Giải pt sau
a) $(x-3)^{2}$ – $(x+1)^{2}$ = 0
b) ($x^{2}$ – 4)(2x+3)=($x^{2}$ – 4)(x-1)
c) $(3x-7)^{2}$ – 4($x+1)^{2}$ = 0
d) (x+1)(x+2)(x+4)(x+5)=40
e) ($x+3)^{4}$ + ($x-3)^{4}$=32
f) ($4-x)^{5}$ + ($x-2)^{5}$ = 32
Giải pt sau
a) $(x-3)^{2}$ – $(x+1)^{2}$ = 0
b) ($x^{2}$ – 4)(2x+3)=($x^{2}$ – 4)(x-1)
c) $(3x-7)^{2}$ – 4($x+1)^{2}$ = 0
d) (x+1)(x+2)(x+4)(x+5)=40
e) ($x+3)^{4}$ + ($x-3)^{4}$=32
f) ($4-x)^{5}$ + ($x-2)^{5}$ = 32
a,(x-3-x+1)(x-3+x+1)=0
⇒-4(2x-2)=0
⇒2x-2=0
⇒2x=2
⇒x=1
vậy=1
b,(x²-4)(2x+3)-(x²-4)(x-1)=0
⇒(x²-4)(2x+3-x+1)=0
⇒(x-2)(x+2)(x+4)=0
⇒x=2;x=-2;x=-4
c,(3x-7)²-(2x+2)²=0
⇒(3x-7-2x-2)(3x-7+2x+2)=0
⇒(x-9)(5x-5)=0
⇒x=9;x=1