\(2x^2+7x-9=0\Leftrightarrow\left(x-1\right)\left(x+\dfrac{9}{2}\right)=0\)

\(x\in\left\{1;\dfrac{-9}{2}\right\}\)

<=> 2x² -2x +9x -9 =0

<=> 2x(x-1) + 9 (x-1) = 0

<=> (2x+9)(x-1) = 0

<=> 2x + 9 =0 hoặc x-1 = 0

<=> x= -9/2 hoặc x= 1

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

\(\Leftrightarrow x\in\left\{3;-3\right\}\)

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

\(TH1\left(x-9\right)=0\Leftrightarrow x=9\)

\(TH2\left(x+9\right)=0\Leftrightarrow x=-9\)

\(\Leftrightarrow x-2=0\)

hay x=2

\(\text{x^2-4x+4=0}\\ \left(x-2\right)^2=0\\ x-2=0\\ x=0+2\\ x=2\)

2

\(a,6\left(x-2\right)=8\left(3x+1\right)\\ \Leftrightarrow6x-12=24x+8\\ \Leftrightarrow18x+20=0\\ \Leftrightarrow x=-\dfrac{10}{9}\\ b,2x-\left(3-7x\right)=5\left(x+3\right)\\ \Leftrightarrow2x-3+7x=5x+15\\ \Leftrightarrow9x-3-5x-15=0\\ \Leftrightarrow4x-18=0\\ \Leftrightarrow x=\dfrac{9}{2}\\ c,\left(x-1\right)^2=\left(x+3\right)\left(x+2\right)\\ \Leftrightarrow x^2-2x+1=x^2+5x+6\\ \Leftrightarrow7x+5=0\\ \Leftrightarrow x=-\dfrac{5}{7}\\ d,\left(3x-9\right)\left(4x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-9=0\\4x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{4}\end{matrix}\right.\)

\(e,x^2-3x+2=0\\ \Leftrightarrow\left(x^2-x\right)-\left(2x-2\right)=0\\ \Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\\ \left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\\ f,x^2-4x+4=0\\ \Leftrightarrow x^2-2.2+2^2=0\\ \Leftrightarrow\left(x-2\right)^2=0\\ \Leftrightarrow x-2=0\\ x=2\)

a, \(6x-12=24x+8\Leftrightarrow18x=-20\Leftrightarrow x=-\dfrac{20}{18}=-\dfrac{10}{9}\)

b, \(2x-3+7x=5x+15\Leftrightarrow4x=18\Leftrightarrow x=\dfrac{9}{2}\)

c, \(x^2-2x+1=x^2+5x+6\Leftrightarrow7x=-5\Leftrightarrow x=-\dfrac{5}{7}\)

d, \(\left[{}\begin{matrix}3x-9=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{4}\end{matrix}\right.\)

e, \(x^2-3x+2=0\Leftrightarrow x^2-2x-x+2=0\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)

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

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

\(x^2\left(2x+3\right)-2x^3=3\\ \Leftrightarrow2x^3+3x^2-2x^3=3\\ \Leftrightarrow3x^2=3\\ \Leftrightarrow x^2=1\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

\(=\dfrac{5x}{2\left(x-2\right)}\cdot\dfrac{x-2}{3x}=\dfrac{5}{6}\)

\(a,\left(x-2\right)^2=x^2-2.x.2+2^2=x^2-4x+4\\ b,\left(x+1\right)^3=x^3+3.x^2.1+3.x.1^2+1^3=x^3+3x^2+3x+1\\ c,x^2-5x^2=-4x^2\)

a) \(2^{x+1}=8^2\)

\(2^{x+1}=\left(2^3\right)^2\)

\(2^{x+1}=2^6\)

\(\Rightarrow x+1=6\)

\(x=5\)

b) \(5^{x-1}=25^2\)

\(5^{x-1}=\left(5^2\right)^2\)

\(5^{x-1}=5^4\)

\(\Rightarrow x-1=4\)

\(x=5\)

a) Ta có : \(2^{x+1}=8^2\Rightarrow2^x.2=\left(2^3\right)^2\Rightarrow2^x=2^6:2\Rightarrow2^x=\) \(2^5\) \(\Rightarrow x=5\)

b) \(5^{x-1}=25^2\Rightarrow5^x:5=\left(5^2\right)^2\Rightarrow5^x:5=5^4\Rightarrow5^x=5^4.5\Rightarrow5^x=5^5\Rightarrow x=5\)

    \(2^{x+1}=8^2\)

=>\(2^{x+1}=\left(2^3\right)^2\)

    \(2^{x+1}=2^6\)

=>x+1=6 =>x=5

    \(5^{x-1}=25\)

=>\(5^{x-1}=5^2\)

=>x-1=2

=>x=3