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\(1)4x^2-25+\left(2x+7\right).\left(5.2x\right)\)
\(=\left(2x\right)^2-5^2-\left(2x+7\right).\left(2x-5\right)\)
\(=\left(2x.5\right)\left(2x+5\right).\left(2x+7\right)\left(2x-5\right)\)
\(=\left(2x-5\right)\left(2x+5-2x+7\right)\)
\(=\left(2x-5\right).12\)
\(2)3x+4-x^2-4x\)
\(=3(x+4)-\left(x+4\right)\)
\(=\left(3-x\right)\left(x+4\right)\)
\(3)5x^2-2y^2-10x+10y\)
\(=5\left(x^2-y^2\right)-10\left(x-4\right)\)
\(=5\left(x-y\right)\left(x+y\right)-10\left(x-y\right)\)
\(=\left(x-y\right)[5(x+y)-10]\)
Còn lại bn lm nốt nha!
a) Ta có: \(x^2\left(x-1\right)+16\left(1-x\right)\)
\(=x^2\left(x-1\right)-16\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-16\right)\)
\(=\left(x-1\right)\left(x-4\right)\left(x+4\right)\)
b) Ta có: \(5x^2-5y^2-10x+10y\)
\(=5\left(x^2-y^2\right)-10\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\cdot2\)
\(=5\left(x-y\right)\left(x+y-2\right)\)
c) Ta có: \(x^2+4x-y^2+4\)
\(=\left(x^2+4x+4\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2-y\right)\left(x+2+y\right)\)
f)\(\left(x-y\right)^2-4=\left(x-y-4\right)\left(x-y+4\right)\)
h) \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)
i)\(10x-x^2-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)
k)\(4x^2-12xy+9y^2=\left(2x\right)^2-2.2x.3y+\left(3y\right)^2=\left(2x-3y\right)^2\)
mấy bài này cơ bản mà, mở sgk toán 8 ra có các dạng đấy, đăng cũng đăng ít chứ, đăng nhiều quá
a)\(6x^3-9y^2=3\left(2x^3-3y^2\right)\)
b)\(4x^2y-8xy^2+18x^2y^2=2xy\left(2x-4y+9xy\right)\)
c)\(18x^2y-12x^3=6x^2\left(3y-2x\right)\)
d) \(5x\left(x-1\right)-3y\left(x-1\right)=\left(x-1\right)\left(5x-3y\right)\)
e)\(\left(x+1\right)^2-3\left(x+1\right)=\left(x+1\right)\left(x+1-3\right)=\left(x+1\right)\left(x-2\right)\)
g)\(\left(4x^2-4x+4\right)-\left(x+1\right)^2=\left(4x^2-4x+4\right)-\left(x^2+2x+1\right)\)
\(=4x^2-4x+4-x^2-2x-1\)\(=3x^2-6x+3\)\(=3\left(x^2-2x+1\right)\)
\(=3\left(x-1\right)^2\)
\(=\left(x^2+4x-3\right)^2-5\left(x^2+4x-3\right)+6x^2\)
\(=x^4+16x^2+9+8x^3-24x-6x^2-5x^2-20x+15+6x^2\)
\(=x^4+8x^3+11x^2-44x+24\)
\(=\left(x^4-x^3\right)+\left(9x^3-9x^2\right)+\left(20x^2-20x\right)-\left(24x-24\right)\)
\(=x^3\left(x-1\right)+9x^2\left(x-1\right)+20x\left(x-1\right)-24\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3+9x^2+20x-24\right)\)
/ (4x−2)(10x+4)(5x+7)(2x+1)+17=0
⇔(4x−2)(5x+7)(10x+4)(2x+1)+17=0
⇔(20x2+18x−14)(20x2+18x+4)+17=0
Đặt t= 20x2+18x+4(t≥0) ta có:
(t-18).t +17=0
⇔t2−18t+17=0
⇔(t−17)(t−1)=0
⇔[t=17(tm)t=1(tm) ⇔[20x2+18x+4=1720x2+18x+4=1⇔[20x2+18x−13=020x2+18+3=0
⇔[(20x+9−341−−−√)(20x+9+341−−−√)=0(20x+9−21−−√)(20x+9+21−−√)=0
⇔⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢x=−9+341−−−√20x=−9−341−−−√20x=−9+21−−√20x=−9−21−−√20
\(a,\)\(\left(4x-2\right)\left(10x+4\right)\left(5x+7\right)\left(2x+1\right)+17\)
\(=\left(4x-2\right)\left(5x+7\right)\left(10x+4\right)\left(2x+1\right)+17\)
\(=\left(20x^2+18x-5\right)\left(20x^2+18x+4\right)+17\)
Đặt ....
1: \(=a\left(x+y\right)-4\left(x+y\right)=\left(x+y\right)\left(a-4\right)\)
2: \(=x\left(x+b\right)+a\left(x+b\right)=\left(x+b\right)\left(x+q\right)\)
3: \(=a\left(x+1\right)-b\left(x+1\right)+c\left(x+1\right)\)
\(=\left(x+1\right)\left(a-b+c\right)\)
6: \(=\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\)
a) \(3\left(x+4\right)-x^2-4x\)
\(=3\left(x+4\right)-x\left(x+4\right)\)
\(=\left(x+4\right)\left(3-x\right)\)
b) \(5x^2-5y^2-10x+10y\)
\(=5\left(x^2-y^2\right)-10\left(x-y\right)\)
\(=5\left(x+y\right)\left(x-y\right)-10\left(x-y\right)\)
\(=\left(x-y\right)\left[5\left(x+y\right)-10\right]\)
\(=5\left(x-y\right)\left(x+y-2\right)\)
c) \(x^2-xy+x-y\)
\(=x\left(x-y\right)+\left(x-y\right)\)
\(=\left(x-y\right)\left(x+1\right)\)
d) \(ax-bx-a^2+2ab-b^2\)
\(=x\left(a-b\right)-\left(a-b\right)^2\)
\(=\left(a-b\right)\left(x-a+b\right)\)
e) \(x^2-4x-y^2+4\)
\(=\left(x^2-2\cdot x\cdot2+2^2\right)-y^2\)
\(=\left(x-2\right)^2-y^2\)
\(=\left(x-2-y\right)\left(x-2+y\right)\)
f) \(x^3-x^2-x+1\)
\(=x^2\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)^2\left(x+1\right)\)
g) \(x^4+6x^2y+9y^2-1\)
\(=\left(x^2\right)^2+2\cdot x^2\cdot3y+\left(3y\right)^2-1^2\)
\(=\left(x^2+3y\right)^2-1^2\)
\(=\left(x^2+3y-1\right)\left(x^2+3y+1\right)\)
1) \(3\left(x+4\right)-x^2-4x=3\left(x+4\right)-x\left(x+4\right)=\left(x+4\right)\left(3-x\right)\)
2) \(5x^2-5y^2-10x+10y=5\left(x^2-y^2\right)-10\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-10\left(x-y\right)=\left(x-y\right)\left(5x+5y-10\right)\)
3) \(x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right)\)
4) \(ax-bx-a^2+2ab-b^2=x\left(a-b\right)-\left(a^2-2ab+b^2\right)\)
\(=x\left(a-b\right)-\left(a-b\right)^2=\left(a-b\right)\left(x-a+b\right)\)
5) \(x^3-x^2-x+1=x^2\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)\left(x-1\right)\left(x+1\right)=\left(x-1\right)^2\left(x+1\right)\)
6) \(x^2+4x-y^2+4=x^2+4x+4-y^2=\left(x+2\right)^2-y^2\)
\(=\left(x+2-y\right)\left(x+2+y\right)\)
Phân tích các đa thức sau thành nhân tử :
1) x^3 + x^2y - 4x - 4y
2) x^3 - 3x^2 +1 - 3x
3) 3x^2 - 6xy + 3y^2 - 12z^2
4) x^2 - 2x - 15
5) 2x^2 +3x - 5
6) 2x^2 - 18
7) x^2 - 7xy + 10y^2
8) x^3 - 2x^2 + x - xy^2
Làm nhanh giúp mình với nhé .....mình đang cần gấp[[[[