\(\sqrt{\frac{9}{25}}+\sqrt{\frac{1}{6}}-\sqrt{\frac{4}{49}}\)\(=\sqrt{\left(\frac{3}{5}\right)^2}+\sqrt{\frac{1}{6}}-\sqrt{\left(\frac{2}{7}\right)^2}\)\(=\frac{3}{5}+\sqrt{\frac{1}{6}}-\frac{2}{7}\)\(=\left(\frac{3}{5}-\frac{2}{7}\right)+\sqrt{\frac{1}{6}}\)\(=\left(\frac{21}{35}-\frac{10}{35}\right)+\sqrt{\frac{1}{6}}\)\(=\frac{11}{35}+\sqrt{\frac{1}{6}}\)\(=\sqrt{\frac{1.6}{6.6}}+\frac{11}{35}\)\(=\frac{\sqrt{6}}{6}+\frac{11}{35}\)\(=\frac{35\sqrt{6}}{210}+\frac{66}{210}\)\(=\frac{35\sqrt{6}+66}{210}\)

=3/5+can1/6-2/7

=1,00824829-2/7

=0,722534004

sorry mình ko có can những số dài lắm

Bài 1:

a) \(\frac{4}{\sqrt{5}-\sqrt{3}}-\sqrt{12}\)

\(=\frac{4}{\sqrt{5}-\sqrt{3}}-2\sqrt{3}\)

\(=\frac{4\sqrt{5}+4\sqrt{3}}{\sqrt{5^2}-\sqrt{3^2}}-2\sqrt{3}\)

\(=\frac{4\left(\sqrt{5}+\sqrt{3}\right)}{5-3}-2\sqrt{3}\)

\(=\frac{4\left(\sqrt{5}+\sqrt{3}\right)}{2}-2\sqrt{3}\)

\(=2\left(\sqrt{5}+\sqrt{3}\right)-2\sqrt{3}\)

\(=2\sqrt{5}+2\sqrt{3}-2\sqrt{3}\)

\(=2\sqrt{5}\)

b) \(\sqrt{\frac{9}{8}}-\sqrt{\frac{49}{2}}+\sqrt{\frac{25}{18}}\)

\(=\frac{3}{2\sqrt{2}}-\frac{7}{\sqrt{2}}+\frac{5}{3\sqrt{2}}\)

\(=\frac{3\sqrt{2}}{2.2}-\frac{7}{\sqrt{2}}+\frac{5\sqrt{2}}{3.2}\)

\(=\frac{3\sqrt{2}}{4}-\frac{7}{\sqrt{2}}+\frac{5\sqrt{2}}{6}\)

\(=-\frac{23\sqrt{2}}{12}\)

chung ta den bai 2 :3

a) \(\frac{x}{\sqrt{x}-2}=-1\)

\(\Leftrightarrow x=-\sqrt{x}+2\)

\(\Leftrightarrow x-2=-\sqrt{x}\)

bình phương 2 vế ta được:

\(\Leftrightarrow x^2-4x+4=x\)

\(\Leftrightarrow x^2-4x+4-x=0\)

\(\Leftrightarrow x^2-5x+4=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=1\end{cases}}\)

b) \(\sqrt{x-2}=x-4\)

chúng ta lại bình phương hai vế như câu a và chúng ta được:

\(\Leftrightarrow x-2=x^2-8x+16\)

\(\Leftrightarrow x-2-x^2+8x-16=0\)

\(\Leftrightarrow9x-18-x^2=0\)

\(\Leftrightarrow\left(x-6\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=3\end{cases}}\)

a) \(\sqrt{49}+\sqrt{25}-4\cdot0,25\)

\(=7+5-1=11\)

b) \(\sqrt{\frac{1}{9}}\cdot\sqrt{0,81}\cdot\sqrt{0,9}\)

\(=\frac{1}{3}\cdot\frac{9}{10}\cdot\frac{3\sqrt{10}}{10}\)

\(=\frac{9\sqrt{10}}{100}\)

c) \(\sqrt{6,4\cdot2400\cdot0,6}\)

\(=\sqrt{64\cdot36\cdot4}\)

\(=8\cdot6\cdot2=96\)

d) \(\sqrt{26^2-24^2}=\sqrt{\left(26-24\right)\left(26+24\right)}\)

\(=\sqrt{2\cdot50}=\sqrt{100}=10\)

\(x=\sqrt[3]{7+\sqrt{\frac{49}{8}}}+\sqrt[3]{7-\sqrt{\frac{49}{8}}}\)

ta lập phương hai vế có

\(x^3=7+\sqrt{\frac{49}{8}}+7-\sqrt{\frac{49}{8}}+3\sqrt[3]{\left(7+\sqrt{\frac{49}{8}}\right)\left(7-\sqrt{\frac{49}{8}}\right)}x\)

\(< =>x^3=14+3\sqrt[3]{7^2-\frac{49}{8}}x\)

\(< =>x^3=14+3\sqrt[3]{\frac{343}{8}}x\)

\(< =>x^3=14+3.\frac{7}{2}x\)

\(< =>2x^3-21x-28=0\)

nên 

\(fx=\left(2x^3-21x-29\right)^3=\left(2x^3-21x-28-1\right)^3=\left(-1\right)^3=-1\)

a) = \(\frac{7}{2}\)

b) = \(\frac{643}{64}\)

c) = 0

\(a,\sqrt{25}-\sqrt{16}+\sqrt{1}=\sqrt{5^2}-\sqrt{4^2}+\sqrt{1^2}=5-4+1=2\)

\(b,\sqrt{\frac{4}{9}}+\sqrt{\frac{25}{4}}+\sqrt{\left(-3\right)^4}=\sqrt{\left(\frac{2}{3}\right)^2}+\sqrt{\left(\frac{5}{2}\right)^2}+\sqrt{\left[\left(-3\right)^2\right]^2}\)

\(=\frac{2}{3}+\frac{5}{2}+\left(-3\right)^2=\frac{2}{3}+\frac{5}{2}+9=\frac{4}{6}+\frac{15}{6}+\frac{54}{6}=\frac{73}{6}\)

\(c,\frac{7}{5}+\sqrt{49}+\sqrt{\left(-3\right)^2}=\frac{7}{5}+\sqrt{7^2}+\sqrt{3^2}=\frac{7}{5}+7+3\)

\(=\frac{7}{5}+\frac{35}{5}+\frac{15}{5}=\frac{57}{5}\)

​con lạy cha nào làm được hết bài này và giải trình tự ra

uk Đinh Tuấn Việt

a) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)

\(=2\sqrt{5}+2+\sqrt{5}-2\)

\(=3\sqrt{5}\)

b) \(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)

\(=3-2\sqrt{2}+2\sqrt{2}-1\)

=2

c) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)

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

\(=2\sqrt{2}\)