Đáp án:a)\(\left[ \begin{array}{l}x=\frac{7}{3}\\x=\frac{-7}{3}\end{array} \right.\)
b)x=0
c)\(\left[ \begin{array}{l}x=-2\\x=2\end{array} \right.\)
d)$x=\frac{-13}{2}$
e)x=2
Giải thích các bước giải:
a)$9x^{2}-49=0$
⇔$(3x)^{2}-7^{2}=0$
⇔$(3x-7)(3x+7)=0$⇔\(\left[ \begin{array}{l}3x-7=0\\3x+7=0\end{array} \right.\)⇔ \(\left[ \begin{array}{l}x=\frac{7}{3}\\x=\frac{-7}{3}\end{array} \right.\)
b) $(x + 3)(x^{2} – 3x + 9) –x(x – 1)(x + 1) – 27 = 0$
⇔$x^{3}+27-x(x^{2}-1)-27=0$
⇔$x^{3}+27-x^{3}+x-27=0$
⇔$x=0$
c)$(x – 1)(x + 2) – x – 2 = 0$
⇔$(x-1)(x+2)-(x+2)=0$
⇔$(x+2)(x-1-1)=0$
⇔$(x+2)(x-2)=0$
⇔\(\left[ \begin{array}{l}x+2=0\\x-2=0\end{array} \right.\)⇔ \(\left[ \begin{array}{l}x=-2\\x=2\end{array} \right.\)
d) $x(3x + 2) + (x + 1)^{2} – (2x – 5)(2x + 5) = 0$
⇔$3x^{2}+2x+x^{2}+2x+1-(4x^{2}-25)=0$
⇔$3x^{2}+2x+x^{2}+2x+1-4x^{2}+25=0$
⇔$4x+26=0⇔x=\frac{-26}{4}=\frac{-13}{2}$
e) $(4x + 1)(x - 2) - (2x -3)(2x + 1) = 7$
⇔$4x^{2}-8x+x-2-(4x^{2}+2x-6x-3)-7=0$
⇔$4x^{2}-7x-2-4x^{2}+4x+3-7=0$
⇔$-3x-6=0⇔ x=2$
