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8.2^3x.7^y=56^2x.5^x-1
=>8.8^x.7^y=(8.7)^2x.5^x-1
=>8^1+x.7^y=8^2x.7^2x.5^x-1
=>8^1+x.7^y / 8^2x.7^2x=5^x-1
=>8^x+1-2x . 7^y-2x = 5^x-1
=>8^1-x.7^y-2x=5^-(1-x)
=>8^1-x.7^y-2x=1/5^1-x
=>8^1-x.7^y-2x.5^1-x=1
=>(8.5)^x-1.7^y-2x=1
=>40^x-1.7^y-2x=1
=>x-1=0 và y-2x=0
=>x=1 và y=2
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\(8.2^{3x}.7^y=56^{2x}.5^{x-1}\)
\(2^3.2^{3x}.7^y=7^{2x}.8^{2x}.5^{x-1}\)
\(2^{3+3x}.7^y=7^{2x}.2^{6x}.5^{x-1}\)
\(7^{2x}:7^y=2^{6x}:2^{3+3x}.5^{x-1}\)
\(7^{2x-y}=2^{6x-3-3x}.5^{x-1}\)
\(7^{2x-y}=2^{3x-3}.5^{x-1}\)
\(7^{2x-y}=2^{3x}:8.5^x:5\)
\(7^{2x-y}=8^x.5^x:40\)
\(7^{2x-y}=40^x:40\)
\(7^{2x-y}=40^{x-1}\)
\(\Rightarrow x=y=1\)
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\(\Leftrightarrow2^{3x+3}\cdot7^y=2^{6x}\cdot7^{2x}\cdot5^{x-1}\)
=>3x+3=6x; y=2x; x-1=0
=>\(\left(x,y\right)\in\varnothing\)
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a) \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7};x+y+z=56\)
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{2+5+7}=\dfrac{56}{14}=4\)
\(\Rightarrow\left\{{}\begin{matrix}x=4.2=8\\y=4.5=20\\z=4.7=28\end{matrix}\right.\)
b) \(\dfrac{x}{1,1}=\dfrac{y}{1,3}=\dfrac{z}{1,4}\left(1\right);2x-y=5,5\)
\(\left(1\right)\Rightarrow\dfrac{2x-y}{1,1.2-1,3}=\dfrac{5,5}{0,9}\)
\(\Rightarrow\left\{{}\begin{matrix}x=1,1.\dfrac{5,5}{0,9}=\dfrac{6,05}{0,9}\\y=1,3.\dfrac{5,5}{0,9}=\dfrac{7,15}{0,9}\\z=\dfrac{1,4}{1,1}.x=\dfrac{1,4}{1,1}.\dfrac{6,05}{0,9}=\dfrac{8,47}{0,99}\end{matrix}\right.\)
d) \(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5};xyz=-30\)
\(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5}=\dfrac{xyz}{2.3.5}=\dfrac{-30}{30}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.\left(-1\right)=-2\\y=3.\left(-1\right)=-3\\z=5.\left(-1\right)=-5\end{matrix}\right.\)
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Ta có: x/3=y/-5 và x-y=32
=> x/3=y/4=x-y/3-(-5)=32/8=4
=> x=4.3=12
y=4.(-5)=-20
Vậy x=12
y=-20