bn vô đây Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học toán với OnlineMath

a) (3²)^2x-1 =  3^-1 => 3^2(2x-1)  = 3^-1 => 2(2x -1)  = -1 => 4x - 2 = -1 => 4x = 1 => x = 1/4. vậy,,,

b) 3.8^x - 2. (2³)^x = 64

=> 3.8^x - 2.8^x = 64

=> 8^x = 8² => x = 2

Vậy...

\(a,3.8^x-2.2^{3x}-16=0\)

\(\Leftrightarrow3.\left(2^3\right)^x-2.2^{3x}=16\)

\(\Leftrightarrow3.2^{3x}-2^{3x}=16\)

\(\Leftrightarrow2^{3x}\left(3-1\right)=16\)

\(\Leftrightarrow2^{3x+1}=2^4\)

\(\Leftrightarrow3x+1=4\)

\(\Leftrightarrow3x=3\)

\(\Leftrightarrow x=1\)

Vậy .....

5.2^3x+1+2.2^3x+2 - 16 = 0

=> 5.2^3x.2+2.2^3x.2² =16

=> 2^3x ( 10 + 8)   = 16

=> 2^3x  = 16:18 = 8/9

=> x khong co gia tri

h) \(\left(x-1\right)^2=25\)

Mà:\(5^2=\left(-5\right)^2=25\)

TH1:\(x-1=5\)

\(x=5+1\)

\(x=6\)

TH2:\(x-1=-5\)

\(x=-5+1\)

\(x=-4\)

Vậy:\(x=6\)hoặc \(x=-4\)

i)\(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)

Mà:\(\left(\frac{2}{5}\right)^2=\left(\frac{-2}{5}\right)^2=\frac{4}{25}\)

TH1:\(x+\frac{1}{2}=\frac{2}{5}\)                                       TH2:\(x+\frac{1}{2}=\frac{-2}{5}\)

\(x=\frac{2}{5}-\frac{1}{2}\)                                                        \(x=\frac{-2}{5}-\frac{1}{2}\)

\(x=\frac{-1}{10}\)                                                              \(x=\frac{-9}{10}\)

Vậy:\(x=\frac{-1}{10}\)hoặc\(x=\frac{-9}{10}\)

k)\(\left|2x-1\right|=\left|x-2\right|\)

\(2x-1=x-2\)

\(2x-x=-2+1\)

\(x=-1\)

Vậy:\(x=-1\)

\(\left\{{}\begin{matrix}x-16=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=16\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left(x-16\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-16=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=16\\x=\dfrac{1}{3}\end{matrix}\right.\)

=>x-16=0 hoặc 3x-1=0

    x     = 16         x   = ⅓