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1.\(\frac{3\sqrt{128}}{\sqrt{2}}=\frac{\sqrt{9.128}}{\sqrt{2}}=\sqrt{\frac{1152}{2}}=\sqrt{576}=24\)
\(1.\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}=2-\sqrt{3}+1+\sqrt{3}=3\) \(2a.\sqrt{x^2-2x+1}=7\)
⇔ \(x^2-2x+1=49\)
⇔ \(x^2-2x-48=0\)
⇔ \(\left(x+6\right)\left(x-8\right)=0\)
⇔ \(x=8orx=-6\)
\(b.\sqrt{4x-20}-3\sqrt{\dfrac{x-5}{9}}=\sqrt{1-x}\)
⇔ \(2\sqrt{x-5}-\sqrt{x-5}=\sqrt{1-x}\)
⇔ \(x-5=1-x\)
⇔ \(x=3\left(KTM\right)\)
KL.............
a) \(A=\frac{-\sqrt{x}+2+4}{\sqrt{x}-2}=-1+\frac{4}{\sqrt{x}-2}\)
Để \(A\in Z\Leftrightarrow\sqrt{x}-2\in\left\{-4;-2;-1;1;2;4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{-2;0;1;3;4;6\right\}\)
Mà \(x\in Z;\sqrt{x}\ge0\Rightarrow x\in\left\{0;1;9;16;36\right\}\)
b)\(A=\frac{4\sqrt{x}-2+3}{2\sqrt{x}-1}=2+\frac{3}{2\sqrt{x}-1}\)
Để \(A\in Z\Leftrightarrow2\sqrt{x}-1\in\left\{-3;-1;1;3\right\}\)
\(\Leftrightarrow2\sqrt{x}\in\left\{-2;0;2;4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{-1;0;1;2\right\}\Leftrightarrow x\in\left\{0;1;4\right\}\)
\(3,\)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
\(VT\ge\sqrt{\left(\sqrt{x}+\sqrt{y}\right)^2+\left(\sqrt{yz}+\sqrt{zx}\right)^2}+\sqrt{z+xy}\)
\(\ge\sqrt{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2+\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}\)
\(=\sqrt{x+y+z+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)+\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}\)
\(=\sqrt{1+2t+t^2}\left(t=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)
\(=\sqrt{\left(t+1\right)^2}=t+1=VP\left(Đpcm\right)\)
\(2,\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\frac{2\sqrt{ab}}{2\sqrt{\sqrt{a}.\sqrt{b}}}=\sqrt{\sqrt{ab}}\left(đpcm\right)\)