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a Đề sai: )
b
\(a^3-a^2x-ay+xy\\ =a^2\left(a-x\right)-y\left(a-x\right)\\ =\left(a-x\right)\left(a^2-y\right)\)
c
\(4x^2-y^2+4x+1\\ =\left(2x\right)^2+2.2x.1+1-y^2\\ =\left(2x+1\right)^2-y^2\\ =\left(2x+1-y\right)\left(2x+1+y\right)\)
d
\(x^4+2x^3+x^2\\ =x^4+x^3+x^3+x^2\\ =x^3\left(x+1\right)+x^2\left(x+1\right)\\ =\left(x^3+x^2\right)\left(x+1\right)\)
e
\(5x^2-10xy+5y^2-5z^2\\ =5\left(x^2-2xy+y^2-z^2\right)\\ =5\left[\left(x-y\right)^2-z^2\right]\\ =5\left(x-y-z\right)\left(x-y+z\right)\)
c: =(2x+1)^2-y^2
=(2x+1+y)(2x+1-y)
d: =x^2(x^2+2x+1)
=x^2(x+1)^2
e: =5(x^2-2xy+y^2-z^2)
=5[(x-y)^2-z^2]
=5(x-y-z)(x-y+z)
a) \(=\left(2x-1\right)^2\)
b) \(=x\left(y^2-x^2+2x-1\right)=x\left[y^2-\left(x-1\right)^2\right]=x\left(y-x+1\right)\left(y+x-1\right)\)
a,x^2-x-y^2-y
=x^2-y^2-(x+y)
=(x-y).(x+y)-(x+y)
=(x+y).(x-y-1)
b, x^2-2xy+y^2-z^2
=(x^2-2xy+y^2)-z^2
=(x-y)^2-z^2
=(x-y-z)(x-y+z)
c,5x-5y+ax-ay( đề bài ở đây phải là -ay ms tính đc)
=(5x-5y)+(ax-ay)
=5(x-y)+a(x-y)
=(x-y).(5+a)
d,a^3-a^2.x-ay+xy
=(a^3-a^2x)-(ay-xy)
=a^2(a-x)-y(a-x)
=(a-x)(a^2-y)
e,4x^2-y^2+4x+1
={(2x)^2+4x+1}-y^2
=(2x+1)^2-y^2
=(2x+1+y^2)(2x+1-y^2)
f,x^3-x+y^3-y
=(x^3+y^3)-(x+y)
=(x+y)(x^2-xy+y^2)-(x+y)
=(x+y)(x^2-xy+y^2-1)
\(1)4x^2-25+\left(2x+7\right).\left(5.2x\right)\)
\(=\left(2x\right)^2-5^2-\left(2x+7\right).\left(2x-5\right)\)
\(=\left(2x.5\right)\left(2x+5\right).\left(2x+7\right)\left(2x-5\right)\)
\(=\left(2x-5\right)\left(2x+5-2x+7\right)\)
\(=\left(2x-5\right).12\)
\(2)3x+4-x^2-4x\)
\(=3(x+4)-\left(x+4\right)\)
\(=\left(3-x\right)\left(x+4\right)\)
\(3)5x^2-2y^2-10x+10y\)
\(=5\left(x^2-y^2\right)-10\left(x-4\right)\)
\(=5\left(x-y\right)\left(x+y\right)-10\left(x-y\right)\)
\(=\left(x-y\right)[5(x+y)-10]\)
Còn lại bn lm nốt nha!
\(a,x^3+x^2-4x-4=x^2\left(x+1\right)-4\left(x+1\right)=\left(x^2-4\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\left(x+1\right)\\ b,x^2-2x-15=\left(x^2-5x\right)+\left(3x-15\right)=x\left(x-5\right)+3\left(x-5\right)=\left(x+3\right)\left(x-5\right)\\ c,x^2-4+\left(x-2\right)^2=\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2=\left(x-2\right)\left(x+2+x-2\right)=2x\left(x-2\right)\)
\(d,x^3-2x^2+x-xy^2=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\)
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: \(x^3-2x+4\)
\(=x^3+2x^2-2x^2-4x+2x+4\)
\(=\left(x+2\right)\left(x^2-2x+2\right)\)
b: \(x^3-4x^2+12x-27\)
\(=\left(x-3\right)\left(x^2+3x+9\right)-4x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
c: \(x^3+2x^2+2x+1\)
\(=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+x+1\right)\)
1) x2 - x - y2 - y = (x - y)(x + y) - (x + y) = (x - y - 1)(x + y)
2. x2 - 2xy + y2 - z2 = (x - y)2 - z2 = (x - y - z)(x - y + z)
3. 5x - 5y + ax - ay = 5(x - y) + a(x - y) = (a + 5)(x - y)
4. a3 - a2x - ay + xy = a2(a - x) - y(a - x) = (a2 - y)(a - x)
5. 4x2 - y2 + 4x + 1 = (2x + 1)2 - y2 = (2x + 1 - y)(2x + y + 1)
6. x3 - x + y3 - y = (x + y)(x2 - xy + y2) - (x + y) = (x + y)(x2 - xy + y2 - 1)
Trả lời:
1, x2 - x - y2 - y
= ( x2 - y2 ) - ( x + y )
= ( x - y ) ( x + y ) - ( x + y )
= ( x + y ) ( x - y - 1 )
2, x2 - 2xy + y2 - z2
= ( x2 - 2xy + y2 ) - z2
= ( x - y )2 - x2
= ( x - y - z ) ( x - y + z )
3, 5x - 5y + ax - ay
= ( 5x + ax ) - ( 5y + ay )
= x ( 5 + a ) - y ( 5 + a )
= ( 5 + a ) ( x - y )
= ( 5 + a ) ( x - y )
4, a3 - a2x - ay + xy
= ( a3 - a2x ) - ( ay - xy )
= a2 ( a - x ) - y ( a - x )
= ( a - x ) ( a2 - y )
5, 4x2 - y2 + 4x + 1
= ( 4x2 + 4x + 1 ) - y2
= ( 2x + 1 )2 - y2
= ( 2x + 1 - y ) ( 2x + 1 + y )
6, x3 - x + y3 - y
= ( x3 + y3 ) - ( x + y )
= ( x + y ) ( x2 - xy + y ) - ( x + y )
= ( x + y ) ( x2 - xy + y - 1 )