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a) ( 5x - y )( 25x2 + 5xy + y2 ) = ( 5x )3 - y3 = 125x3 - y3
b) ( x - 3 )( x2 + 3x + 9 ) - ( 54 + x3 ) = x3 - 33 - 54 - x3 = -27 - 54 = -81
c) ( 2x + y )( 4x2 - 2xy + y2 ) - ( 2x - y )( 4x2 + 2xy + y2 ) = ( 2x )3 + y3 - [ ( 2x )3 - y3 ]= 8x3 + y3 - 8x3 + y3 = 2y3
d) ( x + y )2 + ( x - y )2 + ( x + y )( x - y ) - 3x2 = x2 + 2xy + y2 + x2 - 2xy + y2 + x2 - y2 - 3x2 = y2
e) ( x - 3 )3 - ( x - 3 )( x2 + 3x + 9 ) + 6( x + 1 )2
= x3 - 9x2 + 27x - 27 - ( x3 - 33 ) + 6( x2 + 2x + 1 )
= x3 - 9x2 + 27x - 27 - x3 + 27 + 6x2 + 12x + 6
= -3x2 + 39x + 6
= -3( x2 - 13x - 2 )
f) ( x + y )( x2 - xy + y2 ) + ( x - y )( x2 + xy + y2 ) - 2x3
= x3 + y3 + x3 - y3 - 2x3
= 0
g) x2 + 2x( y + 1 ) + y2 + 2y + 1
= x2 + 2x( y + 1 ) + ( y2 + 2y + 1 )
= x2 + 2x( y + 1 ) + ( y + 1 )2
= ( x + y + 1 )2
= [ ( x + y ) + 1 ]2
= ( x + y )2 + 2( x + y ) + 1
= x2 + 2xy + y2 + 2x + 2y + 1
Câu a : \(4x^3-5x^2+6x+9\)
\(=4x^3+3x^2-8x^2-6x+12x+9\)
\(=\left(4x^3+3x^2\right)-\left(8x^2+6x\right)+\left(12x+9\right)\)
\(=x^2\left(4x+3\right)-2x\left(4x+3\right)+3\left(4x+3\right)\)
\(=\left(4x+3\right)\left(x^2-2x+3\right)\)
Câu b : \(5x^3-12x^2+14x-4\)
\(=5x^3-10x^2-2x^2+10x+4x-4\)
\(=\left(5x^3-2x^2\right)-\left(10x^2-4x\right)+\left(10x-4\right)\)
\(=x^2\left(5x-2\right)-2x\left(5x-2\right)+2\left(5x-2\right)\)
\(=\left(5x-2\right)\left(x^2-2x+2\right)\)
Câu c : \(x^3-5x^2+2x+8\)
\(=x^3+x^2-6x^2-6x+8x+8\)
\(=\left(x^3+x^2\right)-\left(6x^2+6x\right)+\left(8x+8\right)\)
\(=x^2\left(x+1\right)-6x\left(x+1\right)+8\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-6x+8\right)\)
\(=\left(x+1\right)\left[x^2-2x-4x+8\right]\)
\(=\left(x+1\right)\left[x\left(x-2\right)-4\left(x-2\right)\right]\)
\(=\left(x+1\right)\left(x-2\right)\left(x-4\right)\)
Câu d : \(4x^3+5x^2+10x-12\)
\(=4x^3+8x^2-3x^2+16x-6x-12\)
\(=\left(4x^3-3x^2\right)+\left(8x^2-6x\right)+\left(16x-12\right)\)
\(=x^2\left(4x-3\right)+2x\left(4x-3\right)+4\left(4x-3\right)\)
\(=\left(4x-3\right)\left(x^2+2x+4\right)\)
`#3107`
`a.`
`4x^2 - 6x = 2x(2x - 3)`
`b.`
`9x^4y^3 + 3x^2y^4 = 3x^2y^2(3x^2y + y^2)`
`c.`
`x^3 - 2x^2 + 5x`
`= x(x^2 - 2x + 5)`
a) 4x² - 6x
= 2x(2x - 3)
b) 9x⁴y³ + 3x²y⁴
= 3x²y³(3x² + 3y)
c) x³ - 2x² + 5x
= x(x² - 2x + 5)
\(A=\left(5x^5+5x^4\right):5x^2-\left(2x^4-8x^2-6x+12\right):\left(2x-4\right)\)
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
Bài 3 :
a, \(x^3-4x=x\left(x^2-4\right)=x\left(x-2\right)\left(x+2\right)\)
b, \(x^2+2x-y^2+1=\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)
c, \(x^2+y^2-z^2+2xy=\left(x+y\right)^2-z^2=\left(x+y-z\right)\left(x+y+z\right)\)
d, \(x^2-7x+12=x^2-3x-4x+12=\left(x-4\right)\left(x-3\right)\)
e, \(x^2-4x+xy-4y=x\left(x-4\right)+y\left(x-4\right)=\left(x+y\right)\left(x-4\right)\)
g, \(5x^2-10xy+5y^2-20z^2=5\left(x^2-2xy+y^2-4z^2\right)=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
f, \(4x^2-4xy+y^2-9z^2=\left(2x+y\right)^2-\left(3z\right)^2=\left(2x+y-3z\right)\left(2x+y+3z\right)\)
n, \(\left(x+y\right)^3-\left(z-t\right)^3=\left(x+y-z+t\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(z-t\right)+\left(z-t\right)^2\right]\)
Làm nốt nhé, ko phải đi học thì t giải hết cho cậu r :))
Bài 2 :
a, \(x^2-4x=0\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow x=0;4\)
b, \(5x\left(x-2020\right)-x+2020=0\)
\(\Leftrightarrow5x\left(x-2020\right)-\left(x-2020\right)=0\Leftrightarrow\left(5x-1\right)\left(x-2020\right)=0\)
\(\Leftrightarrow x=\frac{1}{5};2020\)
c, \(\left(4x+5\right)^2-\left(2x-1\right)^2=0\)
\(\Leftrightarrow16x^2+40x+25-\left(4x^2-4x+1\right)=0\)
\(\Leftrightarrow12x^2+44x+24=0\Leftrightarrow4\left(x+3\right)\left(3x+2\right)=0\)
\(\Leftrightarrow x=-3;-\frac{2}{3}\)
b: =(x-2)(x+2)
\(a,3x^2y-6x=3x\left(xy-2\right)\)
\(b,x^2-4=x^2-2^2=\left(x-2\right)\left(x+2\right)\)
\(c,5x^2+5xy-x^2-2xy-y^2\)
\(=5x\left(x+y\right)-\left(x+y\right)\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-x+y\right)\)
\(=\left(x+y\right)\left(4x+y\right)\)