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Đặt: \(A=\sqrt{3+\sqrt{8}}\)
=> \(\sqrt{2}A=\sqrt{6+2\sqrt{8}}=\sqrt{\left(2+\sqrt{2}\right)^2}=2+\sqrt{2}=\sqrt{2}\left(\sqrt{2+1}\right)\)
=> \(A=\sqrt{2}+1\)
\(3+\sqrt{18}+\sqrt{3+\sqrt{8}}=3+3\sqrt{2}+\sqrt{2}+1\)
\(=3\left(\sqrt{2}+1\right)+\left(\sqrt{2}+1\right)=4.\left(\sqrt{2}+1\right)\)
\(8-\frac{x\sqrt{x}}{3}\)
\(=8-\frac{\sqrt{x^3}}{3}\)
\(=8-\frac{\left(\sqrt{x}\right)^3}{3}\)
\(=8-\frac{\left(\sqrt{x}\right)^3}{\left(\sqrt[3]{3}\right)^3}\)
\(=2^3-\left(\frac{\sqrt{x}}{\sqrt[3]{3}}\right)^3\)
\(=\left(2-\frac{\sqrt{x}}{\sqrt[3]{3}}\right)\left(4+\frac{2\sqrt{x}}{\sqrt[3]{3}}+\frac{x}{\left(\sqrt[3]{3}\right)^2}\right)\)
\(x-y=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\)
\(a\sqrt{b}+b\sqrt{a}=\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\)
\(xy-y\sqrt{x}+\sqrt{x}-1\)
\(=y\left(x-\sqrt{x}\right)+\left(\sqrt{x}-1\right)\)
\(=y\sqrt{x}\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)\)
\(\left(\sqrt{x}-1\right)\left(y\sqrt{x}+1\right)\)
a, \(1-a\sqrt{a}\)
\(=\left[1-\left(\sqrt{a}\right)^3\right]\)
\(=\left(1-\sqrt{a}\right)\left[\left(\sqrt{a}\right)^2+1.\sqrt{a}+1^2\right]\)
\(=\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)\)
b, \(x-2\sqrt{x-1}\)
\(=\left(x-1\right)-2\sqrt{x-1}+1\)
\(=\left[\left(\sqrt{x-1}\right)-1\right]^2\)
Lời giải :
\(\sqrt{a-b}-\sqrt{a^2-b^2}\)
\(=\sqrt{a-b}-\sqrt{a-b}\cdot\sqrt{a+b}\)
\(=\sqrt{a-b}\left(1-\sqrt{a+b}\right)\)