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21 tháng 7 2019

1. 2^36.2^45=2^81

2. 2^30.5^30=10^30

\(4^{18}.8^{15}=\left(2^2\right)^{18}.\left(2^3\right)^{15}\)

\(=2^{36}.2^{45}\)

\(=2^{81}\)

\(4^{15}.5^{30}=\left(2^2\right)^{15}.5^{30}\)

\(=2^{30}.5^{30}\)

\(=\left(2.5\right)^{30}\)

\(=10^{30}\)

\(\frac{3.72^2.54^2}{108^4}=\frac{3.\left(3^2.2^3\right)^2.\left(3^3.2\right)^2}{\left(3^3.2^2\right)^4}\)

\(=\frac{3.3^4.2^6.3^6.2^2}{3^{12}.2^8}\)

\(=\frac{3^{11}.2^8}{3^{12}.2^8}\)

\(=\frac{1}{3}\)

14 tháng 7 2023

a) \(5\sqrt{48}-4\sqrt{27}-2\sqrt{57}+\sqrt{108}\)

\(=20\sqrt{3}-12\sqrt{3}-2\sqrt{57}+6\sqrt{3}\)

\(=\left(20-12+6\right)\sqrt{3}-2\sqrt{57}\)

\(=14\sqrt{3}-2\sqrt{57}\)

b) \(2\sqrt{24}-2\sqrt{54}+3\sqrt{6}-\sqrt{150}\)

\(=4\sqrt{6}-6\sqrt{6}+3\sqrt{6}-5\sqrt{6}\)

\(=\left(4-6+3-5\right)\sqrt{6}\)

\(=-4\sqrt{6}\)

4 tháng 2 2018

\(A=\frac{3469-54}{6938-108}\)

\(=\frac{3415}{6830}\)

\(=\frac{3415}{6830}=\frac{3415:3415}{6830:3415}=\frac{1}{2}=\frac{3}{6}\)

\(B=\frac{2468-89}{3720-147}\)

\(=\frac{2370}{3555}\)

\(=\frac{2370}{3555}=\frac{2370:1185}{3555:1185}=\frac{2}{3}=\frac{4}{6}\)

4 tháng 2 2018

bn ơi còn C,D

20 tháng 8 2021

1.
A= \(2\sqrt{6}\) + \(6\sqrt{6}\) - \(8\sqrt{6}\)
A= 0
2.
A= \(12\sqrt{3}\) + \(5\sqrt{3}\) - \(12\sqrt{3}\)
A= 0
3.
A= \(3\sqrt{2}\) - \(10\sqrt{2}\) + \(6\sqrt{2}\)
A= -\(\sqrt{2}\)
4.
A= \(3\sqrt{2}\) + \(4\sqrt{2}\) - \(\sqrt{2}\)
A= \(6\sqrt{2}\)
5.
M= \(2\sqrt{5}\) - \(3\sqrt{5}\) + \(\sqrt{5}\)
M= 0
6.
A= 5 - \(3\sqrt{5}\) + \(3\sqrt{5}\)
A= 5

This literally took me a while, pls sub :D
https://www.youtube.com/channel/UC4U1nfBvbS9y_Uu0UjsAyqA/featured

26 tháng 8 2018

3√2 - 5√18 + 6√72 - 4√98 = 3√2-5.3√2+6.2.3√2-4.7/3.3√2

                                          = 3√2(1-5+12-28/3)
                                          = 3√2.(-4/3)
                                          = -4√2

29 tháng 6 2023

\(1,\left(x+y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y.2x=4xy\)

\(2,\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)

\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3\)

\(=6x^2y\)

\(3,\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =4y^2\)

\(4,\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\\ =\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\\ =\left(2x+3-2x-5\right)^2\\ =\left(-2\right)^2\\ =4\)

\(5,9^8.2^8-\left(18^4+1\right)\left(18^4-1\right)\\ =18^8-\left[\left(18^4\right)^2-1\right]\\ =18^8-18^8+1\\ =1\)

1: =x^2+2xy+y^2-x^2+2xy-y^2=4xy

2: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3

=6x^2y

3: =(x+y-x+y)^2=(2y)^2=4y^2

4: =(2x+3-2x-5)^2=(-2)^2=4

5: =18^8-18^8+1=1

3 tháng 10 2018

\(\frac{72^3\times54^2}{108^4}\)

\(=\frac{\left(2^3\times3^2\right)^3\times\left(2\times3^3\right)^2}{\left(2^2\times3^3\right)^4}\)

\(=\frac{2^9\times3^6\times2^2\times3^6}{2^8\times3^{12}}\)

\(=\frac{2^{11}\times3^{12}}{2^8\times3^{12}}\)

\(=\frac{2^3}{1}\)

\(=\frac{8}{1}=8\)

a: Ta có: \(\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\)

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

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

b: Ta có: \(\sqrt{17-2\sqrt{72}}+\sqrt{19+2\sqrt{18}}\)

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

\(=4+\sqrt{2}\)

c: Ta có: \(\sqrt{12-2\sqrt{32}}+\sqrt{9+4\sqrt{2}}\)

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

\(=4\sqrt{2}-1\)

22 tháng 8 2021

a)

\(\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{5+2\sqrt{5}\cdot\sqrt{3}+3}-\sqrt{5+2\sqrt{5}\cdot\sqrt{1}+1}\\ =\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{1}\right)^2}\\ =\sqrt{5}+\sqrt{3}-\sqrt{5}-\sqrt{1}\\ =\sqrt{3}-\sqrt{1}\)

b)

\(\sqrt{17-2\sqrt{72}}+\sqrt{19+2\sqrt{18}}\\ =\sqrt{9-2\sqrt{9}\cdot\sqrt{8}+8}+\sqrt{18+2\sqrt{18}\cdot\sqrt{1}+1}\\ =\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}+1\right)^2}\\ =3-2\sqrt{2}+3\sqrt{2}+1\\ =4+\sqrt{2}\)

c)

\(\sqrt{12-2\sqrt{32}}+\sqrt{9+4\sqrt{2}}\\ =\sqrt{8-2\sqrt{8}\cdot\sqrt{4}+4}+\sqrt{8+2\sqrt{8}\cdot\sqrt{1}+1}\\ =\sqrt{\left(2\sqrt{2}-2\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}\\ =2\sqrt{2}-2+2\sqrt{2}+1\\ =4\sqrt{2}-1\)