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28 tháng 7 2018

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21 tháng 8 2018

A = \(\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}\)\(\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}\)

\(\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}\)\(\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3+1\right)}\)\(\frac{2.6}{3.7}=\frac{4}{7}\)

c, theo đề bài ta có : 

x2 = yz, y2 = xz , z2 = xy

\(\Rightarrow\frac{x}{y}=\frac{z}{x},\frac{y}{x}=\frac{z}{y},\frac{z}{x}=\frac{y}{z}\Rightarrow\frac{x}{y}=\frac{z}{x}=\frac{y}{z}\)

AD t/c DTSBN, ta có 

\(\frac{x}{y}=\frac{z}{x}=\frac{y}{z}\Rightarrow\frac{X+z+y}{y+x+z}=1\)

x= 1y

z= 1x

y= 1z

=> x = y = x

1 tháng 6 2021

\(\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\dfrac{\left(2^4\right)^3.3^{10}+3.5.2^3.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}\)

\(=\dfrac{2^{12}.3^{10}+3.5.2^3.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}+5.2^{12}.3^{10}}{2^{12}.3^{12}+2^{11}.3^{11}}\)

\(=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.\left(2.3+1\right)}\)\(=\dfrac{2^{12}.3^{10}.6}{2^{11}.3^{11}.7}=\dfrac{2.2}{7}=\dfrac{4}{7}\)

30 tháng 8 2017

\(\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)

=\(\dfrac{\left(4^2\right)^3.3^{10}+120.1}{4^6.3^{12}+6^{11-9}}\)

=\(\dfrac{4^{2.3}.1+120}{4^6.3^{12-10}+6^2}\)

=\(\dfrac{4^6+120}{4^6.3^2+6^2}\)

=\(\dfrac{4096+120}{4096.9+36}\)

=\(\dfrac{4216}{36864+36}\)

=\(\dfrac{4216}{36900}=\dfrac{2063}{18450}\)

31 tháng 8 2017

sai rồi ba

6 tháng 9 2015

\(y=\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)

\(y=\frac{2^{12}.3^{10}+2^9.3^9.120}{2^{12}.3^{12}+2^{11}.3^{11}}\)

\(y=\frac{2^9.3^9\left(2^3.3+120\right)}{2^{11}.3^{11}\left(2.3+1\right)}\)

\(y=\frac{6^9\left(2^3.3+120\right)}{6^{11}.7}\)

\(y=\frac{2^3.3+120}{6^2.7}\)

\(y=\frac{144}{252}\)

\(y=\frac{4}{7}\)

 

20 tháng 6 2015

\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+3^{11}.2^{11}}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3+1\right)}=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}\left(2.3+1\right)}=\frac{2.6}{3.7}=\frac{12}{21}=\frac{4}{7}\)

7 tháng 1 2017

\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\frac{4}{7}\)

AH
Akai Haruma
Giáo viên
16 tháng 7

Lời giải:

Gọi biểu thức là $A$.

\(A=\frac{(2^4)^3.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}\\ =\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}(2.3+1)}\\ =\frac{2^{12}.3^{10}(1+5)}{7.2^{11}.3^{11}}=\frac{2^{12}.3^{10}.2.3}{7.2^{11}.3^{11}}\\ =\frac{2^{13}.3^{11}}{7.2^{11}.3^{11}}=\frac{2^2}{7}=\frac{4}{7}\)