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

\(8.2^{3x}.7^y=56^{2x}.5^{x-1}\)

\(2^3.2^{3x}.7^y=7^{2x}.8^{2x}.5^{x-1}\)

\(2^{3+3x}.7^y=7^{2x}.2^{6x}.5^{x-1}\)

\(7^{2x}:7^y=2^{6x}:2^{3+3x}.5^{x-1}\)

\(7^{2x-y}=2^{6x-3-3x}.5^{x-1}\)

\(7^{2x-y}=2^{3x-3}.5^{x-1}\)

\(7^{2x-y}=2^{3x}:8.5^x:5\)

\(7^{2x-y}=8^x.5^x:40\)

\(7^{2x-y}=40^x:40\)

\(7^{2x-y}=40^{x-1}\)

\(\Rightarrow x=y=1\)

\(\Leftrightarrow2^{3x+3}\cdot7^y=2^{6x}\cdot7^{2x}\cdot5^{x-1}\)

=>3x+3=6x; y=2x; x-1=0

=>\(\left(x,y\right)\in\varnothing\)

a, \(\left|x+2\right|-\left|x+7\right|=0\Rightarrow\left|x+2\right|=\left|x+7\right|\Rightarrow\orbr{\begin{cases}x+2=x+7\\x+2=-x-7\end{cases}\Rightarrow\orbr{\begin{cases}0=5\left(loại\right)\\2x=-9\end{cases}\Rightarrow}x=\frac{-9}{2}}\)

b, - Nếu \(2x-1\ge0\Rightarrow x\ge\frac{1}{2}\), ta có: 2x - 1 = 2x - 1 => 2x = 2x (thỏa mãn với mọi x)

- Nếu 2x - 1 < 0 => \(x< \frac{1}{2}\), ta có: 2x - 1 = 1 - 2x => 4x = 2 => x = \(\frac{1}{2}\) (không thỏa mãn điều kiện)

Vậy \(x\ge\frac{1}{2}\)

c,d tương tự b

e, tương tự a

2^x+1.3^y=12^x

=> 2^x+1.3^y=(3.2²)^x

=> 2^x+1.3^y=3^x.(2²)^x

=> 2^x+1.3^y=3^x.2^2x

vậy+) 2^x+1=2^2x

=>x+1=2x

=> x=1

+)3^y=3^x

=> y=x mà x=1=>y=1

Vậy x=1

y=1

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