x3+10x2+25x-xy2
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\(x^3+2x^2y+xy^2-25x\)
\(=x\left(x^2+2xy+y^2-25\right)\)
\(=x\left[\left(x+y\right)^2-5^2\right]\)
\(=x\left(x+y-5\right)\left(x+y+5\right)\)
a) Ta có: \(x^2-3x+xy-3y\)
\(=x\left(x-3\right)+y\left(x-3\right)\)
\(=\left(x-3\right)\left(x+y\right)\)
b) Ta có: \(x^3+10x^2+25x-xy^2\)
\(=x\left(x^2+10x+25-y^2\right)\)
\(=x\left(x+5-y\right)\left(x+5+y\right)\)
c) Ta có: \(x^3+2+3\left(x^3-2\right)\)
\(=4x^3-4\)
\(=4\left(x-1\right)\left(x^2+x+1\right)\)
`a) 8x^2 - 8xy - 4x + 4y`
`= 8x ( x - y ) - 4 ( x - y )`
`= ( x - y ) ( 8x - 4 )`
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`b) x^3 + 10x^2 + 25x - xy^2`
`=x ( x^2 + 10x + 25 ) - xy^2`
`= x ( x + 5 )^2 - xy^2`
`= x [ ( x + 5 )^2 - y^2 ]`
`= x ( x + 5 - y ) ( x + 5 + y )`
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`c) x^2 + x - 6`
`= x^2 + 3x - 2x - 6`
`= x ( x + 3 ) - 2 ( x + 3 )`
`= ( x + 3 ) ( x - 2 )`
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`d) 2x^2 + 4x - 16`
`= 2x^2 - 4x + 8x - 16`
`= 2x ( x - 2 ) + 8 ( x - 2 )`
`= ( x - 2 ) ( 2x + 8 )`
a) x2 + xy –x – y = x(x + y) – (x + y) = (x + y)(x -1 ).
b) a2 – b2 + 8a + 16 = (a2 + 8a + 16) – b2 = (a + 4)2 – b2
= (a + 4 – b)(a + 4 + b).
tui chỉ làm dc này thui
Bài 1:
\(a,6x^2-15x^3y\\ b,=-\dfrac{2}{3}x^2y^3+\dfrac{2}{3}x^4y-\dfrac{8}{3}xy\)
Bài 2:
\(a,=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\\ b,=3x^2-6x-5x+5x^2-8x^2+24=24-11x\\ c,=x^5+x^3-2x^3-2x=x^5-x^3-2x\)
P + Q = (x2y + x3 – xy2 + 3) + (x3 + xy2 – xy – 6)
= x2y + x3 – xy2 + 3 + x3 + xy2 – xy – 6
= (x3 + x3) + x2y + (xy2 – xy2) – xy + (3 – 6)
= 2x3 + x2y – xy – 3
Vậy P + Q = 2x3 + x2y – xy – 3.
1: =(2x+y-2y)(2x+y+2y)
=(2x-y)(2x+3y)
2: =(4-5x)(16+20x+25x^2)
3: =x(x^2-2xy+y^2-4)
=x[(x-y)^2-4]
=x(x-y-2)(x-y+2)
4: =(x-y)(x^2+xy+y^2)+xy(x-y)
=(x-y)(x^2+2xy+y^2)
=(x-y)(x+y)^2
1: =(2x+y-2y)(2x+y+2y)
=(2x-y)(2x+3y)
2: =(4-5x)(16+20x+25x^2)
3: =x(x^2-2xy+y^2-4)
=x[(x-y)^2-4]
=x(x-y-2)(x-y+2)
4: =(x-y)(x^2+xy+y^2)+xy(x-y)
=(x-y)(x^2+2xy+y^2)
=(x-y)(x+y)^2
Ta có:
• P + Q = (x2y + x3 – xy2 + 3) + (x3 + xy2 – xy – 6)
= x2y + x3 – xy2 + 3 + x3 + xy2 – xy – 6
= x2y + (x3 + x3) + (xy2 – xy2) – xy + (3 – 6)
= x2y + 2x3 – xy – 3.
• P – Q = (x2y + x3 – xy2 + 3) – (x3 + xy2 – xy – 6)
= x2y + x3 – xy2 + 3 – x3 – xy2 + xy + 6
= x2y + (x3 – x3) – (xy2 + xy2) + xy + (6 + 3)
= x2y – 2xy2 + xy + 9.
Vậy P + Q = x2y + 2x3 – xy – 3; P – Q = x2y – 2xy2 + xy + 9.
\(\text{ P + Q = (x^2y + x^3 – xy^2 + 3) + (x^3 + xy^2 – xy – 6)}\)
\(\text{= x^2y + x^3 – xy^2 + 3 + x^3 + xy^2 – xy – 6}\)
\(\text{= x^2y + (x^3 + x^3) + (xy^2 – xy^2) – xy + (3 – 6)}\)
\(\text{= x^2y + 2x^3 – xy – 3}\)
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\(\text{P – Q = (x^2y + x^3 – xy^2 + 3) – (x^3 + xy^2 – xy – 6)}\)
\(\text{= x^2y + x^3 – xy^2 + 3 – x^3 – xy^2 + xy + 6}\)
\(\text{= x^2y + (x^3 – x^3) – (xy^2 + xy^2) + xy + (6 + 3)}\)
\(\text{= x^2y – 2xy^2 + xy + 9}\)
\(x^3+10x^2+25x-xy^2\)
\(=x\left(x^2+10x+25-y^2\right)\)
\(=x\left[\left(x+5\right)^2-y^2\right]\)
\(=x\left(x+y+5\right)\left(x+y-5\right)\)