Bài 1 :
\(a,\)\(\left(x-4\right)^2-36=0\)\(\Rightarrow\left(x-4-6\right)\left(x-4+6\right)=0\)
\(\Rightarrow\left(x-10\right)\left(x-2\right)=0\)\(\Rightarrow x\in\left\{10;2\right\}\)
\(b,\)\(\left(x+8\right)^2=121\)\(\Rightarrow\left(x+8\right)^2-11^2=0\)
\(\Rightarrow\left(x+8+11\right)\left(x+8-11\right)=0\)\(\Rightarrow\left(x+19\right)\left(x-3\right)=0\)\(\Rightarrow x\in\left\{-19;3\right\}\)
\(c,x^2+8x+16=0\)\(\Rightarrow\left(x+4\right)^2=0\)
\(\Rightarrow x+4=0\)\(\Leftrightarrow x=-4\)
\(d,4x^2-12x=-9\)\(\Rightarrow4x^2-12x+9=0\)
\(\Rightarrow\left(2x-3\right)^2=0\)\(\Rightarrow2x-3=0\)\(\Rightarrow x=\frac{3}{2}\)
Bài 1 a) (x-4)^2 -36=0
=> (x-4)^2 = 36
=> x-4 = 6
=> x= 10
b) (x+8)^2 = 121
=> x+8 = 11
=> x=3
c) x^2 + 8x +16=0
=> (x+4)^2 =0
=> x+ 4 =0 => x= -4
d) 4x^2 - 12x= -9
=> 4x^2 -12x+9=0
=> ( 2x-3)^2=0
=> 2x-3 =0
=> x= 3/2
Bài 1)
a) (x+4)2 - 36 =0
(x+4)^2 - 6^2 =0
=> (x+2)(x-10) =0 (sử dụng hằng đẳng thức)
=> x =-2 ; x= 10(tự xét)
b) (x+8)^2 =121
(x+8)^2 - 121 =0
(x+8)^2 - 11^2 = 0
=> (x+19)(x-3) = 0
=> x= -19 ; x =3
c) x^2 + 8x + 16 =0
x^2 + 8x + 4^2 = 0
=> (x+4)^2 =0
=> x = -4
d) 4x^2 -12x =-9
(2x)^2 - 12x -9 = 0(chuyển vế)
=> (2x)^2 -12x -(3)^2=0
= (2x-3)^2 =0
=> x = 1,5
Bài 2:
a) (n+2)^2 - ( n-2)^2 = ( n +2+n-2)(n + 2 -n+2)
=> 2n * 4 = 8n : 8 với mọi n
b) (n+7)^2 - ( n-5)^2 = (n+7+n-5)(n+7-n+5)
= (2n +2)*12 = 24(n+1) : 24 với mọi x
Bài 2 :
\(a,\)\(\left(n+2\right)^2-\left(n-2\right)^2=\left(n+2-n+2\right)\left(n+2+n-2\right)\)
\(=4.2n=8n\)\(⋮\)\(8\)
\(\Rightarrow\left(n+2\right)^2-\left(n-2\right)^2\)\(⋮\)\(8\)
\(b,\left(n+7\right)^2-\left(n-5\right)^2=\left(n+7+n-5\right)\left(n+7-n+5\right)\)
\(=\left(2n-2\right).12=2\left(n-1\right).12=24\left(n-1\right)\)\(⋮\)\(24\)
\(\Rightarrow\left(n+7\right)^2-\left(n-5\right)^2\)\(⋮\)\(24\)
Bài 2:
a) Ta có:\(\left(n+2\right)^2-\left(n-2\right)^2=\left(n+2+n-2\right)\left(n+2-n+2\right)\)
\(=2n.4=8n⋮8\)
=>đpcm
b)Ta có: \(\left(n+7\right)^2-\left(n-5\right)^2=\left(n+7+n-5\right)\left(n+7-n+5\right)\)
\(=2n.12=24n⋮24\)
=>đpcm
