Làm theo hằng đẳng thức 3 Dạng 2: tìm x biết a) $x^{2}$ -4 = 0 b) 16$x^{2}$ -9 =0 c) 9$x^{2}$ - 25 = 0 d) (3x +1$)^{2}$ - ( 2x +3$)^{2}$ =0 e) (7 - 2x$)^{2}$ - ( 3x+1$)^{2}$ =0 Dạng 3 : Tính nhanh - tính nhẩm a) 62 . 58; 59 . 61 b) ( 53,2$)^{2}$ - 53,2 . 6,4 + (3,2$)^{2}$ c) (73,7$)^{2}$ + 73,7 . 72,6 + 36,$3^{2}$ Hằng đẳng thức 4 : lập phương 1 tổng hai số a) (x + 1$)^{3}$ b) (x + 2$)^{3}$ c) (x + 3$)^{3}$ d) (x + 4$)^{3}$ e) (2x + 5$)^{3}$ f) (3x + 1$)^{3}$ mn làm giúp mjk trong 45' nx nha. cảm ơn ạ

Đáp án+Giải thích các bước giải:

Dạng `2:`

`a)quadx^2-4=0`

`=>x^2-2^2=0`

`=>(x-2)(x+2)=0`

`=>[(x-2=0),(x+2=0):}=>[(x=2),(x=-2):}`

Vậy `x\in{2;-2}`

`b)quad16x^2-9=0`

`=>(4x)^2-3^3=0`

`=>(4x-3)(4x+3)=0`

`=>[(4x-3=0),(4x+3=0):}=>[(4x=3),(4x=-3):}=>[(x=3/4),(x=-3/4):}`

Vậy `x\in{3/4;-3/4}`

`c)quad9x^2-25=0`

`=>(3x)-5^2=0`

`=>(3x-5)(3x+5)=0`

`=>[(3x-5=0),(3x+5=0):}=>[(3x=5),(3x=-5):}=>[(x=5/3),(x=-5/3):}`

Vậy `x\in{5/3;-5/3}`

`d)quad(3x+1)^2-(2x+3)^2=0`

`=>[(3x+1)-(2x+3)][(3x+1)+(2x+3)]=0`

`=>(3x+1-2x-3)(3x+1+2x+3)=0`

`=>(x-2)(5x+4)=0`

`=>[(x-2=0),(5x+4=0):}=>[(x=2),(5x=-4):}=>[(x=2),(x=-4/5):}`

Vậy `x\in{2;-4/5}`

`e)quad(7-2x)^2-(3x+1)^2=0`

`=>[(7-2x)-(3x+1)][(7-2x)+(3x+1)]=0`

`=>(7-2x-3x-1)(7-2x+3x+1)=0`

`=>(-5x+6)(x+8)=0`

`=>[(-5x+6=0),(x+8=0):}=>[(-5x=-6),(x=-8):}=>[(x=-6/5),(x=-8):}`

Vậy `x\in{-6/5;-8}`

Dạng `3:`

`a)quad62.58=(60+2)(60-2)=60^2-2^2=3600-4=3596`

`59.61=(60-1)(60+1)=60^2-1^2=3600-1=3599`

`b)quad(5,2)^2-53,2.6,4+(3,2)^2=(53,2)^2-2.53,2.3,2+(3,2)^2=(53,2-3,2)^2=50^2=2500`

`c)quad(73,7)^2+73,7.72,6+(36,3)^2=(73,7)^2+2.73,7.36,3+(36,3)^2=(73,7+36,3)^2=110^2=12100`

Dạng `4:`

`a)quad(x+1)^3=x^3+3.x^2 .1+3.x.1^2+1^3=x^3+3x^2+3x+1`

`b)quad(x+2)^3=x^3+3.x^2 .2+3.x.2^2+2^3=x^3+6x^2+12x+8`

`c)quad(x+3)^3=x^3+3.x^2 .3+3.x.3^2+3^3=x^3+9x^2+27x+27`

`d)quad(x+4)^3=x^3+3.x^2 .4+3.x.4^2+4^3=x^3+12x^2+48x+64`

`e)quad(2x+5)^3=(2x)^3+3.(2x)^2 .5+3.2x.5^2+5^3=8x^3+60x^2+150x+125`

`f)quad(3x+1)^3=(3x)^3+3.(3x)^2 .1+3.3x.1^2+1^3=27x^3+27x^2+9x+1`

`#ld`