thu gọn và khai triển theo hằng đẳng thức
a) x^2-2x=24
b) (2x+1)^2-4(x-1)^2=9
c) (x+3)^2-(x-4)(x+8)=1
d) (2x-1)^2+(x+3)^2-5(x+7)(x-7)=0
e) 3(x+2)^2+(2x-1)^2-7(x+3)(x-3)=36
thu gọn và khai triển theo hằng đẳng thức a) x^2-2x=24 b) (2x+1)^2-4(x-1)^2=9 c) (x+3)^2-(x-4)(x+8)=1 d) (2x-1)^2+(x+3)^2-5(x+7)(x-7)=0 e) 3(x+2)^2
By Alaia
`a)x²-2x=24`
`⇔x²-2x-24=0`
`⇔x²-6x+4x-24=0`
`⇔(x²-6x)+(4x-24)=0`
`⇔x(x-6)+4(x-6)=0`
`⇔(x-6)(x+4)=0`
`⇔`\(\left[ \begin{array}{l}x-6=0\\x+4=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=6\\x=-4\end{array} \right.\)
Vậy `S={6;-4}`
`b)(2x+1)²-4(x-1)²=9`
`⇔4x²+4x+1-4(x²-2x+1)=9`
`⇔4x²+4x+1-4x²+8x-4-9=0`
`⇔12x-12=0`
`⇔12x=12`
`⇔x=1`
Vậy `S={1}`
`c)(x+3)²-(x-4)(x+8)=1`
`⇔x²+6x+9-(x²+8x-4x-32)=1`
`⇔x²+6x+9-x²-8x+4x+32=1`
`⇔2x+41=1`
`⇔2x=1-41`
`⇔2x=40`
`⇔x=20`
Vậy `S={20}`
`d)(2x-1)²+(x+3)²-5(x+7)(x-7)=0`
`⇔4x²-4x+1+x²+6x+9-5(x²-49)=0`
`⇔4x²-4x+1+x²+6x+9-5x²+245=0`
`⇔2x+255=0`
`⇔2x=-255`
`⇔x=-255/2`
Vậy `S={-255/2}`
`e)3(x+2)²+(2x-1)²-7(x+3)(x-3)=36`
`⇔3(x²+4x+4)+4x²-4x+1-7(x²-9)=36`
`⇔3x²+12x+12+4x²-4x+1-7x²+63=36`
`⇔8x+76=36`
`⇔8x=36-76`
`⇔8x=-40`
`⇔x=-5`
Vậy `S={-5}`