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\(x+2x+3x+...+9x=459-3^2\)
\(\Rightarrow9x+\left(1+2+3+...+9\right)=450\)
\(\Rightarrow9x+\frac{\left[\left(9+1\right).9\right]}{2}=450\)
\(\Rightarrow9x+45=450\)
\(\Rightarrow9x=450-45\)
\(\Rightarrow x=\frac{450-45}{9}=\frac{405}{9}=45\)
\(a.2^x+2^{x+3}=144\)
\(2^x+2^x.2^3=144\)
\(2^x+2^x.8=144\)
\(2^x.\left(1+8\right)=144\)
\(2^x.9=144\)
\(2^x=144:9\)
\(2^x=16\)
\(2^x=2^4\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
\(b.\left(4.x-1\right)^2=25.9\)
\(\left(4.x-1\right)^2=225^2\)
\(4.x-1=225\)
\(4.x=225+1\)
\(4.x=226\)
\(x=226:4\)
\(x=\frac{226}{4}=\frac{113}{2}\)
Vậy \(x=\frac{113}{2}\)
2x + 2x+3 = 144
<=> 2x + 2x.23 = 144
<=> 2x( 1 + 23 ) = 144
<=> 2x.9 = 144
<=> 2x = 16
<=> 2x = 24
<=> x = 4
( 4x - 1 )2 = 25 . 9
<=> ( 4x - 1 )2 = 52 . 32
<=> ( 4x - 1 )2 = ( 5 . 3 )2
<=> ( 4x - 1 )2 = 152 = (-15)2
<=> \(\hept{\begin{cases}4x-1=15\\4x-1=-15\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\x=-\frac{7}{2}\end{cases}}\)
Ta có: 2x+3 + 2x = 144
\(\Rightarrow\) 2x (23 + 1) = 144
\(\Rightarrow\) 2x.9 = 144
\(\Rightarrow\) 2x = 16
\(\Rightarrow\) x = 4
Vậy x = 4.
2x + 3 + 2x = 144
2x . 23 + 2x . 1 = 144
2x . (8 + 1) = 144
2x . 9 = 144
2x = 16
<=> x = 4
A, (2x+1)3=343
=> (2x+1)3=73
=> 2x + 1 = 7
=> 2x = 6
=> x = 3
B, 2x+2x+3=144
=> 2x+2x . 23 =144
=> 2x ( 1 + 23 ) =144
=> 2x ( 1 + 8 ) =144
=> 2x . 9 =144
=> 2 x = 16
=> 2 x = 2 4
=> x = 4
C, 3x+3x+2=2430
=> 3x+3x . 32 =2430
=> 3x . ( 1 + 32 ) =2430
=> 3x . ( 1 + 9 ) =2430
=> 3x . 10 =2430
=> 3x = 243
=> 3x = 3 5
=> x = 5
2^x+2^x=144-2
2^x.2=142
2^x=142:2
2^x=71
x không tồn tại