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a) \(3x^2-3xy-5x+5y\)
\(=\left(3x^2-3xy\right)-\left(5x-5y\right)\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
b) \(2x^3y-2xy^3-4xy^2-2xy\)
\(=2xy\left(x^2-y^2-2y-1\right)\)
\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=2xy\left[x^2-\left(y+1\right)^2\right]\)
\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)
c) \(x^2+1+2x-y^2\)
\(=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1+y\right)\left(x+1-y\right)\)
d) \(x^2+4x-2xy-4y+y^2\)
\(=\left(x^2-2xy+y^2\right)+\left(4x-4y\right)\)
\(=\left(x-y\right)^2+4\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y+4\right)\)
e) \(x^3-2x^2+x\)
\(=x\left(x^2-2x+1\right)\)
\(=x\left(x-1\right)^2\)
f) \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x^2+2x+1\right)+y^2\right]\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x-y+1\right)\left(x+y+1\right)\)
a: =3x(x-y)-5(x-y)
=(x-y)(3x-5)
b: \(=2xy\left(x^2-y^2-2y-1\right)\)
\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)
d:
Sửa đề: x^2+4x-2xy-4y+y^2
=x^2-2xy+y^2+4x-4y
=(x-y)^2+4(x-y)
=(x-y)(x-y+4)
e: =x(x^2-2x+1)
=x(x-1)^2
f: =2(x^2+2x+1-y^2)
=2[(x+1)^2-y^2]
=2(x+1+y)(x+1-y)
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Ta có:
D=2x2+3y2+4xy−8x−2y+18C=2x2+3y2+4xy−8x−2y+18
D=2(x2+2xy+y2)+y2−8x−2y+18C=2(x2+2xy+y2)+y2−8x−2y+18
D=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1C=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1
D=2(x+y−2)2+(y+3)2+1≥1C=2(x+y−2)2+(y+3)2+1≥1
Dấu "=" xảy ra ⇔x+y=2⇔x+y=2và y=−3y=−3
Hay x = 5 , y = -3
Đc chx bạn
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a: x^2-2x+y^2-8y+17=0
=>x^2-2x+1+y^2-8y+16=0
=>(x-1)^2+(y-4)^2=0
=>x=1 và y=4
b: Sửa đề: 4x^2-4xy+y^2+y^2+4y+4=0
=>(2x-y)^2+(y+2)^2=0
=>y=-2 và x=-1
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a) Xem lại đề
b) x³ - 4x²y + 4xy² - 9x
= x(x² - 4xy + 4y² - 9)
= x[(x² - 4xy + 4y² - 3²]
= x[(x - 2y)² - 3²]
= x(x - 2y - 3)(x - 2y + 3)
c) x³ - y³ + x - y
= (x³ - y³) + (x - y)
= (x - y)(x² + xy + y²) + (x - y)
= (x - y)(x² + xy + y² + 1)
d) 4x² - 4xy + 2x - y + y²
= (4x² - 4xy + y²) + (2x - y)
= (2x - y)² + (2x - y)
= (2x - y)(2x - y + 1)
e) 9x² - 3x + 2y - 4y²
= (9x² - 4y²) - (3x - 2y)
= (3x - 2y)(3x + 2y) - (3x - 2y)
= (3x - 2y)(3x + 2y - 1)
f) 3x² - 6xy + 3y² - 5x + 5y
= (3x² - 6xy + 3y²) - (5x - 5y)
= 3(x² - 2xy + y²) - 5(x - y)
= 3(x - y)² - 5(x - y)
= (x - y)[(3(x - y) - 5]
= (x - y)(3x - 3y - 5)
\(4x^2+4y^2-4xy=2x+2y-1\)
\(\Leftrightarrow2x^2-4xy+2y^2+2x^2-2x+2y^2-2y+1=0\)
\(\Leftrightarrow2\left(x-y\right)^2+2\left(x-\frac{1}{2}\right)^2+2\left(y-\frac{1}{2}\right)^2+\frac{1}{2}=0\)
Suy ra phương trình vô nghiệm.