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Bài 1:
a) Sửa đề: \(\left(\sqrt{12}+3\sqrt{5}-4\sqrt{135}\right)\cdot\sqrt{3}\)
Ta có: \(\left(\sqrt{12}+3\sqrt{5}-4\sqrt{135}\right)\cdot\sqrt{3}\)
\(=\sqrt{12}\cdot\sqrt{3}+3\sqrt{5}\cdot\sqrt{3}-4\sqrt{135}\cdot\sqrt{3}\)
\(=6+3\sqrt{15}-36\sqrt{5}\)
b) Ta có: \(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}\)
\(=3\sqrt{28}-5\sqrt{28}+3\sqrt{112}-2\sqrt{112}\)
\(=-2\sqrt{28}+\sqrt{112}=-\sqrt{112}+\sqrt{112}=0\)
c) Ta có: \(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(=2\cdot4\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}-2\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}-3\cdot2\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}\)
\(=8\sqrt{5}\cdot\sqrt{\sqrt{3}}-2\sqrt{5}\sqrt{\sqrt{3}}-6\sqrt{5}\sqrt{\sqrt{3}}\)
=0
Bài 2:
a) Ta có: \(A=\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)
\(=\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}\)
\(=\frac{1}{\sqrt{2}}\)
b) Ta có: \(B=\frac{9\sqrt{5}+3\sqrt{27}}{\sqrt{5}+\sqrt{3}}\)
\(=\frac{\sqrt{405}+\sqrt{243}}{\sqrt{5}+\sqrt{3}}\)
\(=\frac{9\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}=9\)
c) Ta có: \(C=\frac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)
\(=\frac{\sqrt{72}-\sqrt{48}+\sqrt{20}}{\sqrt{162}-\sqrt{108}+\sqrt{45}}\)
\(=\frac{2\left(\sqrt{18}-\sqrt{12}+\sqrt{5}\right)}{3\left(\sqrt{18}-\sqrt{12}+\sqrt{5}\right)}=\frac{2}{3}\)
a)
\((\sqrt{3}-2\sqrt{12}+2\sqrt{4})(\sqrt{27}+\sqrt{144}-2\sqrt{16})\)
\(=(\sqrt{3}-4\sqrt{3}+4)(3\sqrt{3}+12-8)\)
\(=(-3\sqrt{3}+4)(3\sqrt{3}+4)=4^2-(3\sqrt{3})^2=16-27=-11\)
b)
\((2\sqrt{5}+2\sqrt{3})^2-4\sqrt{60}\)
\(=(2\sqrt{5})^2+2.2\sqrt{5}.2\sqrt{3}+(2\sqrt{3})^2-8\sqrt{15}\)
\(=32+8\sqrt{15}-8\sqrt{15}=32\)
c)
\(\sqrt{6}(3\sqrt{12}-4\sqrt{3}+\sqrt{48}-5\sqrt{6})\)
\(=3\sqrt{72}-4\sqrt{18}+\sqrt{6.48}-5.\sqrt{36}\)
\(=18\sqrt{2}-12\sqrt{2}+12\sqrt{2}-30=18\sqrt{2}-30\)
d)
\((\sqrt{2}-\sqrt{3})(\sqrt{6}+\sqrt{2})(\sqrt{2}+\sqrt{3})\)
\(=(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})(\sqrt{6}+\sqrt{2})\)
\(=(2-3)(\sqrt{6}+\sqrt{2})=-(\sqrt{6}+\sqrt{2})\)
e) Biểu thức bên trong căn lớn âm nên biểu căn bậc 2 không có nghĩa
f)
\((\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}).\frac{1}{\sqrt{3}+5}\)
\(=(\frac{2\sqrt{3}+15}{3-\sqrt{3}}+\frac{3}{\sqrt{3}-2}).\frac{1}{\sqrt{3}+5}\)
\(=\frac{2\sqrt{3}+15)(\sqrt{3}-2)+3(3-\sqrt{3})}{(3-\sqrt{3})(\sqrt{3}-2)}.\frac{1}{\sqrt{3}+5}\)
\(=\frac{-15+8\sqrt{3}}{(-9+5\sqrt{3})(\sqrt{3}+5)}=\frac{-15+8\sqrt{3}}{-30+16\sqrt{3}}=\frac{-15+8\sqrt{3}}{2(-15+8\sqrt{3})}=\frac{1}{2}\)
a)(\(\sqrt{2006}-\sqrt{2005}\)).(\(\sqrt{2006}+\sqrt{2005}\))
=\(\sqrt{2006}^2-\sqrt{2005}^2\)
=2006-2005
=1
Bạn đăng lại đi sai đề òi
đề đúng mà bn thầy mk cho