0,7 * y + 3/10 * y = 6,3

0,7 * y + 0,3 * y = 6,3

y * [0,7+0,3] =6,3

y * 1 =6,3

      y = 6,3 :1

       y=6,3

0,7 x y + \(\frac{3}{10}\)x y = 6,3

( 0, 7 + \(\frac{3}{10}\)) x y = 6,3

          1             x y = 6,3

 0.7*y+3/10*y=6.3

(0.7+3/10)*y=6.3

1*y=6.3

y=6.3:1

y=6.3

Vậy y=6.3

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

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

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

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

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

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

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

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

\(=\dfrac{3^5\cdot2^5\cdot3^6}{2^6\cdot3^{10}}-\dfrac{1}{2}=\dfrac{3}{2}-\dfrac{1}{2}=1\)

A. 0 giờ ngày 5/10/2021.

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