a: \(\Leftrightarrow3^x\cdot\left(\dfrac{2}{3}\cdot3-7\right)=405\)

\(\Leftrightarrow3^x=-81\)(vô lý)

b: \(\left(0,4x-1,3\right)^2=5,29\)

=>0,4x-1,3=2,3 hoặc 0,4x-1,3=-2,3

=>0,4x=3,6 hoặc 0,4x=-1

=>x=9 hoặc x=-2,5

c: \(5\cdot2^{x+1}\cdot2^{-2}-2^x=284\)

\(\Leftrightarrow2^x\cdot5\cdot2\cdot2^{-2}-2^x=284\)

\(\Leftrightarrow2^x\cdot\left(\dfrac{5}{2}-1\right)=284\)

\(\Leftrightarrow2^x=\dfrac{568}{3}\)(vô lý)

d: \(\Leftrightarrow4^x\left(1+4^3\right)=4160\)

\(\Leftrightarrow4^x=64\)

hay x=3

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

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

\(2^x:2+2^x:\frac{4}{5}=\frac{7}{32}\)

\(2^x\cdot\left(\frac{1}{2}+\frac{5}{4}\right)=\frac{7}{32}\)

\(2^x\cdot\frac{7}{4}=\frac{7}{32}\)

\(2^x=\frac{7}{32}:\frac{7}{4}=\frac{1}{8}\)

\(2^x=\frac{2^0}{2^3}=2^{-3}\)

\(\Rightarrow x=-3\)

a) \(4^x+4^{x+3}=4160\)

\(\Rightarrow4^x+4^x.4^3=4160\)

\(\Rightarrow4^x.\left(1+4^3\right)=4160\)

\(\Rightarrow4^x.65=4160\)

\(\Rightarrow4^x=64\)

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

\(\Rightarrow x=4\)

Vậy \(x=4\)

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

\(\Rightarrow2^x.\frac{1}{2}+5.2^x.\frac{1}{4}=\frac{7}{32}\)

\(\Rightarrow2^x.\left(\frac{1}{2}+5.\frac{1}{4}\right)=\frac{7}{32}\)

\(\Rightarrow2^x.\frac{7}{4}=\frac{7}{32}\)

\(\Rightarrow2^x=\frac{7}{32}:\frac{7}{4}\)

\(\Rightarrow2^x=\frac{1}{8}\)

\(\Rightarrow2^x=2^{-3}\)

\(\Rightarrow x=-3\)

Vậy \(x=-3\)

a)\(4^x+4^{x+3}=4160\)

\(4^x+4^x\cdot4^3=4160\)

\(4^x\left(1+4^3\right)=4160\)

\(4^x\cdot65=4160\)

\(4^x=4160:65=64\)

\(4^x=4^3\)

\(\Rightarrow x=3\)

chịch ko

\(a,\left(x-5\right)^2=\left(1-3x\right)^2\)

\(\Rightarrow\left(x-5\right)^2-\left(1-3x\right)^2=0\)

\(\Rightarrow\left(x-5+1-3x\right)\left(x-5-1+3x\right)=0\)

\(\Rightarrow\left(-2x-4\right)\left(4x-6\right)=0\)

\(\Rightarrow\left(x+2\right)\left(2x-3\right)=0\)

\(\Rightarrow\hept{\begin{cases}x+2=0\\2x-3=0\end{cases}\Rightarrow\hept{\begin{cases}x=-2\\x=\frac{3}{2}\end{cases}}}\)

a) \(\left(x-5\right)^2=\left(1-3x\right)^2\)

\(\Rightarrow x-5=1-3x\)

\(\Leftrightarrow x+3x=1+5\)

\(\Leftrightarrow4x=6\)

\(\Leftrightarrow x=\frac{3}{2}\)

a) \(5.2^{x+1}.2^{-2}-2^x=384\Leftrightarrow2^x\left(5.2^{-2}.2-1\right)=384\)\(\Leftrightarrow2^x.1,5=384\Leftrightarrow2^x=384:1,5=256=2^8\)

\(\Rightarrow x=8\)

b) \(3^{x+2}.5^y=45^x\Leftrightarrow3^{x+2}.5^y=3^{2x}.5^x\Leftrightarrow\frac{3^{2x}}{3^{x+2}}=\frac{5^y}{5^x}\)\(\Leftrightarrow3^{2x-x+2}=5^{y-x}\Leftrightarrow3^{x+2}=5^{y-x}\)

\(\Rightarrow x+2=y-x=0\Rightarrow x=y=-2\)

a) \(5.2^{x+1}.2^{-2}-2^x=384\)

\(\Leftrightarrow2^x.2.\frac{5}{4}-2^x=384\)

\(\Leftrightarrow2^x.\left(\frac{5}{2}-1\right)=384\)

\(\Leftrightarrow2^x.\frac{3}{2}=384\)

\(\Leftrightarrow2^x=256\)

\(\Leftrightarrow2^x=2^8\)

\(\Leftrightarrow x=8\)

c) \(\left(x+1\right)^{x+1}=\left(x+1\right)^{x+3}\)

\(\Leftrightarrow\left(x+1\right)^{x+3}-\left(x+1\right)^{x+1}=0\)

\(\Leftrightarrow\left(x+1\right)^{x+1}\left[\left(x+1\right)^2-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^{x+1}=0\\\left(x+1\right)^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x+1=0\\\left(x+1\right)^2=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x\in\left\{0;-2\right\}\end{cases}}}\)

Vậy \(x\in\left\{0;-1;-2\right\}\)