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a) 3x^2 - 6x - x+2=3x(x-2)-(x-2)=(x-2)(3x-1)
b) ax(x-a)-(x-a)=(x-a)(ax-1)
a) \(3x^2-7x+2\)
\(=3x^2-6x-x+2\)
\(=3x\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(3x-1\right)\)
b) \(a\left(x^2+1\right)-x\left(a^2+1\right)\)
\(=ax^2+a-a^2x-x\)
\(=\left(ax^2-a^2x\right)-\left(x-a\right)\)
\(=ax\left(x-a\right)-\left(x-a\right)\)
\(=\left(x-a\right)\left(ax-1\right)\)
\(A=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-20\)
\(=\left(x^2+5x+4\right)\cdot\left(x^2+5x+6\right)-20\)
Đặt: \(x^2+5x+5=a\)Khi đó ta có:
\(A=\left(a-1\right)\left(a+1\right)-20=a^2-21=\left(a-\sqrt{21}\right)\left(a+\sqrt{21}\right)\)
tự thay trở lại
a) \(x^2-5x+6=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\)
b)\(3x^2+9x-30=3x^2-6x+15x-30=3\left(x-2\right)\left(x+5\right)\)
c)\(x^2-7x+12=x^2-3x-4x+12=\left(x-3\right)\left(x-4\right)\)
d)\(x^2-7x+10=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\)
a) \(x^2-5x+6=x^2-2x-3x+6=\left(x^2-2x\right)-\left(3x-6\right)\)
\(=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)
b) \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x^2-2x+5x-10\right)\)
\(=3\left[\left(x^2-2x\right)+\left(5x-10\right)\right]=3\left[x\left(x-2\right)+5\left(x-2\right)\right]\)
\(=3\left(x-2\right)\left(x+5\right)\)
c) \(x^2-7x+12=x^2-3x-4x+12=\left(x^2-3x\right)-\left(4x-12\right)\)
\(=x\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-4\right)\)
d) \(x^2-7x+10=x^2-2x-5x+10=\left(x^2-2x\right)-\left(5x-10\right)\)
\(=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
a, 3x2 - 7x +2 = 3x2 - 6x - x +2 = 3x(x-2) - (x-2) = (3x-1)(x-2);
b, a(x2+1) -x(a2+1) = ax2+a-xa2-x=(ax2-xa2) -(x-a)=ax(x-a)-(x-a)=(ax-1)(x-a)
a) \(3x^2-7x+2=3x^2-3x-4x+2=3x\left(x^2-1\right)-2\left(x-1\right)\)
\(=3x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)=\left(x-1\right)\left[\left(3x.\left(x+1\right)-2\right)\right]\)
\(=\left(x-1\right)\left(3x^2+3x-2\right)\)
b) \(a\left(x^2+1\right)-x\left(a^2+1\right)=ax^2+a-xa^2-x\)
\(=ax\left(x-a\right)+\left(a-x\right)=ax\left(x-a\right)-\left(x-a\right)=\left(x-a\right)\left(ax-1\right)\)
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a) Đăt \(x^2+x=t\) khi đó bt trở thành:
\(t^2-2t-15=t^2+3t-5t-15=t\left(t+3\right)-5\left(t+3\right)\\ =\left(t+3\right)\left(1-5\right)=\left(x^2+x+3\right)\left(x^2+x-5\right)\)
a)\(x^2+7x+12\)
\(=x^2+x+6x+6\)
\(=x\left(x+1\right)+6\left(x+1\right)\)
\(=\left(x+1\right)\left(x+6\right)\)
a) x2 + 7x + 12 = x2 + 3x + 4x + 12
= (x2 + 3x) + (4x + 12)
= x(x + 3) + 4(x + 3)
= (x + 3)(x + 4)
vậy.....