Ta có : \(\frac{3}{2}.4^x+\frac{5}{4}.4^{x+2}=\frac{3}{2}.4^8+\frac{5}{3}.4^{10}\)

\(\Rightarrow4^x\left(\frac{3}{2}+\frac{5}{3}.4^2\right)\)=\(4^8\left(\frac{3}{2}+\frac{5}{3}.4^2\right)\)

\(\Rightarrow4^x=4^8\)

\(\Rightarrow x=8\)

\(Can\) \(you\) \(k\)\(for\)\(me?\)

Dễ lắm

trả lời đi

\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot....\cdot\frac{30}{62}\cdot\frac{31}{64}=2^x\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.....\cdot\frac{30}{31}\cdot\frac{31}{32}\right)=2^x\)

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

Làm nốt.

ko làm được câu này hay câu b ib với tớ nha.khẳng định tối giải.

135 : 12 * x + 108 = 87

a, \(\frac{2}{5}+\frac{1}{4}\times x=\frac{3}{10}\)

\(\Leftrightarrow\frac{1}{4}\times x=\frac{3}{10}-\frac{2}{5}\)

\(\Leftrightarrow\frac{1}{4}\times x=\frac{-1}{10}\)

\(\Leftrightarrow x=\frac{-1}{10}\div\frac{1}{4}\)

\(\Leftrightarrow x=\frac{-2}{5}\)

Vậy \(x=\frac{-2}{5}\)

b, \(\frac{2}{3}+\frac{2}{3}\div x=\frac{4}{15}\)

\(\Leftrightarrow\frac{2}{3}\div x=\frac{4}{15}-\frac{2}{3}\)

\(\Leftrightarrow\frac{2}{3}\div x=\frac{4}{15}-\frac{2}{3}\)

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

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

\(\Leftrightarrow\frac{-5}{3}\)

Vậy \(x=\frac{-5}{3}\)

c, \(2\times\left|\frac{2}{3}-x\right|+\frac{1}{4}=\frac{3}{4}\)

\(\Leftrightarrow2\times\left|\frac{2}{3}-x\right|=\frac{3}{4}-\frac{1}{4}\)

\(\Leftrightarrow2\times\left|\frac{2}{3}-x\right|=\frac{1}{2}\)

\(\Leftrightarrow\left|\frac{2}{3}-x\right|=\frac{1}{2}\div2\)

\(\Leftrightarrow\left|\frac{2}{3}-x\right|=\frac{1}{4}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{2}{3}-x=\frac{1}{4}\\\frac{2}{3}-x=\frac{-1}{4}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{5}{12}\\x=\frac{11}{12}\end{cases}}\)

Vậy \(x\in\left\{\frac{5}{12};\frac{11}{12}\right\}\)

d, \(3\times\left|\frac{5}{4}-x\right|-\frac{1}{8}=\frac{1}{4}\)

\(\Leftrightarrow3\times\left|\frac{5}{4}-x\right|=\frac{1}{4}+\frac{1}{8}\)

\(\Leftrightarrow3\times\left|\frac{5}{4}-x\right|=\frac{3}{8}\)

\(\Leftrightarrow\left|\frac{5}{4}-x\right|=\frac{3}{8}\div3\)

\(\Leftrightarrow\left|\frac{5}{4}-x\right|=\frac{1}{8}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{5}{4}-x=\frac{1}{8}\\\frac{5}{4}-x=\frac{-1}{8}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{9}{8}\\x=\frac{11}{8}\end{cases}}\)

Vậy \(x\in\left\{\frac{9}{8};\frac{11}{8}\right\}\)

b) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=\frac{4^5.\left(1+1+1+1\right)}{3^5.\left(1+1+1\right)}.\frac{6^5.\left(1+1+1+1+1+1\right)}{2^5.\left(1+1\right)}\)

                                                                                                       \(=\frac{4^5.4}{3^5.3}.\frac{6^5.6}{2^5.2}=\frac{4^6}{3^6}.\frac{6^6}{2^6}=\frac{2^{12}.2^6.3^6}{3^6.2^6}=2^{12}\)

   Ta có: \(2^{12}=\left(2^3\right)^4=8^4\)                                                    

Vậy x= 4

a.4^7

b.8^5 

 c.cho x mk sẻ tính kết quả nhưng tìm xmk ko tính đâu