Ẩn danh

tìm x

a, \(x^2+x+\dfrac{1}{4}=\dfrac{9}{4}\)

b, \(25x^2-16\left(x+2\right)^2=0\)

c, \(\left(2x+1\right)^2=\left(x-1\right)^2\)

d, \(9x^2-6x=-1\)

e, \(4x^2-9=0\)

f, \(x^3-9x^2+27x-35=0\)

g, \(x^2-6x-7=0\)

NT

a: \(x^2+x+\dfrac{1}{4}=\dfrac{9}{4}\)

=>\(x^2+x+\dfrac{1}{4}-\dfrac{9}{4}=0\)

=>\(x^2+x-2=0\)

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

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

b: \(25x^2-16\left(x+2\right)^2=0\)

=>\(\left(5x\right)^2-\left(4x+8\right)^2=0\)

=>(5x-4x-8)(5x+4x+8)=0

=>(x-8)(9x+8)=0

=>\(\left[{}\begin{matrix}x-8=0\\9x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{8}{9}\end{matrix}\right.\)

c: \(\left(2x+1\right)^2=\left(x-1\right)^2\)

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

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

=>\(3x\cdot\left(x+2\right)=0\)

=>x(x+2)=0

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

d: \(9x^2-6x=-1\)

=>\(9x^2-6x+1=0\)

=>\(\left(3x-1\right)^2=0\)

=>3x-1=0

=>3x=1

=>\(x=\dfrac{1}{3}\)

e: \(4x^2-9=0\)

=>\(4x^2=9\)

=>\(x^2=\dfrac{9}{4}\)

=>\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

f: \(x^3-9x^2+27x-35=0\)

=>\(x^3-9x^2+27x-27-8=0\)

=>\(\left(x^3-9x^2+27x-27\right)=8\)

=>\(\left(x-2\right)^3=2^3\)

=>x-2=2

=>x=2+2=4

g: \(x^2-6x-7=0\)

=>\(x^2-7x+x-7=0\)

=>x(x-7)+(x-7)=0

=>(x-7)(x+1)=0

=>\(\left[{}\begin{matrix}x-7=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)

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