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do hơi bận nên mk ghi đáp án nha, ko hiểu đâu ib mk
a) \(3xy^2-2xy+12x=x\left(3y^2-2y+12\right)\)
b) \(x^3-10x^2+25x-16xy^2=x\left(x-4y-5\right)\left(x+4y-5\right)\)
c) \(5y^3-10xy^2+5x^2y-20y=5y\left(y-x-2\right)\left(y-x+2\right)\)
d) \(x^2+2xy+y^2-xz-yz=\left(x+y\right)\left(x+y-z\right)\)
e) \(9x^2+y^2+6xy=\left(3x+y\right)^2\)
f) \(8-12x+6x^2-x^3=\left(2-x\right)^3\)
g) \(125x^3-75x^2+15x-1=\left(5x-1\right)^3\)
h) \(x^2-xz-9y^2+3yz=\left(x-3y\right)\left(x+3y-z\right)\)
a) = (x + 3)2 - y2 = (x + 3 - y)(x + 3 + y)
b) = x2(x - 3) -4(x - 3) = (x - 3)(x2 - 4) = (x - 3)(x - 2)(x + 2)
c) = 3x(x - y) - 5(x - y) = (x - y)(3x - y)
d) Nhầm đề. tui sửa lại x3 + y3 + 2x2 - 2xy + 2y2
= x3 + y3 + 2(x2 - xy + y2) = (x + y)(x2 - xy + y2) + 2(x2 - xy + y2) = (x2 - xy + y2)(x + y + 2)
e) = x4 - x3 - x3 + x2 - x2 + x + x - 1 = x3(x - 1) - x2(x - 1) - x(x - 1) + x - 1 = (x - 1)(x3 - x2 - x + 1) = (x - 1)(x - 1)(x2 - 1) = (x - 1)3(x + 1)
f) = x3 - 3x2 - x2 + 3x + 9x - 27 = x2(x - 3) - x(x - 3) + 9(x - 3) = (x-3)(x2 - x + 9)
g) chắc là 3xyz
= x2y + xy2 + y2z + yz2 + x2z + xz2 + 3xyz = x2y + xy2 + xyz + y2z + yz2 + xyz + x2z + xz2 + xyz = (x + y + z)(xy + yz + xz)
h) = 23 -(3x)3 = (2 - 3x)(4 + 6x + 9x2)
i) = (x + y - x + y)(x + y + x - y) = 2y*2x = 4xy
k) = (x3 - y3)(x3 + y3) = (x - y)(x2 + xy +y2)(x + y)(x2 - xy +y2).
a) x2 - 7x + 5 = ( x2 - 2 . 7/2 . x + 49 / 4 ) + 5 - 49 / 4
= (x - 7/2)^2 - 29/4
= (x - 7/2)^2 - (√ 29 / 2 )^2
= ( x - ( 7 + √ 29 / 2 )). ( x + ( 7 - √ 29 / 2 ))
a) \(x^2-xz-9y^2+3yz\)
\(=\left(x^2-9y^2\right)-\left(xz-3yz\right)\)
\(=\left[x^2-\left(3y\right)^2\right]-z\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y-z\right)\)
b) \(x^3-x^2-5x+125\)
\(=\left(x^3+125\right)-\left(x^2+5x\right)\)
\(=\left(x^3+5^3\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+5^2\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+5^2-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
c) \(x^3+2x^2-6x-27\)
\(=\left(x^3-27\right)-\left(2x^2-6x\right)\)
\(=\left(x^3-3^3\right)-2x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+3x+3^2\right)-2x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+3x+3^2-2x\right)\)
\(=\left(x-3\right)\left(x^2+x+9\right)\)
e) \(4x^4+4x^3-x^2-x\)
\(=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=\left(x+1\right)\left(4x^3-x\right)\)
f) \(x^6-x^4-9x^3+9x^2\)
\(=x^4\left(x^2-1\right)-9x^2\left(x-1\right)\)
\(=x^4\left(x-1\right)\left(x+1\right)-9x^2\left(x-1\right)\)
\(=\left(x-1\right)\left[x^4\left(x+1\right)-9x^2\right]\)
\(=\left(x-1\right)\left(x^5+x^4-9x^2\right)\)
5x^2+10xy+5y^2
=5.(x2+2xy+y2)
=5.(x+y)2
x^3-6x^2+9x
=x.(x2-6x+9)
=x.(x-3)2
xy+y^2-x-y
=y.(x+y)-(x+y)
=(x+y)(y-1)
a, -x - y2 + x2 - y = (x2 - y2) - (x + y)
= (x - y)(x + y) - (x + y)
= (x + y)(x - y - 1)
b, x( x + y ) - 5x - 5y = x(x + y) - 5(x + y)
= (x - 5)(x + y)
c, x2 - 5x + 5y - y2 = (x - y)(x + y) - 5(x - y)
= (x - y)(x + y - 5)
d, 5x3 - 5x2y - 10x2 + 10xy = 5x2(x - y) - 10x(x - y)
= 5x(x - y)(x - 2)
e, 27x3 - 8y3 = (3x - 2y)(9x2 + 6xy + 4y2)
f, x2 - y2 - x - y = (x - y)(x + y) - (x + y)
= (x + y)(x - y - 1)
g, x2 - y2 - 2xy + y2 = (x2 - 2xy + y2) - y2
= (x - y)2 - y2
= (x - y - y)(x - y + y) = x(x - 2y)
h, x2 - y2 + 4 - 4x = (x2 - 4x + 4) - y2
= (x - 2)2 - y2
= (x - y - 2)(x + y - 2)
i, x3 + 3x2 + 3x + 1 - 27z3 = (x + 1)3 - 27z3
= (x+1-3z)(x2+2x+1+3xz+3z+9z2)
k, 4x2 + 4x - 9y2 + 1 = (2x + 1)2 - 9y2
= (2x - 3y + 1)(2x + 3y + 1)
m, x2 - 3x + xy - 3y = x(x - 3) + y(x - 3)
= (x - 3)(x + y)
a) Ta có: \(5y^3-10xy^2+5yx^2-20y\)
\(=5y\left(y^2-2xy+x^2-4y\right)\)
b) Ta có: \(x^2+2xy+y^2-xz-yz\)
\(=\left(x+y\right)^2-z\cdot\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-z\right)\)
c) Ta có: \(9x^2+y^2+6xy\)
\(=\left(3x\right)^2+2\cdot3x\cdot y+y^2\)
\(=\left(3x+y\right)^2\)
d) Ta có: \(8-12x+6x^2-x^3\)
\(=2^3-3\cdot2^2\cdot x+3\cdot2\cdot x^2-x^3\)
\(=\left(2-x\right)^3\)
e) Ta có: \(125x^3-75x^2+15x-1\)
\(=\left(5x\right)^3-3\cdot\left(5x\right)^2\cdot1+3\cdot5x\cdot1^2-1^3\)
\(=\left(5x-1\right)^3\)