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

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

<=>\(4^x=\frac{1280}{11}=>saiđề\)

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

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

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........

\(4^{x+1}+4^{x+2}=1280\\ \Leftrightarrow4^{x+1}\left(1+4\right)=1280\\ \Leftrightarrow4^{x+1}=256\\ Mà256=4^4\\ \Rightarrow x+1=4\\ \Leftrightarrow x=3\\ Vậy...\)

\(4^{x+1}+4^{x+2}=1280\)

⇒ \(4^{x+1}.1+4^{x+1}.4^1=1280\)

⇒ \(4^{x+1}.\left(1+4^1\right)=1280\)

⇒ \(4^{x+1}.5=1280\)

⇒ \(4^{x+1}=1280:5\)

⇒ \(4^{x+1}=256\)

⇒ \(4^{x+1}=4^4\)

⇒ \(x+1=4\)

⇒ \(x=4-1\)

⇒ \(x=3\)

Vậy \(x=3.\)

Chúc bạn học tốt!

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!/ :>

137 nha

297+ x + 846 =1280

297 +x = 1280 - 846

297 +x = 434

x= 434 - 297

x=137

297+x+846=1280

 (297+846)+x=1280

1143+x=1280

 x=       1280-1143

x=137

297 + x + 846 = 1280

297 + x          = 1280 - 846

297  + x         =434

x                   =434 - 297

x                   =137

297 + x + 846 = 1280

x + (297 + 846)=1280

x +  1143        = 1280

x                    = 1280 -1143

x                    = 147

297+x+846=1280

=> 297+x=1280-846=434

=> x= 434-297=137

Vậy x=137