I : Phân tích đa thức thành nhân tử
a) a^2+b^2+2ab+2a+2b+1
b)3x(x-2y)+6y(2y-x)
c)16xy+4y^2-9+16x^2
d) x^4+64y^8
3)3x^2-7x+2
K
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28 tháng 12 2018
a, a2+b2+2ab+2a+2b+1=(a+b+1)2
b,3x(x-2y)+6y(2y-x)=3x(x-2y)-6y(x-2y)
=3(x-2y)(x-2y)=3(x-2y)2
c, 16xy +4y2-9 +16x2=(16x2+16xy+4y2)-32
=(4x-2y)2-32=(4x-2y+3)(4x-2y-3)
BD
1
29 tháng 12 2017
B1 :
a, B = (x+1)^2+(y-2)^2 = (99+1)^2+(102-2)^2 = 100^2+100^2 = 20000
b, = (2x^2+16x+32)-2y^2
= 2.(x+4)^2-2y^2
= 2.[(x+4)^2-y^2] = 2.(x+4-y).(x+4+y)
c, <=> (x^2-3x)+(2x-6) = 0
<=> (x-3).(x+2) = 0
<=> x-3=0 hoặc x+2=0
<=> x=3 hoặc x=-2
B2 :
P = (3-x).(x+3)/x.(x-3) = -(x+3)/x = -x-3/x
k mk nha
P
29 tháng 12 2017
Bai 1
a)B=(x+1)2+(y-2)2
Voi x=99,y=102
=>B= 1002+1002
=20000
b)\(2x^2-2y^2+16x+32\)
=\(2\left[\left(x^2+8x+16\right)-y^2\right]\)
=\(2\left[\left(x+4\right)^2-y^2\right]\)
=2(x-y+4)(x+y+4)
c)\(x^2-3x+2x-6=0\)
=>x(x-3)+2(x-3)=0
=>(x-3)(x+2)=0
=>x=-2;3
Bai 2
\(P=\frac{9-x^2}{x^2-3x}\)
=\(-\frac{x^2-9}{x\left(x-3\right)}\)
=\(-\frac{\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)}\)
=\(\frac{-x-3}{x}\)
PT
15 tháng 8 2018
\(1)x^3-x^2y-4x-4y=x^2\left(x-y\right)-4\left(x-y\right)=\left(x^2-2^2\right)\left(x-y\right)=\left(x^2-4x+4\right)\left(x-y\right)\)
\(2)x^3-3x^2+1-3x=\left(x^3+1\right)-3x\left(x-1\right)=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x-1\right)\)
CN
24 tháng 12 2017
a) -y2 + 2xy - x2 + 3x - 3y
= (3x - 3y) - (x2 - 2xy + y2)
= 3(x - y) - (x - y)2
= (x - y)(3 - x + y)
b) x3 - 2x2 - x + 2
= (x3 - x) - (2x2 - 2)
= x(x2 - 1) - 2(x2 - 1)
= (x2 - 1)(x - 2)
= (x - 2)(x - 1)(x + 1)
c) x2(x + 1) - 2x(x + 1) + x + 1
= (x + 1)(x2 - 2x + 1)
= (x + 1)(x - 1)2
d) a2 + b2 + 2a - 2b - 2ab
= (a2 - 2ab + b2) + (2a - 2b)
= (a - b)2 + 2(a - b)
= (a - b)(a - b + 2)
e) 4x2 - 8x + 3
= (4x2 - 2x) - (6x - 3)
= 2x(2x - 1) - 3(2x - 1)
= (2x - 1)(2x - 3)
f) 25 - 16x2
= 52 - (4x)2
= (5 - 4x)(5 + 4x)
24 tháng 12 2017
a, -y2 + 2xy - x2 + 3x - 3y
= - (x2 - 2xy + y2) + 3(x - y)
= - (x - y)2 + 3(x - y)
= (x - y) (3 - x + y)
b, x3 - 2x2 - x + 2
= x2 (x - 2) - (x - 2)
= (x - 2)(x2 - 1)
= (x - 2)(x - 1)(x + 1)
c, x2 (x + 1) - 2x(x + 1) + x + 1
= x2 (x + 1) - 2x(x + 1) + (x + 1)
= (x + 1)(x2 - 2x + 1)
= (x + 1)(x - 1)2
d, a2 + b2 + 2a - 2b - 2ab
= (a2 - 2ab + b2 )+ (2a - 2b)
= (a - b)2 + 2(a - b)
= (a - b)( a - b + 2)
e, 4x2 - 8x + 3
= 4x2 - 2x - 6x + 3
= 2x( 2x - 1) - 3(2x - 1)
= (2x - 1)(2x - 3)
f, 25 - 16x2
= 52 - (4x)2
= (5 - 4x)(5 + 4x)
Chúc bạn học tốt!
NK
1. Phân tích đa thức thành nhân tử
a, 1/4x^2-5xy+25y^2
b, (7x-4)^2-(2x+1)^2
c, (x-2)^2-4y
d, 125-x^6
1
D
25 tháng 6 2018
a) \(\frac{1}{4}x^2-5xy+25y^2=\left(\frac{1}{2}x\right)^2-5xy+\left(5y\right)^2\)
\(=\left(\frac{1}{2}x-5y\right)^2\)
b) \(\left(7x-4\right)^2-\left(2x+1\right)^2\)
\(=\left(7x-4+2x+1\right)\times\left(7x-4-2x-1\right)=\left(9x-3\right)\times\left(5x-5\right)\)
\(=3\times5\times\left(3x-1\right)\times\left(x-1\right)=15\times\left(3x-1\right)\times\left(x-1\right)\)
c)\(\left(x-2\right)^2-4y^2=\left(x-2-2y\right)\left(x-2+2y\right)\)
d) \(125-x^6=5^3-\left(x^2\right)^3=\left(5-x^2\right)\left(25+5x^2+x^4\right)\)
NT
2
HY
14 tháng 4 2020
\(c.x^2y^2+1-x^2-y^2\\ =x^2y^2-x^2+\left(1-y^2\right)\\ =-x^2\left(1-y^2\right)+1\left(1-y^2\right)\\ =\left(1-x^2\right)\left(1-y^2\right)\\ =\left(1-x\right)\left(1+x\right)\left(1-y\right)\left(1+y\right)\)
a)
\(a^2+b^2+2ab+2a+2b+1\)
\(=(a^2+2ab+b^2)+(2a+2b)+1\)
\(=(a+b)^2+2(a+b)+1^2=(a+b+1)^2\)
b)
\(3x(x-2y)+6y(2y-x)\)
\(=3x(x-2y)-6y(x-2y)=(3x-6y)(x-2y)=3(x-2y)(x-2y)\)
\(=3(x-2y)^2\)
c)
\(16xy+4y^2-9+16x^2\)
\(=(16x^2+16xy+4y^2)-9\)
\(=(4x+2y)^2-3^2=(4x+2y-3)(4x+2y+3)\)
d)
\(x^4+64y^8=(x^2)^2+(8y^4)^2=(x^2)^2+(8y^4)^2+2.x^2.8y^4-2x^2.8y^4\)
\(=(x^2+8y^4)^2-16x^2y^4=(x^2+8y^4)^2-(4xy^2)^2\)
\(=(x^2+8y^4-4xy^2)(x^2+8y^4+4xy^2)\)
e)
\(3x^2-7x+2=3x^2-6x-x+2=(3x^2-6x)-(x-2)\)
\(=3x(x-2)-(x-2)=(3x-1)(x-2)\)