1)

\(\frac{7.8^3-5.2^{10}}{\left(-16\right)^2}\)       

=  \(\frac{7.2^8.2-5.2^8.2^2}{16^2}\)

=  \(\frac{2^8.\left(2.7-5.2^2\right)}{2^8}\)

=   \(\frac{2^8.\left(-6\right)}{2^8}\)

=   \(-6\)

2)

b)\(\frac{-28}{4}\le x\le\frac{-21}{7}\)

\(\Rightarrow-7\le x\le-3\)

\(\Rightarrow x=\left\{-7;-6;-5;-4;-3\right\}\)

2)

a)\(\frac{108}{12}\le x\le\frac{91}{7}\)

\(\Rightarrow9\le x\le13\)

\(\Rightarrow x=\left\{9;10;11;12;13\right\}\)

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                           \(x=-3\)

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

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

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

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

a) (2x-1)\(^2\)+\(\left|2y-x\right|\)=0

Vì (2x-1)\(^2\)\(\ge\)0 với mọi x

\(\left|2y-x\right|\)\(\ge\)0 với mọi y

\(\Rightarrow\)\(\left\{\begin{matrix}2x-1=0\\2y-x=0\end{matrix}\right.\)\(\Rightarrow\)\(\left\{\begin{matrix}x=\frac{1}{2}\\2y-\frac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\)\(\left\{\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{4}\end{matrix}\right.\)

Vậy .....

b)\(\left|x-\frac{1}{3}\right|\)+\(\frac{4}{5}\)=\(\frac{14}{5}\)

\(\Rightarrow\)\(\left|x-\frac{1}{3}\right|\)=2

\(\Rightarrow\)\(\left[\begin{matrix}x-\frac{1}{3}=2\\x-\frac{1}{3}=-2\end{matrix}\right.\)\(\Rightarrow\)\(\left[\begin{matrix}x=\frac{7}{3}\\x=\frac{-5}{3}\end{matrix}\right.\)

Vậy ....

\(2^{x-1}+5\cdot2^{x-2=\frac{7}{32}}\)

\(2^{x-1}+5\cdot2^{\frac{71}{32}}\)

\(2^{\frac{39}{32}}+23.2744\)

\(25.60194\)

=2^x-1 + 5.(2^x : 4) = 2^x-1 + 5[(2^x-1)/2]=2^x-1 + 2,5.2^x-1=3,5.(2^x-1)=7/32

từ đó suy ra dpcm