135 - 3 ( x + 1 ) = 30
3 ( x + 1 ) = 105
x + 1 = 35
x = 34
\(1005^2\cdot1005^x=1005^7\)
\(1005^{2+x}=1005^7\)
=> x + 2 = 7
=> x = 5
Vậy,........
\(135-3.\left(x+1\right)=30\)
\(\Rightarrow3.\left(x+1\right)=105\)
\(\Rightarrow x+1=35\)
\(\Rightarrow x=34\)
\(1005^2.1005^x=1005^7\)
\(\Rightarrow1005^x=1005^7\div1005^2\)
\(\Rightarrow1005^x=1005^{7-2}\)
\(\Rightarrow1005^x=1005^2\)
\(\Rightarrow x=2\)
a)135-3 .(x+1)=30 b)10052 . 1005x = 10057
3.(x+1)=135-30 2 . x = 7
x+1=105:3 x = 7:2
x=35-1 x=\(\frac{7}{2}\)
x=24
Vội quá
\(1005^x=1005^{7-2}\)
\(\Rightarrow1005^x=1005^5\)
\(\Rightarrow x=5\)
Vậy x = 5
1005\(^x\)=1005\(^7\)−2
⇒1005\(^x\)=1005\(^5\)
⇒x=5