a, $(2x-7)(x + 3)$ = $x^{2}$ $-$ $9$

$\Longleftrightarrow$ $x^{2}$ $-$ $9$ $-$ $(2x-7)(x + 3)$ = 0

$\Longleftrightarrow$ $x^{2}$ $-$ $9$ $-$ $(2x-7)(x + 3)$ $= 0$

$\Longleftrightarrow$ $(x - 3)(x + 3)$ $-$ $(2x-7)(x + 3)$ $= 0$

$\Longleftrightarrow$ $(x + 3)(x - 3 - (2x - 7))$ = 0$

$\Longleftrightarrow$ $(x + 3)(x - 3 - 2x + 7)$ = 0$

$\Longleftrightarrow$ $(x + 3)(- x + 4)$ = 0$

$\Longleftrightarrow$ $\left[\begin{matrix} x+3=0\\ -x+4=0\end{matrix}\right.$

$\Longleftrightarrow$ $\left[\begin{matrix} x=-3\\  x=4\end{matrix}\right.$

Vậy ...........

b, $(3x+4)(x - 4)$ = $(x - 4)^{2}$

$\Longleftrightarrow$ $(3x + 4)(x - 4)0$ - $(x - 4)^{2}$  = 0$

$\Longleftrightarrow$ $(x - 4)(3x + 4 - (x - 4)$ = 0$

$\Longleftrightarrow$ $(x - 4)(3x + 4 - x + 4)$ = 0$

$\Longleftrightarrow$ $(x - 4)(2x + 8)$ = 0$

$\Longleftrightarrow$ $\left[\begin{matrix} x-4=0\\ 2x+8=0\end{matrix}\right.$

$\Longleftrightarrow$ $\left[\begin{matrix} x=4\\  x=-4\end{matrix}\right.$

Vậy .............

c, $(3x - 1)^{2}$ = $(x + 3)^{2}$

$\Longleftrightarrow$ $(3x - 1)^{2}$ - $(x + 3)^{2}$ = 0

$\Longleftrightarrow$ $((3x - 1) - (x + 3)) . ((3x - 1) + (x + 3))$ = 0

$\Longleftrightarrow$ $(3x - 1 - x - 3)) . (3x - 1 + x + 3))$ = 0

$\Longleftrightarrow$ $(2x - 4)) . (4x + 2))$ = 0

$\Longleftrightarrow$ $\left[\begin{matrix} 2x-4=0\\ 4x+2=0\end{matrix}\right.$

$\Longleftrightarrow$ $\left[\begin{matrix} x=2\\ x=\frac{-1}{2}\end{matrix}\right.$

Vậy ..............

d, $(5x^{2} + 3x - 2)^2$ = $(4x^{2} - 3x - 2)^2$

$\Longleftrightarrow$ $(5x^{2} + 3x - 2)^2$ - $(4x^{2} - 3x - 2)^2$   = 0$

$\Longleftrightarrow$ $(5x^{2} + 3x - 2)^2$ - $(4x^{2} - 3x - 2)^2$   = 0$

$\Longleftrightarrow$ $(5x^{2} + 3x - 2 - (4x^{2} - 3x - 2))(5x^{2} + 3x - 2 + (4x^{2} - 3x - 2))$ = 0

$\Longleftrightarrow$ $(5x^{2} + 3x - 2 - 4x^{2} + 3x + 2)(5x^{2} + 3x - 2 + 4x^{2} - 3x - 2)$ = 0

$\Longleftrightarrow$ $(x^{2} + 6x)(9x^{2} - 4)$ = 0

$\Longleftrightarrow$ $\left[\begin{matrix} x^{2} + 6x=0\\ 9x^{2} - 4=0\end{matrix}\right.$

$\Longleftrightarrow$ $\left[\begin{matrix} x(x + 6)=0\\ (3x - 2)(3x + 2)=0\end{matrix}\right.$

$\Longleftrightarrow$ $\left[\begin{matrix} x=0\\ x + 6=0\\3x - 2 = 0\\ 3x + 2 = 0\end{matrix}\right.$

$\Longleftrightarrow$ $\left[\begin{matrix} x=0\\ x =-6\\x = \frac{2}{3} \\ x = \frac{-2}{3} \end{matrix}\right.$

Vậy .......

e, $(4x+3)(x^2 - 9)$ = $(x + 3)(16x^2 - 9)$

$\Longleftrightarrow$ $(4x+3)(x^2 - 9)$ - $(x + 3)(16x^2 - 9)$ = 0

$\Longleftrightarrow$ $(4x+3)(x - 3)(x + 3)$ - $(x + 3)(4x - 3)(4x+3)$ = 0

$\Longleftrightarrow$ $(4x+3)(x + 3)(x - 3 - (4x - 3))$ = 0

$\Longleftrightarrow$ $(4x+3)(x + 3)(x - 3 - 4x + 3))$ = 0

$\Longleftrightarrow$ $(4x+3)(x + 3)(-3x)$ = 0

$\Longleftrightarrow$ $\left[\begin{matrix} 4x+3=0\\ x+3=0\\-3x=0\end{matrix}\right.$

$\Longleftrightarrow$ $\left[\begin{matrix} x=\frac{-3}{4}\\ x=-3\\x=0\end{matrix}\right.$

Vậy ...........

Câu 2: Giải phương trình:

$y^{2}(y -1)(y + 1)$ = $72$

Ta có: $y^{2}(y -1)(y + 1)$ = $72$

$\Longleftrightarrow$ $y^2(y^2 - 1)$ = 72

Đặt $k$ = $y^{2}$ ($k > 0$) ta được:

$\Longleftrightarrow$ $k^2 - k = 72$

$\Longleftrightarrow$ $4k^2 - 4k = 288$

$\Longleftrightarrow$ $4k^2 - 4k + 1 = 289$

$\Longleftrightarrow$ $(2k - 1)^2 = 17^2$

$\Longleftrightarrow$ $(2k - 1)^2 - 17^2$ = 0

$\Longleftrightarrow$ $(2k - 1 - 17)(2k - 1 + 17)$ = 0

$\Longleftrightarrow$ $(2k - 18)(2k + 16)$= 0

$\Longleftrightarrow$ $\left[\begin{matrix} 2k-18=0\\ 2k+16=0\end{matrix}\right.$

$\Longleftrightarrow$ $\left[\begin{matrix} k=9 (nhận)\\ k=-8 (loại)\end{matrix}\right.$

Với $k = 9$ ta có: 

$\Longleftrightarrow$ $y^{2}$ = 9

$\Longleftrightarrow$ $\left[\begin{matrix} y=3\\ y=-3\end{matrix}\right.$

Vậy .................

Chúc bn học tốt !

$a)(2x-7)(x+3)=x^2-9$

$⇔(2x-7)(x+3)=(x-3)(x+3)$

$⇔2x-7=x-3$

$⇔2x-x=-3+7$

$⇔x=4$

$Vậy....$

$b)(3x+4)(x-4)=(x-4)^2$

$⇔ 3x+4=x-4$

$⇔ 3x-x=-4-4$

$⇔ 2x = -8$

$⇔ x = -4$

$Vậy.....$

$c) (3x-1)^2 = (x+3)^2$

$⇔ 3x-1=x+3$

$⇔ 3x-x=3+1$

$⇔ 2x=4$

$⇔ x=2$

$Vậy.......$

$d) (5x^2 + 3x -2)^2 = (4x^2 - 3x-2)^2$

$⇔ 5x^2 + 3x - 2 = 4x^2-3x-2$

$⇔ 5x^2 - 4x^2 + 3x + 3x = 0$

$⇔ x^2 + 6x = 0$

$⇔ x(x+6) = 0$

$⇔$\(\left[ \begin{array}{l}x=0\\x+6=0\end{array} \right.\)

$⇔$\(\left[ \begin{array}{l}x=0\\x=-6\end{array} \right.\)
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