Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\sqrt{45}\cdot\sqrt{15}\cdot\sqrt{27}=\sqrt{45\cdot15\cdot27}=135\)
b) \(\frac{\sqrt{17}}{\sqrt{12}}+\frac{1}{2}=\frac{\sqrt{51}}{6}+\frac{3}{6}=\frac{\sqrt{51}+3}{6}\)
c) \(\sqrt{\frac{1}{3}}:\sqrt{\frac{27}{50}}\cdot\sqrt{2}=\sqrt{\frac{1}{3}\cdot\frac{50}{27}\cdot2}=\frac{10}{9}\)
d) \(\sqrt{117^2-108^2}=\sqrt{\left(117-108\right)\left(117+108\right)}=\sqrt{9\cdot225}=45\)
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 = \(\sqrt[3]{10+6\sqrt{3}}+\sqrt[3]{10-6\sqrt{3}}\)
<=> A3 = 20 - 3×2A
<=> A3 + 6A - 20 = 0
<=> A = 2
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)
\((\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}).\frac{1}{\sqrt{3}+5}\)
\(=(\frac{2\sqrt{3}+15}{3-\sqrt{3}}+\frac{3}{\sqrt{3}-2}).\frac{1}{\sqrt{3}+5}\)
\(=\frac{2\sqrt{3}+15)(\sqrt{3}-2)+3(3-\sqrt{3})}{(3-\sqrt{3})(\sqrt{3}-2)}.\frac{1}{\sqrt{3}+5}\)
\(=\frac{-15+8\sqrt{3}}{(-9+5\sqrt{3})(\sqrt{3}+5)}=\frac{-15+8\sqrt{3}}{-30+16\sqrt{3}}=\frac{-15+8\sqrt{3}}{2(-15+8\sqrt{3})}=\frac{1}{2}\)
1,\(4\sqrt{5}+2\sqrt{5}-\sqrt{5}-15\sqrt{5}=-10\sqrt{5}\)
2,\(8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}=5\sqrt{5}\)
3,\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right):\sqrt{3}=33\)
4,\(7\sqrt{7a}+3\sqrt{7a}-2\sqrt{7a}=8\sqrt{7a}\)
5,\(-6\sqrt{a}-\sqrt{6a}+\sqrt{6a}=-6\sqrt{a}\)
6,\(8\sqrt{3}-12\sqrt{3}+5\sqrt{3}+2\sqrt{3}=3\sqrt{3}\)
b)\(\frac{\sqrt{27}}{\sqrt{12}}+\frac{1}{2}\)
\(=\frac{\sqrt{3}.\sqrt{9}}{\sqrt{3}.\sqrt{4}}+\frac{1}{2}\)
\(=\frac{\sqrt{9}}{\sqrt{4}}+\frac{1}{2}\)
\(=\frac{3}{2}+\frac{1}{2}\)
\(\frac{4}{2}=2\)
a) \(\sqrt{45}.\sqrt{15}.\sqrt{27}\)
\(=\left(\sqrt{15}\right)^2.\left(\sqrt{3}\right)^2.\sqrt{9}\)
\(=15.3.3\)
\(=135\)