a) \(\sqrt{\left(1+2\sqrt{3}\right)^2}-5\sqrt{3}\)

\(=\left|1+2\sqrt{3}\right|-5\sqrt{3}\)

\(=1+2\sqrt{3}-5\sqrt{3}\)

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

b) \(3\sqrt{2x}+2\sqrt{4x}-6\sqrt{8x}\) (Điều kiện: \(x\ge0\))

\(=3\sqrt{2x}+2.2\sqrt{x}-6.2\sqrt{2x}\)

\(=-9\sqrt{2x}+4\sqrt{x}\)

Ta có :

\(\left(x+\frac{1}{x}\right)\cdot\left(y+\frac{1}{y}\right)=3.5\)

\(\Leftrightarrow xy+\frac{x}{x}+\frac{y}{y}+\frac{1}{xy}=15\)

\(\Leftrightarrow xy+\frac{1}{xy}=15-2\)

\(\Leftrightarrow xy+\frac{1}{xy}=13\)

Hay A = 13

có sin B = AC/AB = 6,5/5,7

=> ^B = 76^o

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thuỷ  

ok

a) \(\sqrt{144}-\sqrt{9}-\sqrt{250}\cdot\sqrt{1,6}=12-3-20=-11\)

a) \(\frac{\sqrt{144}}{\sqrt{9}}-\sqrt{250}.\sqrt{1,6}\) = \(\frac{12}{3}-\sqrt{250.1,6}\) =  \(4-\sqrt{25.16}\)  = \(4-20=-16\)

b) \(\frac{\sqrt{3,6}}{\sqrt{4,9}}-\sqrt{3.2^3.24}\) = \(\sqrt{\frac{36}{49}}-\sqrt{576}\) = \(\frac{6}{7}-24=-\frac{162}{7}\)

c) \(\sqrt{2+\sqrt{3}}\) = \(\sqrt{3+2.\frac{\sqrt{3}}{2}+\frac{1}{4}-\frac{3}{4}}\) = \(\sqrt{\left(\sqrt{3}+\frac{1}{2}\right)^2-\left(\frac{\sqrt{3}}{2}\right)^2}\) 

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

thôi dẹp đi em lớp 4 ko biết