Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) (2x - 1)2 - (x + 3)2
= (2x - 1 - x - 3).(2x - 1 + x + 3)
= (x - 4).(3x + 2)
b) x2.(x - 3) + 12 - 4x
= x2.(x - 3) - 4x + 12
= x2.(x - 3) - 4.(x - 3)
= (x - 3).(x2 - 4)
= (x - 3).(x - 2).(x + 2)
Áp dụng HĐT:
a2 - b2 = (a - b)(a + b)
\(\left(2x-1\right)^2-\left(x+3\right)^2\)
\(=\left(2x-1-x-3\right)\left(2x-1+x+3\right)\)
\(=\left(x-4\right)\left(3x+2\right)\)
x2-8x + 16 - 4 = ( x - 4 )2-22= ( x -4-2 ) . ( x-4+2 ) = ( x - 6 ) .( x -2 )
\(\left(x+2\right)\left(x+3\right)\left(x+8\right)\left(x+12\right)-4x^2\)
\(=\left(x+2\right)\left(x+12\right)\left(x+3\right)\left(x+8\right)-4x^2\)
\(=\left(x^2+14x+24\right)\left(x^2+11x+24\right)-\left(2x\right)^2\)
Đặt \(x^2+11x+24=a\)
\(=a\left(a+3x\right)-4x^2=a^2+3ax-4x^2=a^2-ax+4ax-4x^2=\left(a-x\right)\left(a+4x\right)\)
a) \(x^2\left(x-3\right)+12-4x=x^2\left(x-3\right)-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-4\right)\)
\(=\left(x-3\right)\left(x^2-2^2\right)\)
\(=\left(x+3\right)\left(x-2\right)\left(x+2\right)\)
b)\(x^2-4+\left(x-2\right)^2=x^2-2^2+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2+x-2\right)\)
\(=\left(x-2\right)2x\)
c)\(x^3-4x^2-12x+27=x^3+3x^2-7x^2-21x+9x+27\)
\(=x^2\left(x+3\right)-7x\left(x+3\right)+9\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
a) => x2.(x-3)-4(x-3)=(x-3)(x2-4)=(x-3)(x-2)(x+2)
b) => (x+2)(x-2)+(x-2)2=(x-2)(x+2+x-2)=2x(x-2)
c) => x3+27-(4x2+12x)=(x+3)(x2-3x+3)-4x(x+3)=(x+3)(x2-3x+3-4x)=(x-3)(x2-7x+3)
a) Ta có : x2 + 7x + 12
= x2 + 3x + 4x + 12
= (x2 + 3x) + (4x + 12)
= x(x + 3) + 4(x + 3)
= (x + 4)(x + 3)
Bạn ơi mk nhầm đề rồi số 30 thay bằng số 60 còn 36 thay bằng 72 và 39 thay bằng 75 nha
ta có
x3-3x2-4x+12
=x2(x-3) -4(x-3)
=(x-3)(x2-4)
=(x-3)(x-2)(x+2)
bn k mk nha
\(\left(x^2-x-3\right)\left(x^2-x-4\right)-12\)
\(=x^4-x^3-4x^2-x^3+x^2+4x-3x^2+3x+12-12=x^4-2x^3-6x^2+7x\)
\(=x.\left(x^3-2x^2-6x+7\right)=x.\left(x^3-x^2-x^2+x-7x+7\right)\)
\(=x.\left[x^2.\left(x-1\right)-x.\left(x-1\right)-7.\left(x-1\right)\right]=x\left(x-1\right)\left(x^2-x-7\right)\)
=x^2-4x+3x-12
=x[x-4]+3[x-4]
=[x+3][x-4]
C2:=x^2-16-[x-4]
=[x-4][x+4]-[x-4]
=[x-4][x+4-1]
=[x-4][x+3]
Cách nữa nè !
x^2-x-12
=(x^2-9)-(x+3)
=(x-3)(x+3)-(x+3)
=(x+3)(x-4)