xl chuyển hộ mk "bình phương" thành "lập phương" nha

\(=\left(\dfrac{2}{3}a\right)^3-3.\left(\dfrac{2}{3}\right)^2a^2.2b+3.\dfrac{2}{3}a.4b^2-\left(2b\right)^3=\left(\dfrac{2}{3}a-2b\right)^3\)

\(C=\dfrac{2^6\cdot3^{10}}{3^9\cdot2^6}=3\\ D=\dfrac{3^{24}\cdot3^{10}}{3^{21}\cdot3^{11}}=\dfrac{3^{34}}{3^{32}}=3^2=9\\ F=\dfrac{2^{45}\cdot5^{14}}{5^{15}\cdot2^{47}}=\dfrac{1}{2^2\cdot5}=\dfrac{1}{20}\\ G=\dfrac{2^2\cdot5^2\cdot5^3}{2^3\cdot5^4}=\dfrac{1\cdot5}{2}=\dfrac{5}{2}\)

C=3

D=9

F=1/20

G=5/2

Em ko giải chi tiết vì nó lâu

Mong thông cảm!

C= \(\dfrac{8^2.9^5}{3^9.4^3}\)

   = \(\dfrac{\left(2^3\right)^2.\left(3^2\right)^5}{3^9.\left(2^2\right)^3}\)

   = \(\dfrac{2^6.3^{10}}{3^9.2^6}\)

   = \(\dfrac{3^1}{1}\)

   = \(\dfrac{3}{1}\)

\(a.\sqrt{72}-5\sqrt{2}+3\sqrt{12}\\ =6\sqrt{2}-5\sqrt{2}+6\sqrt{3}\\ =\sqrt{2}+6\sqrt{3}\\ b.6\sqrt{\dfrac{1}{2}}-\dfrac{2}{\sqrt{2}}-5\sqrt{2}\\ =3\sqrt{2}-\sqrt{2}-5\sqrt{2}\\ =-3\sqrt{2}\\ c.\dfrac{\sqrt{8}-2}{\sqrt{2}-1}+\dfrac{2}{\sqrt{3}-1}-\dfrac{3}{\sqrt{3}}\\ =2+1+\sqrt{3}-\sqrt{3}\\ =3\\ d.\sqrt[3]{64}+\sqrt[3]{27}-2\sqrt[3]{-8}\\ =4+3+4\\ =11\)

a) \(\dfrac{2}{5}-\dfrac{3}{15}\)

\(=\dfrac{2}{5}-\dfrac{3:3}{15:3}\)

\(=\dfrac{2}{5}-\dfrac{1}{5}\)

\(=\dfrac{1}{5}\)

b) \(\dfrac{9}{27}-\dfrac{2}{9}\)

\(=\dfrac{9:3}{27:3}-\dfrac{2}{9}\)

\(=\dfrac{3}{9}-\dfrac{2}{9}\)

\(=\dfrac{1}{9}\)

c) \(\dfrac{18}{24}-\dfrac{4}{8}\)

\(=\dfrac{18:6}{24:6}-\dfrac{4:2}{8:2}\)

\(=\dfrac{3}{4}-\dfrac{2}{4}\)

\(=\dfrac{1}{4}\)

d) \(\dfrac{6}{16}-\dfrac{10}{64}\)

\(=\dfrac{6\times2}{16\times2}-\dfrac{10:2}{64:2}\)

\(=\dfrac{12}{32}-\dfrac{5}{32}\)

\(=\dfrac{7}{32}\)

a) Ta có: \(\dfrac{x^2}{y^2}:\sqrt{\dfrac{x^2}{y^4}}\)

\(=\dfrac{x^2}{y^2}:\dfrac{x}{y^2}\)

=x

b) Ta có: \(\sqrt{\dfrac{27\left(x-1\right)^2}{12}}+\dfrac{3}{2}-\left(x-2\right)\sqrt{\dfrac{50x^2}{8\left(x-2\right)^2}}\)

\(=\sqrt{\dfrac{9}{4}}\cdot\sqrt{\left(x-1\right)^2}+\dfrac{3}{2}-\left(x-2\right)\cdot\sqrt{\dfrac{25}{4}}\cdot\sqrt{\dfrac{x^2}{\left(x-2\right)^2}}\)

\(=\dfrac{3}{2}\cdot\left(x-1\right)+\dfrac{3}{2}-\left(x-2\right)\cdot\dfrac{5}{2}\cdot\dfrac{x}{2-x}\)

\(=\dfrac{3}{2}x-\dfrac{3}{2}+\dfrac{3}{2}-\dfrac{5}{2}\left(x-2\right)\cdot\dfrac{-x}{x-2}\)

\(=\dfrac{3}{2}x+\dfrac{5}{2}\cdot\left(x\right)\)

=4x

Lời giải:

Gọi biểu thức là $A$

\(A=\frac{2^{10}.3^8+5.(2^2)^5.3^8}{2^{10}.(3^3)^3-2^{10}.(3^2)^4}=\frac{2^{10}.3^8+5.2^{10}.3^8}{2^{10}.3^9-2^{10}.3^{8}}\)

\(=\frac{2^{10}.3^8(1+5)}{2^{10}.3^8(3-1)}=\frac{6}{2}=3\)