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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)
\(a,x^2-y^2-2x+2y=\left(x^2-y^2\right)-\left(2x-2y\right)=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)=\left(x-y\right)\left(x+y-2\right).\) \(b,2x+2y-x^2-xy=2\left(x+y\right)-x\left(x+y\right)=\left(x+y\right)\left(2-x\right)\)
\(c,3a^2-6ab+3b^2-12c^2=3\left(a^2-2ab+b^2-4c^2\right)=3.\left(\left(a-b\right)^2-\left(2c\right)^2\right)\)
\(=3\left(a-b-2c\right).\left(a-b+2c\right)\)
\(d,x^2-25+y^2-2xy=\left(x^2-2xy+y^2\right)-5^2=\left(x-y\right)^2-5^2\)
\(=\left(x-y+5\right)\left(x-y-5\right)\)
\(e,a^2+2ab+b^2-ac-bc=\left(a+b\right)^2-c\left(a+b\right)=\left(a+b\right)\left(a+b-c\right)\)
\(f,x^2-2x-4y^2-4y=\left(x^2-4y^2\right)-\left(2x+4y\right)=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\(h,x^2\left(x-1\right)+16\left(1-x\right)=x^2\left(x-1\right)-16\left(x-1\right)=\left(x-1\right)\left(x^2-16\right)=\)
\(=\left(x-1\right)\left(x-4\right)\left(x+4\right)\)
a) \(x-xy+y-y^2=x\left(1-y\right)+y\left(1-y\right)=\left(x+y\right)\left(1-y\right)\)
b) \(x^2-2x-y^2+1=\left(x^2-2x+1\right)-y^2=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\)
c) \(4x^2-4xy+y^2=\left(2x\right)^2-2.2x.y+y^2=\left(2x-y\right)^2\)
d) \(9x^3-9x^2y-4x+4y=9x^2\left(x-y\right)-4\left(x-y\right)=\left(9x^2-4\right)\left(x-y\right)=\left(3x-2\right)\left(3x+2\right)\left(x-y\right)\)
e) \(x^3+2+3\left(x^3-2\right)=x^3+2+3x^3-6=4x^3-4=4\left(x^3-1\right)=4\left(x-1\right)\left(x^2+x+1\right)\)
\(3,\)Nhẩm nghiệm của đa thức trên ta đc : -1
Ta có lược đồ sau :
1 | 1 | -4 | -4 | |
-1 | 1 | 0 | -4 | 0 |
Phân tích thành nhân tử ta có :\(\left(x+1\right)\left(x^2-4\right)\)
\(a.10x\left(x-y\right)-6y\left(y-x\right)\\ =10x\left(x-y\right)+6y\left(x-y\right)\\ =\left(10x-6y\right)\left(x-y\right)\\ =2\left(5x-3y\right)\left(x-y\right)\)
\(b.14x^2y-21xy^2+28x^3y^2\\ =7xy\left(x-y+xy\right)\)
\(c.x^2-4+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2+x-2\right)\\ =2x\left(x-2\right)\)
\(d.\left(x+1\right)^2-25\\ =\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)
a: \(=y^2\left(2y-3\right)+\left(2y-3\right)=\left(2y-3\right)\left(y^2+1\right)\)
b: \(=2b\left(a+3\right)+c\left(a+3\right)=\left(a+3\right)\left(2b+c\right)\)
c: \(=25-\left(x-2y\right)^2\)
\(=\left(5-x+2y\right)\left(5+x-2y\right)\)
e: \(=\left(a+b\right)\left(a^2-ab+b^2\right)-\left(a+b\right)\)
\(=\left(a+b\right)\left(a^2-ab+b^2-1\right)\)