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a, 5x2 + 10xy + 5y2 = 5 ( x2 + 2xy + y2 ) = 5 ( x + y )2
b, x2 + 3x - y2 + 3y = ( x2 - y2 ) + ( 3x + 3y ) = ( x - y )( x + y ) + 3 ( x + y ) = ( x + y )( x - y + 3 )
c, x2 + 5x - y2 + 5y = ( x2 - y2 ) + ( 5x + 5y ) = ( x - y )( x + y ) + 5 ( x + y ) = ( x + y )( x - y + 5 )
d, 3x2 - 3y2 - 2 ( x - y )2 = 3 ( x2 - y2 ) - 2 ( x - y )2 = 3 ( x - y )( x + y ) - 2 ( x - y )2 = ( x - y )[ 3 ( x + y ) - 2 ] = ( x - y )( 3x + 3y - 2 )
e, x2 - 2x - 4y2 - 4y = ( x2 - 4y2 ) - ( 2x + 4y ) = ( x - 2y )( x + 2y ) - 2 ( x + 2y ) = ( x + 2y )( x - 2y - 2 )
a) 5x2+10xy+5y2
=5(x2+2xy+y2)
=5(x+y)2
b) x2+3x-y2+3y
=(x2-y2)+(3x+3y)
=(x-y)(x+y)+3(x+y)
=(x+y)(x-y+3)
c) x2+5x-y2+5y
=(x2-y2)+(5x+5y)
=(x-y)(x+y)+5(x+y)
=(x+y)(x-y+5)
d) 3x2-3y2-2(x-y)2
=3(x2-y2)-2(x-y)2
=3(x-y)(x+y)-2(x-y)2
=(x-y)[3(x+y)-2(x-y)]
e) x2-2x-4y2-4y
=(x2-4y2)-(2x+4y)
=(x-2y)(x+2y)-2(x+2y)
=(x+2y)(x-2y-2)
#H
1, \(x^2\left(x-3\right)-4x+12=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)
2, \(2a\left(x+y\right)-x-y=2a\left(x+y\right)-\left(x+y\right)=\left(2a-1\right)\left(x+y\right)\)
3, \(2x-4+5x^2-10x=2\left(x-2\right)+5x\left(x-2\right)=\left(2+5x\right)\left(x-2\right)\)
4, sửa đề :
\(6x^2-12x-7x+14=6x\left(x-2\right)-7\left(x-2\right)=\left(6x-7\right)\left(x-2\right)\)
5, \(xy-y^2-3x+3y=y\left(x-y\right)-3\left(x-y\right)=\left(y-3\right)\left(x-y\right)\)
a) x2(x-3)-4x+12
=x2(x-3)-4(x-3)
=(x-3)(x2-4)
=(x-3)(x-2)(x+2)
b) 2a(x+y)-x-y
=2a(x+y)-(x+y)
=(x+y)(2a-1)
c) 2x-4+5x2-10x
=2(x-2)+5x(x-2)
=(x-2)(2+5x)
d) 5x2-12x-7x+14
=5x2-19x+14
e) xy-y2-3x+3y
=y(x-y)-3(x-y)
=(x-y)(y-3)
#H
Bài 1 :
a) \(3x\left(5x^2-2x-1\right)=3x\cdot5x^2+3x\left(-2x\right)+3x\left(-1\right)\)
\(=15x^3-6x^2-3x\)
b) \(\left(x^2-2xy+3\right)\left(-xy\right)\)
\(=x^2\left(-xy\right)-2xy\left(-xy\right)+3\left(-xy\right)\)
\(=-x^3y+2x^2y^2-3xy\)
c) \(\frac{1}{2}x^2y\left(2x^3-\frac{2}{5}xy-1\right)\)
\(=\frac{1}{2}x^2y\cdot2x^3+\frac{1}{2}x^2y\cdot\left(-\frac{2}{5}xy\right)+\frac{1}{2}x^2y\left(-1\right)\)
\(=x^5y-\frac{1}{5}x^3y^2-\frac{1}{2}x^2y\)
d) \(\frac{1}{2}xy\left(\frac{2}{3}x^2-\frac{3}{4}xy+\frac{4}{5}y^2\right)\)
\(=\frac{1}{2}xy\cdot\frac{2}{3}x^2+\frac{1}{2}xy\cdot\left(-\frac{3}{4}xy\right)+\frac{1}{2}xy\cdot\frac{4}{5}y^2\)
\(=\frac{1}{3}x^3y-\frac{3}{8}x^2y^2+\frac{2}{5}xy^3\)
e) \(\left(x^2y-xy+xy^2+y^3\right)\left(3xy^3\right)\)
= \(x^2y\cdot3xy^3-xy\cdot3xy^3+xy^2\cdot3xy^3+y^3\cdot3xy^3\)
\(=3x^3y^4-3x^2y^4+3x^2y^5+3xy^6\)
Bài 2 :
3(2x - 1) + 3(5 - x) = 6x - 3 + 15 - x = (6x - x) - 3 + 15 = 5x - 3 + 15
Thay x = -3/2 vào biểu thức trên ta có : \(5\cdot\left(-\frac{3}{2}\right)-3+15\)
\(=-\frac{15}{2}-3+15=\frac{9}{2}\)
b) 25x - 4(3x - 1) + 7(5 - 2x)
= 25x - 12x + 4 + 35 - 14x
= (25x - 12x - 14x) + 4 + 35 = -x + 4 + 35 = -x + 39
Thay \(x=2\)vào biểu thức trên ta có : -2 + 39 = 37
c) 4x - 2(10x + 1) + 8(x - 2)
= 4x - 20x - 2 + 8x - 16
= (4x - 20x + 8x) - 2 - 16 = -8x - 2 - 16 = -8x - 18
Thay x = 1/2 vào biểu thức trên ta có \(-8\cdot\frac{1}{2}-18=-4-18=-22\)
d) Tương tự
Bài 3:
a) \(2x\left(x-4\right)-x\left(2x+3\right)=4\)
=> 2x2 - 8x - 2x2 - 3x = 4
=> (2x2 - 2x2) + (-8x - 3x) = 4
=> -11x = 4
=> x = \(-\frac{4}{11}\)
b) x(5 - 2x) + 2x(x - 7) = 18
=> 5x - 2x2 + 2x2 - 14x = 18
=> 5x - 14x = 18
=> -9x = 18
=> x = -2
Còn 2 câu làm tương tự
\(1,x^2-y^2+4x-4y\)
\(\left(x-y\right)\left(x+y\right)+4\left(x-y\right)\)
\(\left(x-y\right)\left(x+y+4\right)\)
\(x^2+2x-4y^2-4y\)
\(\left(x-2y\right)\left(x+2y\right)+2\left(x-2y\right)\)
\(\left(x-2y\right)\left(x+2y+2\right)\)
\(3,3x^2-4y+4x-3y^2\)
\(3\left(x^2-y^2\right)-4\left(x-y\right)\)
\(3\left(x-y\right)\left(x+y\right)-4\left(x-y\right)\)
\(\left(x-y\right)\left(3x+3y-4\right)\)
\(x^4-6x^3+54x-81\)
\(x^4+3x^3-9x^3+27x^2-27x^2+81x-27x-81\)
\(\left(x^4+3x^3\right)-\left(9x^3+27x^2\right)+\left(27x^2+81x\right)-\left(27x+81\right)\)
\(x^3\left(x+3\right)-9x^2\left(x+3\right)+27x\left(x+3\right)-27\left(x+3\right)\)
\(\left(x+3\right)\left(x^3-9x^2+27x-27\right)\)
\(\left(x+3\right)\left(x-3\right)^3\)
\(A=\left(5x^5+5x^4\right):5x^2-\left(2x^4-8x^2-6x+12\right):\left(2x-4\right)\)
Phép chia thứ nhất:
\(\left(5x^5+5x^4\right):5x^2=x^3+x^2\)
Phép chia thứ hai:
2x^4 - 4x^3 - 2x^4 - 8x^2 - 6x + 12 - 4x^3 - 8x^2 4x^3 - 8x^2 - 6x + 12 - -6x + 12 -6x + 12 0 2x - 4 x^3 - 2x^2 - 3
Vậy A = ( x^3 + x^2 ) - ( x^3 + 2x^2 - 3 ) = -x^2 + 3
Với x = -2 thì: A = -(-2)^2 + 3 = -4 + 3 = -1
B) bạn làm tương tự nhé
a, \(5x^2-10xy+5y^2=5\left(x^2-2xy+y^2\right)=5.\left(x-y\right)^2\)
b, \(x^2-4x+4-y^2=\left(x^2-4x+4\right)-y^2=\left(x-2\right)^2-y^2\)
\(=\left(x-2-y\right)\left(x-2+y\right)\)
c, \(3x^2-2x-5=3x^2-5x+3x-5=x\left(3x-5\right)+3x-5\)
\(=\left(3x-5\right)\left(x+1\right)\)