tìm x biết a, x^2-x=20 b,4x^2+19x+9x=0 c, (4x-3)*(4x+3)-5 x(3x-1)=-7

Đáp án:

Giải thích các bước giải:

a/ $x^2-x=20$
=> $x^2-x-20=0$
=> $x^2+4x-5x-20=0$
=> $(x+4)(x-5)=0$
=> \(\left[ \begin{array}{l}x+4=0\\x-5=0\end{array} \right.\)
=> \(\left[ \begin{array}{l}x=-4\\x=5\end{array} \right.\)

b/ $4x^2+19x+9x=0$
=> $4x^2+28x=0$
=> $4x(x+7)=0$
=> \(\left[ \begin{array}{l}x=0\\x+7=0\end{array} \right.\)
=> \(\left[ \begin{array}{l}x=0\\x=-7\end{array} \right.\)

c/ $(4x-3)(4x+3)-5x(3x-1)=-7$
=> $16x^2-9-15x^2+5x+7=0$
=> $x^2+5x-2=0$
=> $4x^2+20x-8=0$
=> $4x^2+20x+25-33=0$
=> $(2x+5)^2-33=0$
=> $(2x+5+\sqrt{33})(2x+5-\sqrt{33})=0$
=> \(\left[ \begin{array}{l}2x+5+\sqrt{33}=0\\2x+5-\sqrt{33}=0\end{array} \right.\)

=> \(\left[ \begin{array}{l}x=\frac{-5-\sqrt{33}}{2}\\x=\frac{-5+\sqrt{33}}{2}\end{array} \right.\)
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