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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)
B1:
a) \(5\left(x^2+y^2\right)-20x^2y^2\)
\(=5\left(x^2-4x^2y^2+y^2\right)\)
b) \(=2\left(x^8-16\right)=2\left(x^4-4\right)\left(x^4+4\right)=2\left(x^2-2\right)\left(x^2+2\right)\left(x^4+4\right)\)
B2:
a) Đặt \(x^2-3x+1=y\)
=> \(y^2-12y+27\)
\(=\left(y^2-12y+36\right)-9\)
\(=\left(y-6\right)^2-3^2\)
\(=\left(y-9\right)\left(y-3\right)\)
\(=\left(x^2-3x-10\right)\left(x^2-3x-4\right)\)
\(=\left(x+1\right)\left(x-4\right)\left(x^2-3x-10\right)\)
b) Đặt \(x^2+7x+11=t\)
Ta có: \(\left[\left(x+2\right)\left(x+5\right)\right]\cdot\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
a/ \(x^{12}-3x^6+1\)
= \(\left(x^6\right)^2-2x^6+1-x^6\)
= \(\left(x^6-1\right)^2-\left(x^3\right)^2\)
= \(\left(x^6-x^3-1\right)\left(x^6+x^3-1\right)\)
b/ \(x^8-3x^4+1\)
= \(\left(x^4\right)^2-2x^4+1-x^4\)
= \(\left(x^4-1\right)^2-\left(x^2\right)^2\)
= \(\left(x^4-x^2-1\right)\left(x^4+x^2-1\right)\)
a) x8 + x + 1 = (x^2+x+1)*(x^6-x^5+x^3-x^2+1)
b) x^8 + 3x^4 + 4 = (x^4-x^2+2)*(x^4+x^2+2)
\(x^4+3x^2+36\)
\(=\left(x^2\right)^2+2.x^2.6+6^2-9x^2\)
\(=\left(x^2+6\right)^2-\left(3x\right)^2=\left(x^2-3x+6\right)\left(x^2+3x+6\right)\)
\(2x^4-3x^3-7x^2+6x+8\)
\(=2x^4+2x^3-5x^3-5x^2-2x^2-2x+8x+8\)
\(=2x^3\left(x+1\right)-5x^2\left(x+1\right)-2x\left(x+1\right)+8\left(x+1\right)\)
\(=\left(x+1\right)\left(2x^3-5x^2-2x+8\right)\)
\(=\left(x+1\right)\left[2x^2\left(x-2\right)-x\left(x-2\right)-4\left(x-2\right)\right]\)
\(=\left(x+1\right)\left(x-2\right)\left(2x^2-x-4\right)\)
Chúc bạn học tốt.
a) -7x2 + 5xy +12y2
=-7x2-7xy+12xy+12y2
=-7x(x+y)+12y(x+y)
=(x+y)(12y-7x)
b) x8 + 3x4 + 4
=x8+4x4+4-x4
=(x4+2)2-x4
=(x4-x2+2)(x4+x2+2)
=(x4+x2-2x2+2)(x4+x2+2)
=[x2(x+1)-2(x+1)](x4+x2+1)
=(x+1)(x2-2)(x4+x2+1)
\(x^8+3x^4+4\)
\(=\left(x^8-x^6+2x^4\right)+\left(x^6-x^4+2x^2\right)+\left(2x^4-2x^2+4\right)\)
\(=x^4\left(x^4-x^2+2\right)+x^2\left(x^4-x^2+2\right)+2\left(x^4-x^2+2\right)\)
\(=\left(x^4+x^2+2\right)\left(x^4-x^2+2\right)\)
\(4x^4+4x^3+5x^2+2x+1\)
\(=\left(4x^4+2x^3+2x^2\right)+\left(2x^3+x^2+x\right)+\left(2x^2+x+1\right)\)
\(=2x^2\left(2x^2+x+1\right)+x\left(2x^2+x+1\right)+\left(2x^2+x+1\right)\)
\(=\left(2x^2+x+1\right)^2\)
a) ta có : x^2 -x-12 =( x^2 -4x) +(3x-12)=x(x-4) + 3(x-4) =(x+3)(x-4)
b)ta có: x^8 +3x^4 -4= x^4(x^4 +4) - (x^4 +4) =( x^4 -1)(x^4 +4) =(x^2 -1)(x^2 +1)(x^4 +4)
\(x^8+3x^4+4\)
\(=x^8+4x^4+4-x^4\)
\(=\left(x^4+2\right)^2-x^4\)
\(=\left(x^4+x^2+2\right)\left(x^4-x^2+2\right)\)