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a, \(\left(x^2-2x\right)\left(x^2-2x-1\right)-6\)
Đặt \(x^2-2x=a\)
Thay vào biểu thức ta đc:
\(a.\left(a-1\right)-6=a^2-a-6\) \(=a^2-3a+2a-6=a\left(a-3\right)+2\left(a-3\right)\)
\(=\left(a-3\right).\left(a+2\right)\)
\(\Rightarrow\left(x^2-2x\right)\left(x^2-2x-1\right)-6=\left(x^2-2x-3\right)\left(x^2-2x+2\right)\)
b, \(\left(x^2+x+4\right)^2+8x\left(x^2+x+4\right)+15x^2\)
\(=\left[\left(x^2+x+4\right)^2+6x\left(x^2+x+4\right)+9x^2\right]+\left[2x\left(x^2+x+4\right)+6x^2\right]\)
\(=\left(x^2+x+4+3x\right)^2+2x\left(3x+x^2+x+4\right)\)
\(=\left(x^2+4x+4\right)\left(x^2+4x+4+2x\right)\) \(=\left(x+2\right)^2\left(x^2+6x+4\right)\)
\(x^3-5x^2+8x-4\)
\(=x^3-x^2-4x^2+4x+4x-4\)
\(=x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-4x+4\right)\)
\(=\left(x-1\right)\left(x-2\right)^2\)
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)
1)x2-8x-9
= x^2 - 9x +x -9
= x(x+1) - 9 (x+1)
= (x-9) (x+1)
2)x2+3x-18
3)x3-5x2+4x
=x^3 - 4x^2 - x^2 + 4x
= x^2 (x-1) - 4x(x-1)
= (x^2 - 4x) (x-1)
= x(x-4)(x-1)
4)x3-11x2+30x
5)x3-7x-6
6)x16-64
\(=\left(x^8\right)^2-8^2\)
\(=\left(x^8-8\right)\left(x^8+8\right)\)
7)x3-5x2+8x-4
8)x2-3x+2
= x^2 - 2x - x +2
= x(x-1) -2(x-1)
= (x-2)(x-1)
1) \(\left(x-9\right)\left(x+1\right)\) 2) \(\left(x-3\right)\left(x+6\right)\) 3) \(x\left(x-4\right)\left(x-1\right)\)
4) \(x\left(x-6\right)\left(x-5\right)\) 5)\(\left(x-3\right)\left(x+1\right)\left(x+2\right)\) 6) ........
7) \(\left(x-1\right)\left(x-2\right)\left(x-2\right)\) 8) \(\left(x-2\right)\left(x-1\right)\)
\(\left(x^2-5x+6\right)\left(x^2-5x+2\right)-5\)
\(\text{Phần tích thành nhân tử :}\)
\(\left(x^2-5x+2\right)\left(x^2-5x+7\right)\)
\(\left(x^2+8x-5\right)\left(x^2+8x+1\right)-16\)
\(\text{Phần tích thành nhân tử :}\)
\(\left(x^2+8x-7\right)\left(x^2+8x+3\right)\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)-3\backslash2.x^2\)
\(\text{Phần tích thành nhân tử :}\)
Lười lắm
x6+3x4y2-8x3y3+3x2y4+y6= x6+3x4y2+3x2y4+y6-8x3y3=(x2+y2)3-(2xy)3
= (x2+y2-2xy)[(x2+y2)2+2xy(x2+y2)+(2xy)2]= (x-y)2(x4+6x2y2+y4+2x3y+2xy3)
(x2+y2-5)2-4x2y2-16xy-16=(x2+y2-5)2-(4x2y2+16xy+16)=(x2+y2-5)2-(2xy+4)2
=(x2+y2-5+2xy+4)(x2+y2-5-2xy-4)=(x2+2xy+y2-1)(x2-2xy+y2-9)=[(x+y)2-1][(x-y)2-32]=(x+y-1)(x+y+1)(x-y-3)(x-y+3)
x4+324=x4+36x2+324-36x2=(x2+18)2-(6x)2=(x2+18-6x)(x2+18+6x)
a, \(3x^2-9x-2x+6=3x\left(x-3\right)-2\left(x-3\right)=\left(x-3\right)\left(3x-2\right)\)
b. \(8x^2-2x+12x-3=2x\left(4x-1\right)+3\left(4x-1\right)=\left(4x-1\right)\left(2x+3\right)\)
c. đề kiểu gì vậy? -2x-x để thành -3x à? xem lại đi nha
d. \(\left(x^2+10x+25\right)-\left(y^2+6y+9\right)=\left(x+5\right)^2-\left(y+3\right)^2=\left(x+5-y-3\right)\left(x+5+y+3\right)=\left(x-y+2\right)\left(x+y+8\right)\)
e. \(=x^4+2x^2y^2+y^4-x^2y^2=\left(x^2+y^2\right)^2-x^2y^2=\left(x^2+y^2-xy\right)\left(x^2+y^2+xy\right)\)
nhớ L I K E
(x^2+5x+4)(x^2+5x+6)-24
Đặt x^2+5x+5 = a
Do đó (a-1)(a+1)-24
= a^2- 25
= a^2-5^2 =(a-5)(a+5)
= ( x^2+5x+5-5)( x^2+5x+5+5)
= ( x^2+5x)( x^2+5x+10)
1/ \(\left(x^2+x+4\right)^2+8x\left(x^2+x+4\right)+15x^2=x^4+10x^3+32x^2+40x+16\)(làm tắt nhưng chắc bạn tự hiểu đc)
\(=\left(x^4+2x^3\right)+\left(4x^2+2x^3\right)+\left(12x^2+6x^3\right)+\left(4x^2+8x\right)+\left(12x^2+24x\right)+\left(8x+16\right)\)
\(=x^3\left(x+2\right)+2x^2\left(2+x\right)+6x^2\left(2+x\right)+4x\left(x+2\right)+12x\left(x+2\right)+8\left(x+2\right)\)
\(=\left(x+2\right)\left(x^3+2x^2+6x^2+4x+12x+8\right)=\left(x+2\right)\left(x^3+8x^2+16x+8\right)\)
\(=\left(x+2\right)\left[\left(x^3+2x^2\right)+\left(6x^2+12x\right)+\left(4x+8\right)\right]=\left(x+2\right)\left[x^2\left(x+2\right)+6x\left(x+2\right)+4\left(x+2\right)\right]\)
\(=\left(x+2\right)\left(x+2\right)\left(x^2+6x+4\right)\)
2/ \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16=x^4+20x^3+140x^2+400x+400\)
\(=\left(x^4+10x^3+20x^2\right)+\left(10x^3+100x^2+200x\right)+\left(20x^2+200x+400\right)\)
\(=x^2\left(x^2+10x+20\right)+10x\left(x^2+10x+20\right)+20\left(x^2+10x+20\right)\)
\(=\left(x^2+10x+20\right)\left(x^2+10x+20\right)=\left(x^2+10x+20\right)^2\)