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a) \(\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
\(=\sqrt{4-4\sqrt{3}+3}-\sqrt{4+4\sqrt{3}+3}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\left|2-\sqrt{3}\right|-\left|2+\sqrt{3}\right|\)
\(=2-\sqrt{3}-2-\sqrt{3}\)
\(=-2\sqrt{3}\)
\(P=\frac{\frac{1}{a^2}}{\frac{1}{b}+\frac{1}{c}}+\frac{\frac{1}{b^2}}{\frac{1}{a}+\frac{1}{c}}+\frac{\frac{1}{c^2}}{\frac{1}{a}+\frac{1}{b}}\)
Đặt \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\Rightarrow xyz=1\Rightarrow P=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(P\ge\frac{\left(x+y+z\right)^2}{y+z+x+z+x+y}=\frac{x+y+z}{2}\ge\frac{3\sqrt[3]{xyz}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z\Leftrightarrow a=b=c=1\)
Cần cách khác thì nhắn cái
đk: \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
Ta có:
\(P=\left(\frac{1}{1-\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\div\left(\frac{2x+\sqrt{x}-1}{1-x}+\frac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right)\)
\(P=\frac{\sqrt{x}-1+\sqrt{x}}{\left(1-\sqrt{x}\right)\sqrt{x}}\div\frac{\left(2x+\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)+\left(2x+\sqrt{x}-1\right)\left(\sqrt{x}-x\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}\)
\(P=\frac{2\sqrt{x}-1}{\left(1-\sqrt{x}\right)\sqrt{x}}\cdot\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}{2x+\sqrt{x}-1}\)
\(P=\frac{\left(2\sqrt{x}-1\right)\left(1+\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)\left(1+\sqrt{x}\right)\sqrt{x}}\)
\(P=\frac{x-\sqrt{x}+1}{\sqrt{x}}=\frac{x\sqrt{x}-x+\sqrt{x}}{x}\)