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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

12 tháng 11 2021

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

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: \(=2\cdot2\sqrt{6}-2\cdot3\sqrt{6}+3\sqrt{6}-5\sqrt{6}=-4\sqrt{6}\)

13 tháng 11 2021

\(a.\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}=\left|\sqrt{2}-\sqrt{3}\right|+3\sqrt{2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}\left(\sqrt{3}>\sqrt{2}\right)=\sqrt{3}+2\sqrt{2}\)\(b.3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}=-6\sqrt{2}\)

29 tháng 6 2021

\(a,A=2\sqrt{2}-9\sqrt{2}+16\sqrt{2}-5\sqrt{2}\)

\(=4\sqrt{2}\)

\(b,B=\left|1-\sqrt{5}\right|+\sqrt{5+2\sqrt{5}+1}\)

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

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

\(c,C=\dfrac{2+\sqrt{6}+2-\sqrt{6}}{\left(2+\sqrt{6}\right)\left(2-\sqrt{6}\right)}=\dfrac{4}{4-6}=-2\)
 

AH
Akai Haruma
Giáo viên
29 tháng 6 2021

Lời giải:

a. 

\(A=2\sqrt{2}-3\sqrt{18}+4\sqrt{32}-\sqrt{50}=2\sqrt{2}-9\sqrt{2}+16\sqrt{2}-5\sqrt{2}\)

\(=(2-9+16-5)\sqrt{2}=4\sqrt{2}\)

b.

\(B=\sqrt{(1-\sqrt{5})^2}+\sqrt{(\sqrt{5}+1)^2}=|1-\sqrt{5}|+|\sqrt{5}+1|=\sqrt{5}-1+\sqrt{5}+1=2\sqrt{5}\)

c.

\(C=\frac{2+\sqrt{6}+2-\sqrt{6}}{(2-\sqrt{6})(2+\sqrt{6})}=\frac{4}{2^2-6}=-2\)

a: \(=\left(2\sqrt{3}-12\sqrt{3}+15\sqrt{3}\right)\cdot\sqrt{3}=5\sqrt{3}\cdot\sqrt{3}=15\)

b: \(=\left(6\sqrt{2}-16\sqrt{2}+15\sqrt{2}\right):5=\sqrt{2}\)

c: \(=\dfrac{\left(2\sqrt{5}-6\sqrt{5}+15\sqrt{5}\right)}{\sqrt{5}}=17-6=11\)

4 tháng 7 2021

\(a,=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}\)

\(=\sqrt{2}\left(3-12+8-5\right)=-6\sqrt{2}\)

\(b,=\left|\sqrt{2}-\sqrt{3}\right|+3\sqrt{2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}=\sqrt{3}+2\sqrt{2}\)

\(c,=\sqrt{5}+\sqrt{5}+\dfrac{5}{\sqrt{5}}-1=3\sqrt{5}-1\)

\(d,=\sqrt{3-2.2\sqrt{3}+4}+\sqrt{\left(1+\sqrt{3}\right)^2}\)

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

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

4 tháng 7 2021

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: \(A=\left(1-\sqrt{7}\right)\cdot\left(1+\sqrt{7}\right)=1-7=-6\)

b: \(B=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}=-4\sqrt{3}\)

c: \(C=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)

30 tháng 5 2022

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) \(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)

\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\dfrac{10}{3}\sqrt{3}\)

\(=-\dfrac{17}{3}\sqrt{3}\)