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B) Ta có: 2x-2y-x2+2xy-y2
⇔ 2(x-y)-(x2-2xy+y2)
⇔ 2(x-y)-(x-y)2
⇔ (x-y)(2-x+y)
Đúng thì tick nhé
1: \(6x^2y-9xy^2+3xy\)
\(=3xy\left(2x-3y+1\right)\)
2: \(\left(4-x\right)^2-16\)
\(=\left(4-x-4\right)\left(4-x+4\right)\)
\(=-x\cdot\left(8-x\right)\)
3: \(x^3+9x^2-4x-36\)
\(=x^2\left(x+9\right)-4\left(x+9\right)\)
\(=\left(x+9\right)\left(x-2\right)\left(x+2\right)\)
1) \(6x^2y-9xy^2+3xy=3xy\left(2x-3y+1\right)\)
2) \(\left(4-x\right)^2-16=\left(4-x\right)^2-4^2=\left(4-x-4\right)\left(4-x+4\right)=-x\left(8-x\right)\)
3) \(x^3+9x^2-4x-36\\ =\left(x^3-2x^2\right)+\left(11x^2-22x\right)+\left(18x-36\right)\\ =x^2\left(x-2\right)+11x\left(x-2\right)+18\left(x-2\right)\\ =\left(x^2+11x+18\right)\left(x-2\right)\\ =\left[\left(x^2+2x\right)+\left(9x+18\right)\right]\left(x-2\right)\\ =\left[x\left(x+2\right)+9\left(x+2\right)\right]\left(x-2\right)\\ =\left(x+2\right)\left(x+9\right)\left(x-2\right)\)
\(a,3x-15xy=3x\left(1-5y\right)\\ ---\\ 8x^2+6x-4=2\left(4x^2+3x-2\right)\\ ---\\ 5x^2+25xy+10y^2=5\left(x^2+5xy+2y^2\right)\\ ---\\ 9x^2y^2+6x^2y-\dfrac{1}{2}xy^2=\dfrac{1}{2}xy\left(18xy+12x-y\right)\)
a, \(=12x^5+9x^3y^2-6x^2y^3-20x^4y-15x^2y^3-10xy^4-24x^3y^2-18xy^4+12y^5\)
(tự rút gọn cái :P)
b, \(8x^3+4x^2y-2xy^2-y^3\)
\(=4x^2\left(2x+y\right)-y^2\left(2x+y\right)=\left(2x+y\right)^2\left(2x-y\right)\)
\(4x^2y^2-4x^2-4xy-y^2=4x^2y^2-\left(2x+y\right)^2\)
\(=\left(2x+y+2xy\right)\left(2xy-2x+y\right)\)
Mấy cái còn lại nhân tung ra là được mà :))))
\(5x^3-5x=5x\left(x^2-1\right)\)
\(3x^2+5x-3xy-5x=x\left(3x+5\right)-x\left(3y+5\right)=x\left(3x-3y\right)=3x\left(x-y\right)\)
câu 1.
P= 2(x+y)(x-y)+(x-y)^2+(x+y)^2-4y^2
P= (x+y+x-y)^2-(2y)^2
P=(2x-2y)(2x+2y)
P=4(x^2-y^2)
câu 2.
a, x^3-2x^2-4xy^2+x= x(x^2-2x+1)-4xy^2
=x(x-1)^2-4xy^2
=x(x-1-2y)(x-1+2y)
b, (x+1)(x+2)(x+3)(x+4)-24= (x^2+5x+4)(x^2+5x+6)-24
Đặt x^2+5x+4= a
Lúc đó: (x+1)(x+2)(x+3)(x+4)-24= a(a+2)-24
= a^2+2a-24
=a^2+2a+1-25
= (a+1)^2-5^2
= (a+1-5)(a+1+5)
= (a-4)(a+6)
mà ta đặt x^2+5x+4=a => (x+1)(x+2)(x+3)(x+4)-24= (x^2+5x+4-4)(x^2+5x+4+6)
= (x^2+5x)(x^2+5x+10)
câu3. (x+2)^2= 4-x^2
=> (x+2)^2-4+x^2=0
=>. (x+2)^2-(2-x)(2+x)=0
=> (x+2)(x+2-2+x)=0
=> (x+2)2x=0
=> x+2=0 hoặc 2x=0
=> x=-2 hoặc x=0
1)P=2(x^2-y^2)+x^2-2xy+y^2+x^2+2xy+y^2-4y^2=2x^2-2y^2+2x^2+2y^2-4y^2=4x^2-4y^2 . 3) <=> x^2+4x+4-4+x^2=0
<=> 2x^2+4x=0 <=>2x(x+2)=0 <=>2x=0 hay x+2=0 <=>x=0 hay x=-2
1/
x2 - 3x - 4
= \(x^2-3x+\frac{9}{4}-\frac{9}{4}-4\)
\(=\left(x^2-3x+\frac{9}{4}\right)-\frac{25}{4}\)
\(=\left(x-\frac{3}{2}\right)^2-\left(\frac{5}{2}\right)^2\)
\(=\left(x-\frac{3}{2}-\frac{5}{2}\right)\left(x-\frac{3}{2}+\frac{5}{2}\right)\)
\(=\left(x-4\right)\left(x+1\right)\)
Bài 1 :
\(x^2-3x-4\)
\(=x^2+x-4x-4\)
\(=x\left(x+1\right)-4\left(x+1\right)\)
\(=\left(x+1\right)\left(x-4\right)\)