Giải phương trình :
a,(x+2)(x+3) = 0
b,(x-1)(x+5)(-3x+8) =0
c,(2x -7)^2-6(2x-7)(x-3)=0
d,(x+3)(x-5)+(x+3)(3x-4)=0
e,(x+6)(3x-1)+x+6= 0
f,(x-2)(x+1)=x^2-4
Giải phương trình :
a,(x+2)(x+3) = 0
b,(x-1)(x+5)(-3x+8) =0
c,(2x -7)^2-6(2x-7)(x-3)=0
d,(x+3)(x-5)+(x+3)(3x-4)=0
e,(x+6)(3x-1)+x+6= 0
f,(x-2)(x+1)=x^2-4
`a)` `(x+2)(x+3)=0`
`<=>` \(\left[ \begin{array}{l}x+2=0\\x+3=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-2\\x=-3\end{array} \right.\)
Vậy phương trình có nghiệm `S={-3;-2}`
`b)` `(x-1)(x+5)(-3x+8)=0`
`<=>` \(\left[ \begin{array}{l}x-1=0\\x+5=0\\-3x+8=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=1\\x=-5\\x=\dfrac{8}{3}\end{array} \right.\)
Vậy phương trình có nghiệm `S={1;-5;8/3}`
`c)` `(2x-7)^2-6(2x-7)(x-3)=0`
`<=>(2x-7)[2x-7-6(x-3)]=0`
`<=>(2x-7)(2x-7-6x+18)=0`
`<=>(2x-7)(11-4x)=0`
`<=>` \(\left[ \begin{array}{l}2x-7=0\\11-4x=0\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x=\dfrac{7}{2}\\x=\dfrac{11}{4}\end{array} \right.\)
Vậy phương trình có nghiệm `S={7/2;11/4}`
`d)` `(x+3)(x-5)+(x+3)(3x-4)=0`
`<=>(x+3)(x-5+3x-4)=0`
`<=>(x+3)(4x-9)=0`
`<=>` \(\left[ \begin{array}{l}x+3=0\\4x-9=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-3\\x=\dfrac{9}{4}\end{array} \right.\)
Vậy phương trình có nghiệm `S={-3;9/4}`
`e)` `(x+6)(3x-1)+x+6=0`
`<=>(x+6)(3x-1)+(x+6)=0`
`<=>(x+6)(3x-1+1)=0`
`<=>3x(x+6)=0`
`<=>` \(\left[ \begin{array}{l}3x=0\\x+6=0\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x=0\\x=-6\end{array} \right.\)
Vậy phương trình có nghiệm `S={-6;0}`
`f)` `(x-2)(x+1)=x^2-4`
`<=>x^2+x-2x-2=x^2-4`
`<=>x^2-x-2-x^2+4=0`
`<=>-x+2=0`
`<=>-x=-2`
`<=>x=2`
Vậy phương trình có nghiệm `S={2}`
a.
`(x+2)(x+3)=0`
`⇔` \(\left[ \begin{array}{l}x+2=0\\x+3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-2\\x=-3\end{array} \right.\)
Vậy phương trình có TN `S = {-2;-3}`
b.
`(x-1)(x+5)(-3x+8)=0`
`⇔x-1=0` hoặc `x+5=0` hoặc `-3x+8=0`
+) `x-1=0⇔x=1`
+) `x+5=0⇔x=-5`
+) `-3x+8=0⇔-3x=-8⇔x=8/3`
Vậy phương trình có TN `S={1;-5;8/3}`
c.
`(2x-7)^2-6(2x-7)(x-3)=0`
`⇔(2x-7)(2x-7)-6(2x-7)(x-3)=0`
`⇔(2x-7)(2x-7)-(2x-7)(6x-18)=0`
`⇔(2x-7)(2x-7-6x+18)=0`
`⇔(2x-7)(-4x+11)=0`
`⇔` \(\left[ \begin{array}{l}2x-7=0\\-4x+11=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}2x=7\\-4x=-11\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{7}{2}\\x=\dfrac{11}{4}\end{array} \right.\)
Vậy phương trình có TN `S={7/2;{11}/4}`
d.
`(x+3)(x-5)+(x+3)(3x-4)=0`
`⇔(x+3)(x-5+3x-4)=0`
`⇔(x+3)(4x-9)=0`
`⇔` \(\left[ \begin{array}{l}x+3=0\\4x-9=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-3\\4x=9\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-3\\x=\dfrac{9}{4}\end{array} \right.\)
Vậy phương trình có TN `S={-3;9/4}`
e.
`(x+6)(3x-1)+x+6=0`
`⇔(x+6)(3x-1)+(x+6)=0`
`⇔(x+6)(3x-1+1)=0`
`⇔3x(x+6)=0`
`⇔` \(\left[ \begin{array}{l}3x=0\\x+6=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\x=-6\end{array} \right.\)
Vậy phương trình có TN `S={0;-6}`
f.
`(x-2)(x+1)=x^2-4`
`⇔(x-2)(x+1)=(x-2)(x+2)`
`⇔(x-2)(x+1)-(x-2)(x+2)=0`
`⇔(x-2)(x+1-x-2)=0`
`⇔(x-2).(-1)=0`
`⇔x-2=0`
`⇔x=2`
Vậy phương trình có nghiệm `x=2`