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a,x2-y2-2x+2y
= (x+y)(x-y) - 2(x-y)
= (x-y)(x+y-2)
b,2x+2y-x2-xy
= 2(x+y) - x(x+y)
= (x+y)(2-x)
c,3a2-6ab+3b2-12c2
= 3(a2 - 2ab + b2 - 4c2)
= 3[(a-b)2 - 4c2)
= 3(a-b-2c)(a-b+2c)
d,x2-25+y2+2xy
= (x+y)2 - 25
= (x+y+5)(x+y-5)
e) a2+2ab+b2-ac-bc
= (a+b)2-c(a+b)
= (a+b)( a+b-c)
f) x2-2x-4x2-4y
= -3x2-2x-4y
= -(3x2+2x+4y)
g)x2y-x3-9y+9x
= x2(y-x)-9(y-x)
= (y-x)(x2-9)
h) x2(x-1)+16(1-x)
= x2(x-1)-16(x-1)
= (x-1)(x2-16)
= (x-1)(x-4)(x+4)
n) 81x2-6yz-9y2-z2
= (9x)2-[(3y)2+6yz+z2]
=(9x)2-(3y+z)2
=(9x+3y+z)(9x-3y-z)
m) xz- yz-x2+2xy-y2
= z(x-y)-(x2-2xy+y2)
= z(x-y)-(x-y)2
= (x-y)(z-x+y)
p) x2 + 8x + 15
= x2 + 3x + 5x + 15
= x(x+3) + 5(x+3)
= (x+3)(x+5)
k) x2 - x - 12
= x2 + 3x - 4x - 12
= x(x+3) - 4(x+3)
= (x+3)(x-4)
A= -x2+2x+3
=>A= -(x2-2x+3)
=>A= -(x2-2.x.1+1+3-1)
=>A=-[(x-1)2+2]
=>A= -(x+1)2-2
Vì -(x+1)2 ≤0=> A≤-2
Dấu "=" xảy ra khi
-(x+1)2=0 => x=-1
Vây A lớn nhất= -2 khi x= -1
B=x2-2x+4y2-4y+8
=> B= (x2-2x+1)+(4y2-4y+1)+6
=> B=(x-1)2+(2y+1)2+6
=> B lớn nhất=6 khi x=1 và y=-1/2
a) \(x^2-y^2-3x+3y\)
\(=\left(x-y\right)\left(x+y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-3\right)\)
b) \(2x+2y-x^2+y^2\)
\(=2\left(x+y\right)-\left(x^2-y^2\right)\)
\(=2\left(x+y\right)-\left(x-y\right)\left(x+y\right)\)
\(=\left(x+y\right)\left(2-x+y\right)\)
c) \(x^2-16+y^2+2xy\)
\(=x^2+y^2+2xy-16\)
\(=\left(x+y\right)^2-16\)
\(=\left(x+y+4\right)\left(x+y-4\right)\)
a) \(x^2-y^2-3x+3y\)
\(=\left(ax+y\right)\left(ax-y\right)-3.\left(x-y\right)\)
b) \(2x+2y-x^2+y^2\)
\(=2\left(x+y\right)-\left(x+y\right)\left(x-y\right)\)
c) \(x^2-16+y^2+2xy\)
\(=\left(x+y\right)\left(x-y\right)+2xy-16\)
a) x2-2x-y2+2y
=(x2-y2)-(2x-2y)
=(x-y)(x+y)-2(x-y)
=(x-y)(x+y-2)
2x – 2y – x2 + 2xy – y2
(Có x2 ; 2xy ; y2 ta liên tưởng đến HĐT (1) hoặc (2))
= (2x – 2y) – (x2 – 2xy + y2)
= 2(x – y) – (x – y)2
(Có x – y là nhân tử chung)
= (x – y)[2 – (x – y)]
= (x – y)(2 – x + y)
a. = \(\left(x^3+x^2\right)+\left(7x^2+7x\right)+\left(10x+10\right)\)
= \(x^2\left(x+1\right)+7x\left(x+1\right)+10x\left(x+1\right)\)
= \(\left(x+1\right)\left(x^2+7x+10x\right)\)
= \(\left(x+1\right)\left(x+2\right)\left(x+5\right)\)
c) \(5x^2+3y+15x+xy=5x\left(x+3\right)+y\left(x+3\right)=\left(x+3\right)\left(5x+y\right)\)
d) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+3-y\right)\left(x+3+y\right)\)
e) \(x^2-y^2+2x+1=\left(x^2+2x+1\right)-y^2=\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)
f) \(x^2-2xy-9+y^2=\left(x^2-2xy+y^2\right)-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
c: \(5x^2+15x+3y+xy\)
\(=5x\left(x+3\right)+y\left(x+3\right)\)
\(=\left(x+3\right)\left(5x+y\right)\)
d: \(x^2+6x+9-y^2\)
\(=\left(x+3\right)^2-y^2\)
\(=\left(x+3-y\right)\left(x+3+y\right)\)
e: \(x^2+2x+1-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1-y\right)\left(x+1+y\right)\)
f: \(x^2-2xy+y^2-9\)
\(=\left(x-y\right)^2-9\)
\(=\left(x-y-3\right)\left(x-y+3\right)\)
Lời giải:
$\frac{x}{y}$ không phải đơn thức bạn nhé.
a. $x^2-2x+1=(x-1)^2$
b. $x^2+2xy-25+y^2=(x^2+2xy+y^2)-25=(x+y)^2-5^2=(x+y-5)(x+y+5)$
c. $5x^2-10xy=5x(x-2y)$
d. $x^2-y^2+x-y=(x^2-y^2)+(x-y)=(x-y)(x+y)+(x-y)$
$=(x-y)(x+y+1)$
Bài làm:
a) x2 - y2 - 2x + 2y = (x - y)(x + y) - (2x - 2y)
= (x - y)(x + y) - 2(x - y)
= [(x + y) - 2].(x - y)
= (x + y - 2)(x - y)
c)3a2 - 6ab + 3b2 - 12c2 = (3a2 - 6ab + 3b2) - 12c2
= 3(a2 - 2ab + b2) - 12c2
= 3[(a - b)2] - 12c2
= 3[(a - b)2 - 4c2]
= 3[(a - b)2 - (2c)2]
= 3[(a - b - c) - (a - b + c)]
= 3(a - b - c - a + b - c)
= 3(-2c)
= -6c
d)x2 - 5 + y2 + 2xy = (x2 + 2xy + y2) - 5
= (x + y)2 - 5
= (x + y)2 -(\(\sqrt{5}\))2
= (x + y - \(\sqrt{5}\)) - (x + y + \(\sqrt{5}\))
= x + y - \(\sqrt{5}\) - x - y -\(\sqrt{5}\)
= -2\(\sqrt{5}\)
e) a2 + 2ab + b2 - ac - bc = (a2 + 2ab + b2) - (ac + bc)
= (a + b)2 - c(a + b)
= [(a + b) - c].(a + b)
= (a + b - c)(a + b)
Còn câu b) và câu f) Vàng sẽ nghĩ sau :v
Tiếp câu f luôn !
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)