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\(=\frac{a\sqrt{ab}+ab-ab}{a+\sqrt{ab}}.\frac{a-b}{\sqrt[4]{ab}-\sqrt{b}}.\frac{1}{\sqrt{b}+\sqrt[4]{ab}}\)
\(=\frac{a\sqrt{ab}}{a+\sqrt{ab}}.\frac{a-b}{\sqrt{ab}-b}=\frac{a\sqrt{ab}}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}.\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}=a\)
\(I=\frac{a^{\frac{4}{3}}-8a^{\frac{2}{3}}b}{a^{\frac{2}{3}}+2\sqrt[3]{ab}+4b^{\frac{2}{3}}}\left(1-2\sqrt[3]{\frac{b}{a}}\right)^{-1}-a^{\frac{2}{3}}=\frac{a^{\frac{1}{3}}\left(a-8b\right)}{a^{\frac{2}{3}}+2a^{\frac{1}{3}}.b^{\frac{1}{3}}+4b^{\frac{2}{3}}}\left(\frac{\sqrt[3]{a}-2\sqrt[3]{b}}{\sqrt[3]{a}}\right)^{-1}-a^{\frac{2}{3}}\)
\(=\frac{\sqrt[3]{a}\left[\left(\sqrt[3]{a}\right)^3-\left(2\sqrt[3]{b}\right)^3\right]}{a^{\frac{2}{3}}+2\sqrt[3]{ab}+4b^{\frac{2}{3}}}.\frac{\sqrt[3]{a}}{\sqrt[3]{a}-2\sqrt[3]{b}}-a^{\frac{2}{3}}\)
\(=\frac{\left(\sqrt[3]{a}\right)^2\left(\sqrt[3]{a}-2\sqrt[3]{b}\right)\left[\left(\sqrt[3]{a}\right)^2+2\sqrt[3]{ab}+\left(2\sqrt[3]{b}\right)^2\right]}{\left(\sqrt[3]{a}-a\sqrt[3]{b}\right)\left[\left(\sqrt[3]{a}\right)^2+2\sqrt[3]{ab}+\left(2\sqrt[3]{b}\right)^2\right]}-a^{\frac{2}{3}}=a^{\frac{2}{3}}-a^{\frac{2}{3}}=0\)
a) \(A=\left[\left(\frac{1}{5}\right)^2\right]^{\frac{-3}{2}}-\left[2^{-3}\right]^{\frac{-2}{3}}=5^3-2^2=121\)
b) \(B=6^2+\left[\left(\frac{1}{5}\right)^{\frac{3}{4}}\right]^{-4}=6^2+5^3=161\)
c) \(C=\frac{a^{\sqrt{5}+3}.a^{\sqrt{5}\left(\sqrt{5}-1\right)}}{\left(a^{2\sqrt{2}-1}\right)^{2\sqrt{2}+1}}=\frac{a^{\sqrt{5}+3}.a^{5-\sqrt{5}}}{a^{\left(2\sqrt{2}\right)^2-1^2}}\)
\(=\frac{a^{\sqrt{5}+3+5-\sqrt{5}}}{a^{8-1}}=\frac{a^8}{a^7}=a\)
d) \(D=\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)^2:\left(b-2b\sqrt{\frac{b}{a}}+\frac{b^2}{a}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right)^2:b\left[1-2\sqrt{\frac{b}{a}}+\left(\sqrt{\frac{b}{a}}\right)^2\right]\)
\(=\left(\sqrt{a}-\sqrt{b}\right)^2:b\left(1-\sqrt{b}a\right)^2\)
Với mọi \(k\ge2\) thì \(\frac{2k+\sqrt{k^2-1}}{\sqrt{k-1}+\sqrt{k+1}}=\frac{\left[\left(\sqrt{k-1}\right)^2+\left(\sqrt{k+1}\right)^2+\sqrt{\left(k-1\right)\left(k+1\right)}\right]\left(\sqrt{k+1}-\sqrt{k-1}\right)}{\left(\sqrt{k-1}+\sqrt{k+1}\right)\left(\sqrt{k+1}-\sqrt{k-1}\right)}\)
\(=\frac{\sqrt{\left(k+1\right)^3}-\sqrt{\left(k-1\right)^3}}{2}\)
Suy ra tổng đã cho có thể viết là :
\(A=\frac{1}{2}\left[\sqrt{3^3}-\sqrt{1^3}+\sqrt{4^3}-\sqrt{2^3}+\sqrt{5^3}-\sqrt{3^3}+\sqrt{6^3}-\sqrt{4^3}+...+\sqrt{101^3}-\sqrt{99^3}\right]\)
\(=\frac{1}{2}\left[-1-\sqrt{2^3}+\sqrt{101^3}+\sqrt{100^3}\right]\)
\(=\frac{999+\sqrt{101^3}-\sqrt{8}}{2}\)
\(M=\frac{\left(a^{\frac{1}{3}}+b^{\frac{1}{3}}\right)^2}{\sqrt[3]{ab}}:\left(2+\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}}\right)=\frac{\left(a^{\frac{1}{3}}+b^{\frac{1}{3}}\right)^2}{\sqrt[3]{ab}}:\frac{2\sqrt[3]{ab}+\left(\sqrt[3]{a}\right)^2+\left(\sqrt[3]{a}\right)^2}{\sqrt[3]{ab}}\)
\(=\frac{\left(\sqrt[3]{a}+\sqrt[3]{b}\right)^2}{\sqrt[3]{ab}}-\frac{\sqrt[3]{ab}}{\left(\sqrt[3]{a}+\sqrt[3]{b}\right)^2}=1\)
a) \(A=\frac{a^{\frac{5}{2}}\left(a^{\frac{1}{2}}-a^{\frac{-3}{2}}\right)}{a^{\frac{1}{2}}\left(a^{\frac{-1}{2}}-a^{\frac{3}{2}}\right)}=\frac{a^3-a}{1-a^2}=-a\)
Do đó : \(A=-\left(\pi-3\sqrt{2}\right)=3\sqrt{2}-\pi\)
b) Rút gọn B ta có :
\(B=\left(a^{\frac{1}{3}}+b^{\frac{1}{3}}\right)\left[\left(a^{\frac{1}{3}}\right)^2+\left(b^{\frac{1}{3}}\right)^2\right]=\left(a^{\frac{1}{3}}\right)^3+\left(b^{\frac{1}{3}}\right)^3=a+b\)
Do đó :
\(B=\left(7-\sqrt{2}\right)+\left(\sqrt{2}+3\right)=10\)
\(ĐK:a\ge0,a^2\pm b\ge0,a\ge\sqrt{a^2-b}\)
\(A=\frac{\sqrt{a+\sqrt{a^2-b}}+\sqrt{a-\sqrt{a^2-b}}}{\sqrt{2}}\)
\(A^2=\frac{2a+2\sqrt{a^2+b}}{2}=a+\sqrt{a^2+b}\)
\(A=\sqrt{a+\sqrt{a^2+b}}\)
A2=\(\frac{2a+2\sqrt{b}}{2}\) mới đúng ạ. Nhưng cảm ơn vì đã giải ạ
\(D=\left(\frac{a-b}{a^{\frac{3}{4}}+a^{\frac{1}{2}}.b^{\frac{1}{4}}}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}\right):\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)^{-1}\sqrt{\frac{a}{b}}\)
\(=\left[\frac{a-b}{a^{\frac{1}{2}}\left(a^{\frac{1}{4}}+b^{\frac{1}{4}}\right)}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}\right]:\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)^{-1}\sqrt{\frac{b}{a}}\)
\(=\frac{a-b-a+a^{\frac{1}{2}}.b^{\frac{1}{2}}}{a^{\frac{1}{2}}\left(a^{\frac{1}{4}}+b^{\frac{1}{4}}\right)}.\frac{1}{\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)}=\frac{b^{\frac{1}{2}}}{a^{\frac{1}{2}}}\frac{\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)}{\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)}\sqrt{\frac{a}{b}}.\sqrt{\frac{a}{b}}=1\)
Biểu thức này không rút gọn được nữa bạn ạ.