\(a) x^2-6x+9=4\)
\((x-3)^2-4=0\)
\((x-3-2)(x-3+2)=0\)
th1 x-5=0
x=5
th2 x-1=0
x=1
ptrình có tập nghiệm S={5;1}
b)\(x^2(x-3)-(4x-12)=0\)
\(x^2(x-3)-4(x-3)=0\)
\((x^2-4)(x-3)=0\)
\((x-2)(x+2)(x-3)=0\)
th1 x-2=0
x=2
th2 x+2=0
x=-2
th3 x-3=0
x=3
ptrình có tập nghiệm S={2;-2;3}
c)\((2x-3)^2-4(x+2)^2=12\)
\(4x^2-12x+9-4(x^2+4x+4)-12=0\)
\(4x^2-12x+9-4x^2-16x-16-12=0\)
\(-28x-19=0\)
x= -19/28
ptrình có tập nghiệm S={-19/28}
a/ \(x^2-6x+9=4\)
\(\Leftrightarrow\left(x-3\right)^2=4\)
Vậy: \(x=5;x=1\)
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b/ \(x^2\left(x-3\right)-\left(4x-12\right)=0\)
\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\x+2=0\Leftrightarrow x=-2\\x-2=0\Leftrightarrow x=2\end{matrix}\right.\)
Vậy: \(x=3;x=\pm2\)
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c/ \(\left(2x+3\right)^2-4\left(x+2\right)^2=12\)
\(\Leftrightarrow4x^2+12x+9-4x^2-16x-16=12\)
\(\Leftrightarrow-4x-7=12\)
\(\Leftrightarrow-4x=19\Leftrightarrow x=-\dfrac{19}{4}\)
Vậy: \(x=-\dfrac{19}{4}\)
a: Ta có: \(x^2-6x+9=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
b: Ta có: \(x^2\left(x-3\right)-\left(4x-12\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=2\end{matrix}\right.\)
c: Ta có: \(\left(2x+3\right)^2-4\left(x+2\right)^2=12\)
\(\Leftrightarrow4x^2+12x+9-4x^2-16x-16-12=0\)
\(\Leftrightarrow-4x=19\)
hay \(x=-\dfrac{19}{4}\)
