqrt{b}}}=\sqrt{\sqrt{ab}}\left(đpcm\right)\)