Giải Giúp Mik nhék
x^2-7xy+10y^2
x^2-5x-10
5x^2+6xy+y^2
Phân tích đa thức thành nhân tử:
a) x2 - 7xy + 10y2
b) 5x2 + 6xy + y2
c) x2 - 5x - 14
d) x2 + 2x - 15
a) Ta có: \(x^2-7xy+10y^2\)
\(=x^2-2xy-5xy+10y^2\)
\(=x\left(x-2y\right)-5y\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x-5y\right)\)
b) Ta có: \(5x^2+6xy+y^2\)
\(=5x^2+5xy+xy+y^2\)
\(=5x\left(x+y\right)+y\left(x+y\right)\)
\(=\left(x+y\right)\left(5x+y\right)\)
c) Ta có: \(x^2-5x-14\)
\(=x^2-7x+2x-14\)
\(=x\left(x-7\right)+2\left(x-7\right)\)
\(=\left(x-7\right)\left(x+2\right)\)
d) Ta có: \(x^2+2x-15\)
\(=x^2+5x-3x-15\)
\(=x\left(x+5\right)-3\left(x+5\right)\)
\(=\left(x+5\right)\left(x-3\right)\)
x^2 + 4xy - 21y^2
5x^2 + 6xy + y^2
x^2 + 2xy - 15y^2
(x - y)^2 + 4(x - y) - 12
x^2v- 7xy + 10y^2
x^2yz + 5xyz - 14yz
2) \(5x^2+6xy+y^2\)
\(=9x^2+6xy+y^2-4x^2\)
\(=\left(3x+y\right)^2-\left(2x\right)^2\)
\(=\left(3x+y+2x\right)\left(3x+y-2x\right)\)
\(=\left(5x+y\right)\left(x+y\right)\)
3) \(x^2+2xy-15y^2=x^2+2xy+y^2-16y^2\)
\(=\left(x+y\right)^2-\left(4y\right)^2\)
\(=\left(x+y+4y\right)\left(x+y-4y\right)\)
\(=\left(x+5y\right)\left(x-3y\right)\)
hc tốt
BT2: Phân tích các đa thức sau thành nhân tử bằng phương pháp tách hạng tử. a, x^2 + 4xy - 21y^2 b, 5x^2 + 6xy + y^2 c, x^2 + 2xy - 15y^2 d, x^2 - 7xy + 10y^2
a: x^2+4xy-21y^2
\(=x^2+7xy-3xy-21y^2\)
\(=x\left(x+7y\right)-3y\left(x+7y\right)\)
\(=\left(x+7y\right)\left(x-3y\right)\)
b: \(5x^2+6xy+y^2\)
\(=5x^2+5xy+xy+y^2\)
=5x(x+y)+y(x+y)
=(x+y)(5x+y)
c: \(x^2+2xy-15y^2\)
\(=x^2+5xy-3xy-15y^2\)
=x(x+5y)-3y(x+5y)
=(x+5y)(x-3y)
d: \(x^2-7xy+10y^2\)
\(=x^2-2xy-5xy+10y^2\)
=x(x-2y)-5y(x-2y)
=(x-2y)(x-5y)
a) \(x^2+4xy-21y^2\)
\(=x^2+7xy-3xy-21y^2\)
\(=x\left(x+7y\right)-3y\left(x+7y\right)\)
\(=\left(x+7y\right)\left(x-3y\right)\)
b) \(5x^2+6xy+y^2\)
\(=5x^2+5xy+xy+y^2\)
\(=5x\left(x+y\right)+y\left(x+y\right)\)
\(=\left(5x+y\right)\left(x+y\right)\)
c) \(x^2+2xy-15y^2\)
\(=x^2+5xy-3xy-15y^2\)
\(=x\left(x+5y\right)-3y\left(x+5y\right)\)
\(=\left(x+5y\right)\left(x-3y\right)\)
d) \(x^2-7xy+10y^2\)
\(=x^2-2xy-5xy+10y^2\)
\(=x\left(x-2y\right)-5y\left(x-2y\right)\)
\(=\left(x-5y\right)\left(x-2y\right)\)
a) x² + 4xy - 21y²
= x² - 3xy + 7xy - 21y²
= (x² - 3xy) + (7xy - 21y²)
= x(x - 3y) + 7y(x - 3y)
= (x - 3y)(x + 7y)
b) 5x² + 6xy + y²
= 5x² + 5xy + xy + y²
= (5x² + 5xy) + (xy + y²)
= 5x(x + y) + y(x + y)
= (x + y)(5x + y)
c) x² + 2xy - 15y²
= x² + 2xy + y² - 16y²
= (x² + 2xy + y²) - 16y²
= (x + y)² - (4y)²
= (x + y - 4y)(x + y + 4y)
= (x - 3y)(x + 5y)
d) x² - 7xy + 10y²
= x² - 2xy - 5xy + 10y²
= (x² - 2xy) - (5xy + 10y²)
= x(x - 2y) - 5y(x - 2y)
= (x - 2y)(x - 5y)
Phân tích đa thức thành nhân tử bằng cách cách phối hợp nhiều phương pháp
x2-7xy+10y2
5x2+6xy+y2
x2-5x-14
X2 + 4xy - 21y2 5x2 + 6xy + y2
( x-y)2 +4 ( x-y) - 12 X2 - 7xy + 10y2
X2 + 2xy - 15y2
(Tách 1 hạng tử thành nhiều hạng tử)
Phân tích các đa tức sao thành nhân tử
M.n giải giùm mik nhé mik cần gấp ^^
X2+4xy-21y2=(x2+4xy+4y2)-25y2=(x+2)2-(5y)2=(x+2-5y)(x+2+5y)
5x2+6xy+y2=9x2+6xy+y2-4x2=(3x+y)2-4x2=(3x+y+2x)(3x+y-2x)
(x-y)2+4(x-y)-12=(x-y+2)2-16=(x-y+2+4)(x-y+2-4)
x2-7xy+10y2=x2-7xy+\(\frac{49y^2}{4}-\frac{9y^2}{4}\)= \(\left(x-\frac{7}{2}\right)^2-\left(\frac{3y}{2}\right)^2\)=\(\left(x-\frac{7}{2}-\frac{3y}{2}\right)\left(x-\frac{7}{2}+\frac{3y}{2}\right)\)
x2+2xy-15y2=(x+y)2-16y2=(x+y-4y)(x+y+4y
Chứng minh rằng vs mọi x và y ta có:
5x^2+10y^2-6xy-4x-2y+3>0
CÁC BN GIÚP MIK VS NHA!!! CẢM MƠN NHÌU NHÌU!!!^,^!!!
\(5x^2+10y^2-6xy-4x-2y+3\)
\(=\left(x^2-6xy+9y^2\right)+\left(4x^2-4x+1\right)+\left(y^2-2y+1\right)+1\)
\(=\left(x-3y\right)^2+\left(2x-1\right)^2+\left(y-1\right)^2+1>0\) (đpcm)
p/s: chúc bạn học tốt
Hi hi mik cảm ơn bn và mik cx chúc bn học tốt nha!!!^-^!!!
a) (x - 3)(2x ^ 2 - 3x + 4)
b) (4x ^ 2 * y - 5x * y ^ 2 + 6xy) 2xy:
c ) x/(2x + 4) - 2/(x ^ 3 + 2x)
Ai giúp mik vs ạ🥲
a: \(\left(x-3\right)\left(2x^2-3x+4\right)\)
\(=2x^3-3x^2+4x-6x^2+9x-12\)
\(=2x^3-9x^2+13x-12\)
b: \(\left(4x^2y-5xy^2+6xy\right):2xy\)
\(=\dfrac{4x^2y-5xy^2+6xy}{2xy}\)
\(=\dfrac{2xy\cdot2x-2xy\cdot2,5y+2xy\cdot3}{2xy}\)
\(=2x-2,5y+3\)
c: \(\dfrac{x}{2x+4}-\dfrac{2}{x^3+2x}\)
\(=\dfrac{x\left(x^3+2x\right)-2\left(2x+4\right)}{x\left(x^2+2\right)\cdot2\cdot\left(x+2\right)}\)
\(=\dfrac{x^4+2x^2-4x-8}{2x\left(x^2+2\right)\left(x+2\right)}\)
CMR:
a,\(x^2+5y^2+2x-4xy-10y+10>0\forall x,y\)
b,\(5x^2+10y^2-6xy-4x-2y+3>0\forall x,y\)
tính gtnn: d= x^2 +xy +y^2 +1
e= 5x^2 + 10y^2 - 6xy - 4x - 2y +3
g= (2x-1)^2 + (x+2)^2