Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a.\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}=\left|\sqrt{2}-\sqrt{3}\right|+3\sqrt{2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}\left(\sqrt{3}>\sqrt{2}\right)=\sqrt{3}+2\sqrt{2}\)\(b.3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}=-6\sqrt{2}\)
a: \(=\sqrt{3}-\sqrt{2}+3\sqrt{2}=2\sqrt{2}+\sqrt{3}\)
a: \(=12\sqrt{2}+5\cdot3\sqrt{2}-3\cdot5\sqrt{2}-2\cdot4\sqrt{2}\)
\(=12\sqrt{2}-8\sqrt{2}=4\sqrt{2}\)
b: \(=\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(5-\sqrt{3}\right)^2}\)
\(=\sqrt{3}+1+5-\sqrt{3}\)
=6
\(2\sqrt{20}-\sqrt{45}+3\sqrt{18}+3\sqrt{32}-\sqrt{50}\\ =4\sqrt{5}-3\sqrt{5}+9\sqrt{2}+12\sqrt{2}-5\sqrt{2}\\ =\sqrt{5}+16\sqrt{2}\)
6: \(=3\cdot2\sqrt{3}-4\cdot3\sqrt{3}+5\cdot4\sqrt{3}=14\sqrt{3}\)
7: \(=2\sqrt{3}+5\sqrt{3}-4\sqrt{3}=3\sqrt{3}\)
8: \(=2\cdot4\sqrt{2}+4\cdot2\sqrt{2}-5\cdot3\sqrt{2}=\sqrt{2}\)
9: \(=3\cdot2\sqrt{5}-2\cdot3\sqrt{5}+4\sqrt{5}=4\sqrt{5}\)
10: \(=2\cdot2\sqrt{6}-2\cdot3\sqrt{6}+3\sqrt{6}-5\sqrt{6}=-4\sqrt{6}\)
\(\sqrt{32}+\sqrt{50}-2\sqrt{8}+\dfrac{1}{3}\sqrt{18}\)
\(=\sqrt{4^2\cdot2}+\sqrt{5^2\cdot2}-2\cdot2\sqrt{2}+\dfrac{1}{3}\cdot\sqrt{3^2\cdot2}\)
\(=4\sqrt{2}+5\sqrt{2}-4\sqrt{2}+\dfrac{1}{3}\cdot3\sqrt{2}\)
\(=\left(4\sqrt{2}-4\sqrt{2}\right)+5\sqrt{2}+\sqrt{2}\)
\(=5\sqrt{2}+\sqrt{2}\)
\(=6\sqrt{2}\)
1) \(3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}\)
\(=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}\)
\(=-6\sqrt{2}\)
2) \(\sqrt{50}-\sqrt{18}+\sqrt{200}-\sqrt{162}\)
\(=5\sqrt{2}-3\sqrt{2}+10\sqrt{2}-9\sqrt{2}\)
\(=3\sqrt{2}\)
3) \(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
\(=5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
\(=-2\sqrt{5}\)
4) \(5\sqrt{48}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)
\(=20\sqrt{3}-12\sqrt{3}-10\sqrt{3}+6\sqrt{3}\)
\(=4\sqrt{3}\)
5) \(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\dfrac{10}{3}\sqrt{3}\)
\(=-\dfrac{17}{3}\sqrt{3}\)
\(a,=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}\)
\(=\sqrt{2}\left(3-12+8-5\right)=-6\sqrt{2}\)
\(b,=\left|\sqrt{2}-\sqrt{3}\right|+3\sqrt{2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}=\sqrt{3}+2\sqrt{2}\)
\(c,=\sqrt{5}+\sqrt{5}+\dfrac{5}{\sqrt{5}}-1=3\sqrt{5}-1\)
\(d,=\sqrt{3-2.2\sqrt{3}+4}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+1+\sqrt{3}=2\)
a) \(3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}=3\sqrt{2}-4\sqrt{9.2}+2\sqrt{16.2}-\sqrt{25.2}\)
\(=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}=-6\sqrt{2}\)
b) \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}=\left|\sqrt{2}-\sqrt{3}\right|+\sqrt{9.2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}\)
\(=2\sqrt{2}+\sqrt{3}\)
c) \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}=\sqrt{25.\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{9.5}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}\)
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-1=3\sqrt{5}-1\)
d) \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}=\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}+\left|\sqrt{3}+1\right|\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{3}+1=\left|2-\sqrt{3}\right|+\sqrt{3}+1=2-\sqrt{3}+\sqrt{3}+1=3\)
a) 2√18 - 4√50 + 3√32
= 6√2 - 20√2 + 12√2
= -2√2
b) √(√8 - 4)² + √8
= 4 - √8 + √8
= 4
c) √(14 - 6√5) + √(6 + 2√5)
= √(3 - √5)² + √(√5 + 1)²
= 3 - √5 + √5 + 1
= 4
\(a,2\sqrt{18}-4\sqrt{50}+3\sqrt{32}\\ =6\sqrt{2}-20\sqrt{2}+12\sqrt{2}=-2\sqrt{2}\\ b,\sqrt{\left(\sqrt{8}-4\right)^2}+\sqrt{8}\\ =4-\sqrt{8}+\sqrt{8}\\ =4\\ c,\sqrt{14-6\sqrt{5}}+\sqrt{6+2\sqrt{5}}\\ =\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}=3+\sqrt{5}+\sqrt{5}+1\\ =4+2\sqrt{5}\)
a: \(A=\left(1-\sqrt{7}\right)\cdot\left(1+\sqrt{7}\right)=1-7=-6\)
b: \(B=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}=-4\sqrt{3}\)
c: \(C=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)
Nah I'd Win
-6\(\sqrt{ }\)2