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Ẩn danh

Tìm x, biết :

a, x\(^2\) - 4 = 0

b, x\(^2\) + 6x + 9 = 25

c, x\(^2\) - 4x + 4 = 9

d, x\(^3\) + x\(^2\) + x + 1 = 0

e, ( x - 2 ).( 3 + 4x ) + ( x - 2 )\(^2\) = 0

f, 4x\(^2\) - 9 + ( 2x - 3 ).( 4 - x ) = 0

Nguyễn Hữu Phước
16 tháng 11 2024 lúc 9:28

a) \(x^2-4=0\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

b) \(x^2+6x+9=25\Leftrightarrow\left(x+3\right)^2=25\Leftrightarrow\left|x+3\right|=5\Leftrightarrow\left[{}\begin{matrix}x+3=5\\x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)

c) \(x^2-4x+4=9\Leftrightarrow\left(x-2\right)^2=9\Leftrightarrow\left|x-2\right|=3\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

d) \(x^3+x^2+x+1=0\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)

Vì \(x^2+1>0\forall x\in R\Rightarrow x+1=0\Leftrightarrow x=-1\)

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

\(\Leftrightarrow\left(x-2\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\5x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{5}\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left(2x-3\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-7\end{matrix}\right.\)