Question

[TFBoys]

tính

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

c, Cho x,y,z là các số # 0 và x² = yz, y² = xz , z² = xy. Chứng minh rằng x = y= z

Answer 1

[Hotel del Luna]

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