Đề bị lỗi. Bạn coi lại đề.

1) \(2^x=4^3 \Leftrightarrow2^x=2^6\Leftrightarrow x=6\)

2) \(2^x=4^6\Leftrightarrow2^x=2^{12}\Leftrightarrow x=12\)

3) \(3^x=9^{10}\Leftrightarrow3^x=3^{20}\Leftrightarrow x=20\)

1) $2^x=4^3\iff 2^x=2^6\iff x=6$

2) $2^x=4^6\iff 2^x=2^{12}\iff x=12$

3) $3^x=9^{10}\iff 3^x=3^{20}\iff x=20$

a/

\(x^3-4x^2-\left(x-4\right)=0\)

\(\Leftrightarrow x^2\left(x-4\right)-\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=-1\end{matrix}\right.\)

b/

\(x^5-9x=0\)

\(\Leftrightarrow x\left(x^4-9\right)=x\left(x^2-3\right)\left(x^2+3\right)=0\)

\(\Leftrightarrow x\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=0\)

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

c/

\(\left(x^3-x^2\right)^2-4x^2+8x-4=0\)

\(\Leftrightarrow x^4\left(x-1\right)^2-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x^4-4\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x^2-2\right)\left(x^2+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\pm\sqrt{2}\end{matrix}\right.\)

Sos

\(3^{x+2}+3^x=270\\\Rightarrow 3^x\cdot\cdot3^2+3^x=270\\\Rightarrow 3^x\cdot(3^2+1)=270\\\Rightarrow 3^x\cdot(9+1)=270\\\Rightarrow 3^x\cdot10=270\\\Rightarrow 3^x=270:10\\\Rightarrow 3^x=27\\\Rightarrow 3^x=3^3\\\Rightarrow x=3\)

( 3x - 2⁴ ) . 7³ = 2 . 7⁴

3x - 16 = 2 . 7⁴ : 7³ = 14

3x = 14 + 16 = 30

x = 10

\(\left(3\times x-2^4\right)\times7^3=2\times7^4\)

 \(\left(3\times x-2^4\right)\div2=7^4\div7^3\) 

  \(\left(3\times x-16\right)\div2=7\)

           \(3\times x-16=7\times2\) 

           \(3\times x-16=14\) 

                     \(3\times x=14+16\)                      

                      \(3\times x=30\)

                           \(x=30\div3\) 

                           \(x=10\)

  Bài 1:

2^\(x\) = 4

2\(^x\) = 2²

 \(x=2\)

Vậy \(x=2\)

Bài 2:

2\(^x\) = 8

2\(^x\) = 2³

\(x=3\)

Vậy \(x=3\)

a. x mũ 2 - 2x + 1 = 25 

= x^2 + 2.x.1 + 1^2

= ( x + 1 ) ^2

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Sos

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