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\(a,6x^2-9x=3x\left(x-3\right)\)
\(b,x^3-2x^2-3x+6\)
\(=\left(x^3-2x^2\right)-\left(3x-6\right)\)
\(=x^2\left(x-2\right)-3\left(x-2\right)\)
\(=\left(x^2-3\right)\left(x-2\right)\)
\(e,2x\left(x-y\right)-3y\left(x-y\right)\)
\(=\left(2x-3y\right)\left(x-y\right)\)
a) 6x2 - 9x
= 3x (2x - 3)
b) x3 - 2x2 - 3x + 6
= x2(x - 2) - 3 (x - 2)
=(x - 2) (x2 - 3)
c) x2 - 4x + 4 - 9y2
= (x - 2)2 - 9y2
=(x - 2 - 3y)(x - 2 + 3y)
e) 2x(x - y) - 3y(x - y)
= (x - y)(2x - 3y)
xin lỗi mình học ngu nên không biết làm nhìu nha
b, A=[(a+1)(a+7)][(a+3)(a+5)]+15
=>A=(a2+8a+7)(a2+8a+15)+15
Đặt a2+8a+11= t
=>a2+8a+7= t-4 và a2+8a+15= t+4
=>A=(t-4)(t+4)+15
=>A=t2-16+15
=t2-1=(t-1)(t+1)
Thay t = a2+8a+11
=>A=(a2+8a+11-1)(a2+8a+11+1)
=>A=(a2+8a+10)(a2+8a+12)
a) \(x^2+2xy+7x+7y+y^2+10\)
\(=\left(x+y\right)^2+7\left(x+y\right)+\frac{49}{4}-\frac{9}{4}\)
\(=\left(x+y+\frac{7}{2}\right)^2-\frac{9}{4}\)
\(=\left(x+y+\frac{7}{2}-\frac{3}{2}\right)\left(x+y+\frac{7}{2}+\frac{3}{2}\right)\)
\(=\left(x+y-2\right)\left(x+y+5\right)\)
Bài 2
\(a,x^3+2x^2+x\)
\(=x.\left(x^2+2x+1\right)\)
\(b,xy+y^2-x-y\)
\(=y.\left(x+y\right)-\left(x+y\right)\)
\(=\left(y-1\right).\left(x+y\right)\)
bài 3
\(a,3x.\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\\x^2=4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=2,x=-2\end{cases}}\)
vậy x=0,x=2 hay x=-2
\(b,xy+y^2-x-y=0\)
\(y.\left(x+y\right)-\left(x+y\right)=0\)
\(\left(y-1\right).\left(x+y\right)=0\)
\(\Rightarrow\orbr{\begin{cases}y-1=0\\x+y=0\end{cases}\Rightarrow\orbr{\begin{cases}y=1\\x=-1\end{cases}}}\)
vậy x=-1, y=1
1. \(4x^2-2x-3y-9y^2\)
\(=\left(2x\right)^2-\left(3y\right)^2-\left(2x+3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y\right)-\left(2x+3y\right)\)
\(=\left(2x+3y\right)\left(2x-3y-1\right)\)
2. \(x^2-25=6x-9\)
\(\Rightarrow x^2-6x+9=25\)
\(\Rightarrow\left(x-3\right)^2=25\)
\(\Rightarrow\orbr{\begin{cases}x-3=5\\x-3=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=-2\end{cases}}\)
1. Ta có: \(3xy\left(a^2+b^2\right)+ab\left(x^2-9y^2\right)\)
\(=3xya^2+3xyb^2+abx^2+ab9y^2\)
\(=\left(3xya^2+abx^2\right)+\left(3xyb^2+ab9y^2\right)\)
\(=ax\left(3ya+bx\right)+3by\left(xb+3ya\right)\)
\(=\left(3ya+xb\right)\left(3yb+ax\right)\)
2.Check lại đề hộ mình nha:((
Câu 2 nên sủa lại đề nha
2. xy(a2+2b2)+ab(2x2+y2)
=xya2+xy2b2+ab2x2+aby2
=(xya2+aby2)+(xy2b2+ab2x2)
=ay(ax+by)+2bx(by+ax)
=(ax+by(ay+2bx)
x(y - z) + 2(z - y)
= x(y - z) - 2(y - z)
= (x - 2)(y - z)
(2x - 3y)(x - 2) - (x + 3)(3y - 2x)
= (2x - 3y)(x - 2) + (x + 2)(2x - 3y)
= (2x - 3y)(x - 2 + x + 2)
= 2x(2x - 3y)
a) \(x^2-25-4xy+4y^2\)
\(=\left(x^2-4xy+4y^2\right)-25\)
\(=\left(x-2y\right)^2-5^2\)
\(=\left(x-2y-5\right)\left(x-2y+5\right)\)
b) \(x^2-8x+15\)
\(=x^2-3x-5x+15\)
\(=x\left(x-3\right)-5\left(x-3\right)\)
\(=\left(x-3\right)\left(x-5\right)\)
a)\(x^2-25-4xy+4y^2\Leftrightarrow\left(x^2-4xy+4y^2\right)-25\)
\(\Leftrightarrow\left(x-2y\right)^2-5^2\)
\(\Leftrightarrow\left(x-2y-5\right)\left(x-2y+5\right)\)
b)\(x^2-8x+15\Leftrightarrow\left(x-3\right)\left(x-5\right)\)
Bài làm:
1) Ta có: \(2x^2+5xy+2y^2\)
\(=\left(2x^2+4xy\right)+\left(xy+2y^2\right)\)
\(=2x\left(x+2y\right)+y\left(x+2y\right)\)
\(=\left(2x+y\right)\left(x+2y\right)\)
2) Ta có: \(2x^2+2xy-4y^2\)
\(=\left(2x^2-2xy\right)+\left(4xy-4y^2\right)\)
\(=2x\left(x-y\right)+4y\left(x-y\right)\)
\(=2\left(x+2y\right)\left(x-y\right)\)
\(1)2x^2+5xy+2y^2=2x^2+4xy+xy+2y^2=\left(2x^2+4xy\right)+\left(xy+2y^2\right)=2x\left(x+2y\right)+y\left(x+2y\right)=\left(2x+y\right)\left(x+2y\right)\)\(2)2x^2+2xy-4y^2=2x^2+4xy-2xy-4y^2=\left(2x^2-2xy\right)+\left(4xy-4y^2\right)=2x\left(x-y\right)+4y\left(x-y\right)=\left(2x+4y\right)\left(x-y\right)\)
\(x^2-2x+\left(x-2\right)^2\)
\(=x^2-2x+x^2-4x+4\)
\(=2x^2-6x+4\)
\(=2.\left(x^2-3x+2\right)\)
\(=2.\left[\left(x^2-x\right)-\left(2x-2\right)\right]\)
\(=2.\left[x.\left(x-1\right)-2.\left(x-1\right)\right]\)
\(=2.\left(x-1\right)\left(x-2\right)\)
\(a,x^2-2x+\left(x-2\right)^2\)
\(=x\left(x-2\right)+\left(x-2\right)^2\)
\(=\left(x+x-2\right)\left(x-2\right)\)
\(b,x^2-6xy-16+9y^2\)
\(=\left(x^2-6xy+9y^2\right)-16\)
\(=\left(x+3y\right)^2-4^2\)
\(=\left(x+3y-4\right)\left(x+3y+4\right)\)