\(\Leftrightarrow2.2^{x-2}+5.2^{x-2}=7.2^{-5}\Leftrightarrow7.2^{x-2}=7.2^{-5}\)

\(\Leftrightarrow x^{x-2}=2^{-5}\Leftrightarrow x-2=-5\Leftrightarrow x=-3\)

<=> \(2.2^{x-2}+5.2^{x-2}=\frac{7}{32}\) <=> \(\left(2+5\right).2^{x-2}=\frac{7}{32}\)

<=> \(7.2^{x-2}=\frac{7}{32}\)<=> \(2^{x-2}=\frac{1}{32}=2^{-5}\) => x - 2 = -5 => x = -3

2^x-1+5.2^x-2=7/32

=>2.2^x-2+5.2^x-2=7/32

=7.2^x-2=7/32

=>2^x-2=1/32

=>2^x-2=2^-5

=>x-2=-5

=>x=-3

vậy x=-3

2^x-1+5.2^x-2=7/32

=>2.2^x-2+5.2^x-2=7/32

=7.2^x-2=7/32

=>2^x-2=1/32

=>2^x-2=2^-5

=>x-2=-5

=>x=-3

vậy x=-3

k nha

\(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)

\(\Rightarrow2.2^{x-2}+5.2^{x-2}=\frac{7}{32}\)

\(=7.2^{x-2}=\frac{7}{32}\)

\(\Rightarrow2^{x-2}=\frac{1}{32}\)

\(=2^{x-2}=2^{-5}\)

\(\Rightarrow x-2=-5\)

\(\Rightarrow x=-3\)

2
x-1+5.2
x-2=7/32
=>2.2
x-2+5.2
x-2=7/32
=7.2
x-2=7/32
=>2
x-2=1/32
=>2
x-2=2
-5
=>x-2=-5
=>x=-3
vậy x=-3

1) \(4x=7y\Leftrightarrow\dfrac{x}{7}=\dfrac{y}{4}\Rightarrow\dfrac{x^2}{49}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{49+16}=\dfrac{260}{65}=4\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2=4.49=196\\y^2=4.16=64\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=14,y=8\\x=-14,y=-8\end{matrix}\right.\) (vì \(\dfrac{x}{7}=\dfrac{y}{4}\) nên \(x,y\) cùng dấu) 

2) \(2^{x-1}+5.2^{x-2}=\dfrac{7}{32}\)

\(\Leftrightarrow2^{x-1}+\dfrac{5}{2}.2^{x-1}=\dfrac{7}{32}\)

\(\Leftrightarrow2^{x-1}=\dfrac{1}{16}=2^{-4}\)

\(\Leftrightarrow x-1=-4\)

\(\Leftrightarrow x=-3\)

3) \(\left|x+5\right|+\left(3y-4\right)^{2016}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\3y-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=\dfrac{4}{3}\end{matrix}\right.\)

a. 2^x-1+ 5.2^x-1:2=7/32

=> 2^x+1.(1+5/2)=7/32

=>2^x+1.7/2=7/32

=> 2^x+1=1/16=1/2⁴

=> x+1=-4=>x=-5

\(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)

\(\Rightarrow2^{x-1}+5.2^{x-1}.2^3=\frac{7}{32}\)

\(\Rightarrow2^{x-1}.\left(1+5.2^3\right)=\frac{7}{32}\)

\(\Rightarrow2^{x-1}.41=\frac{7}{32}\)

\(\Rightarrow2^{x-1}=\frac{7}{1312}\)

\(\Rightarrow\)   Ko có x thỏa mãn

ai ko bt = -3 giải rõ ràng đi              

\(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)

\(2.2^{x-1}+5.2^{x-2}=\frac{7}{32}\)

      \(2^{x-2}.\left(5+2\right)=\frac{7}{32}\)

                   \(2^{x-2}.7=\frac{7}{32}\)

                       \(2^{x-2}=\frac{7}{32}:7\)

                       \(2^{x-2}=\frac{1}{32}\)

               \(\Rightarrow x-2=-5\)

                              \(x=-5+2\)

Vậy                           \(x=-3\)

\(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)

\(\Rightarrow2^{x-1}+2^{x-1}.\frac{5}{2}=\frac{7}{32}\Rightarrow2^{x-1}.\left(1+\frac{5}{2}\right)=\frac{7}{32}\Rightarrow2^{x-1}.\frac{7}{2}=\frac{7}{32}\)

\(\Rightarrow2^{x-1}=\frac{7}{32}:\frac{7}{2}=\frac{7}{32}.\frac{2}{7}=\frac{1}{16}\)

\(\Rightarrow2^{x-1}=2^{-4}\Rightarrow x-1=-4\Rightarrow x=-4+1=-3\)