\(4^{x+2}.5.4=1280\Leftrightarrow4^{x+2}=\frac{1280}{5.4}\Leftrightarrow4^{x+2}=64\Leftrightarrow4^{x+2}=4^3\)

\(\Rightarrow x+2=3\Leftrightarrow x=3-2\Leftrightarrow x=1\)

4^x+2.5.4=1280

4^x+2.20=1280

4^x+2     =1280:20

4^x.4² =64

4^x.16=64

4^x  =64:16

4^x=4

=>x=1

giúp mình câu này luôn ạ 
4^x+2.5.4^x=1280

\(4^{x+2}\cdot5\cdot4^x=1280\)

\(\Leftrightarrow4^x\cdot16\cdot5\cdot4^x=1280\)

\(\Leftrightarrow4^{2x}\cdot80=1280\)

\(\Leftrightarrow4^{2x}=16\)

\(\Leftrightarrow4^{2x}=4^2\)

\(\Leftrightarrow2x=2\)

\(\Leftrightarrow x=1\)

vậy........

\(4^x+10.4^x=1280\)

<=>\(11.4^x=1280\)

<=>\(4^x=\frac{1280}{11}=>saiđề\)

\(2^x+2^{x+1}=96\)

\(\Rightarrow2^x\left(1+2\right)=96\)

\(\Rightarrow2^x.3=96\)

\(2^x=96:3\)

\(2^x=32\)

\(\Rightarrow2^x=2^5\)

\(x=5\)

a)2^x+2^x+1=96

2^x+2^x.2=96

2^x.(1+2)=96

2^x=32=2⁵

      Vậy x=5

b)4^x+2 × 5 × 4^x = 1280

4^x.4² × 4^x = 256

4^x.(4²+1)=256

4^x.17=256

4^x=\(\frac{256}{17}\)

Theo đầu bài ta có:
a) \(4^{x+2}+4^{x+3}=1280\)
\(\Rightarrow4^x\cdot4^2+4^x\cdot4^3=1280\)
\(\Rightarrow4^x\cdot80=1280\)
\(\Rightarrow4^x=16\)
\(\Rightarrow x=2\)

b) \(6^{x+2}+5\cdot6^{x+2}=216\)
\(\Rightarrow6\cdot6^{x+2}=216\)
\(\Rightarrow6^{x+3}=216\)
\(\Rightarrow x+3=3\)
\(\Rightarrow x=0\)

Làm hơi tắt tí, nhở!/ :>

2.(2x+3+1)=1280

2.(2x+4)=1280

4x+8=1280

4x   =1280-8

4x   =1272

x     =318