Câu hỏi đâu em ơi

a: \(\sqrt{\frac47}-10\sqrt{\frac{7}{25}}-6\cdot\sqrt{\frac{1}{28}}\)

\(=\frac{2}{\sqrt7}-10\cdot\frac{\sqrt7}{5}-6\cdot\frac{1}{2\sqrt7}\)

\(=\frac27\sqrt7-2\sqrt7-\frac{3\sqrt7}{7}=-\frac{15}{7}\sqrt7\)

b: \(\left(\sqrt{10}+\sqrt2\right)\cdot\sqrt{3-\sqrt5}\)

\(=\left(\sqrt5+1\right)\cdot\sqrt{6-2\sqrt5}\)

\(=\left(\sqrt5+1\right)\cdot\sqrt{\left(\sqrt5-1\right)^2}=\left(\sqrt5+1\right)\left(\sqrt5-1\right)\)

=5-1

=4

c: \(\frac{\sqrt{6+\sqrt{11}}-\sqrt{7-\sqrt{33}}}{\sqrt6+\sqrt2}\)

\(=\frac{\sqrt{12+2\sqrt{11}}-\sqrt{14-2\sqrt{33}}}{2\left(\sqrt3+1\right)}\)

\(=\frac{\sqrt{\left(\sqrt{11}+1\right)^2}-\sqrt{\left(\sqrt{11}-\sqrt3\right)^2}}{2\left(\sqrt3+1\right)}=\frac{\sqrt{11}+1-\sqrt{11}+\sqrt3}{2\left(1+\sqrt3\right)}\)

\(=\frac{1+\sqrt3}{2\left(1+\sqrt3\right)}=\frac12\)

d: \(\frac{5\sqrt3-3\sqrt5}{\sqrt5-\sqrt3}+\frac{2}{4+\sqrt{15}}-\frac{5\sqrt5+3\sqrt3}{\sqrt5+\sqrt3}\)

\(=\frac{\sqrt{15}\left(\sqrt5-\sqrt3\right)}{\sqrt5-\sqrt3}+\frac{2\left(4-\sqrt{15}\right)}{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}-\frac{\left(\sqrt5+\sqrt3\right)\left(8-\sqrt{15}\right)}{\sqrt5+\sqrt3}\)

\(=\sqrt{15}+2\left(4-\sqrt{15}\right)-\left(8-\sqrt{15}\right)\)

\(=2\sqrt{15}-8+8-2\sqrt{15}\)

=0

ĐKXĐ: \(x\ge0\)
\(\dfrac{2\sqrt{x}+x^2+1}{x+2}=\dfrac{\left(\sqrt{x}+1\right)^2}{x+2}\)

b: \(\sqrt{1\frac{9}{16}}+\frac{\sqrt{15}-\sqrt{12}}{\sqrt5-2}-\frac{3}{\sqrt3}\)

\(=\sqrt{\frac{25}{16}}+\frac{\sqrt3\left(\sqrt5-2\right)}{\sqrt5-2}-\sqrt3\)

\(=\frac54+\sqrt3-\sqrt3=\frac54\)

a) \(\sqrt{0,64.a^2}\left(a>0\right)=0,8.\left|a\right|=0,8a\)

b) \(\sqrt{a^2\left(a-2\right)^2}\left(a>2\right)=\left|a\left(a-2\right)\right|=a\left(a-2\right)=a^2-2a\)

c) \(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}\left(a\ge0,a\ne1\right)=\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}=1+\sqrt{a}+a\)

\(P=\left(\dfrac{x-1}{\sqrt{x}+1}-\dfrac{x-2\sqrt{x}+1}{x-\sqrt{x}}+1\right).\dfrac{1}{x\sqrt{x}+1}\)

\(=\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}-\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}+1\right).\dfrac{1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\left(\sqrt{x}-1-\dfrac{\sqrt{x}-1}{\sqrt{x}}+1\right).\dfrac{1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)+\sqrt{x}}{\sqrt{x}}.\dfrac{1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}.\dfrac{1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

Bài 2: 

Ta có: \(P=\left(\dfrac{x-1}{\sqrt{x}+1}-\dfrac{x-2\sqrt{x}+1}{x-\sqrt{x}}+1\right)\cdot\dfrac{1}{x\sqrt{x}+1}\)

\(=\left(\sqrt{x}-1-\dfrac{\sqrt{x}-1}{\sqrt{x}}+1\right)\cdot\dfrac{1}{x\sqrt{x}+1}\)

\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\cdot\dfrac{1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\dfrac{1}{x+\sqrt{x}}\)

d: \(\dfrac{-\left(\sqrt{3}-\sqrt{6}\right)}{1-\sqrt{2}}+\dfrac{6\sqrt{3}+3}{\sqrt{3}}-\dfrac{13}{4+\sqrt{3}}\)

\(=-\sqrt{3}+6+\sqrt{3}-4+\sqrt{3}\)

\(=2+\sqrt{3}\)

ĐKXĐ: x khác 1

\(M=\frac{x-\sqrt[3]{x}}{x-1}+\frac{1}{\sqrt[3]{x}-1}+\frac{1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}\)

\(=\frac{x-\sqrt[3]{x}}{x-1}+\frac{\sqrt[3]{x^2}+\sqrt[3]{x}+1+\sqrt[3]{x}-1}{\left(\sqrt[3]{x}-1\right)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}=\frac{x-\sqrt[3]{x}}{x-1}+\frac{\sqrt[3]{x^2}+2\sqrt[3]{x}}{\left(\sqrt[3]{x}\right)^3-1}\)

\(=\frac{x-\sqrt[3]{x}}{x-1}+\frac{\sqrt[3]{x^2}+2\sqrt[3]{x}}{x-1}=\frac{x+\sqrt[3]{x^2}+\sqrt[3]{x}}{x-1}=\frac{\sqrt[3]{x}\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}{\left(\sqrt[3]{x}-1\right)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}\)

\(=\frac{\sqrt[3]{x}}{\sqrt[3]{x}-1}\)

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