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Bài làm:
1) Ta có: \(2x^2+5xy+2y^2\)
\(=\left(2x^2+4xy\right)+\left(xy+2y^2\right)\)
\(=2x\left(x+2y\right)+y\left(x+2y\right)\)
\(=\left(2x+y\right)\left(x+2y\right)\)
2) Ta có: \(2x^2+2xy-4y^2\)
\(=\left(2x^2-2xy\right)+\left(4xy-4y^2\right)\)
\(=2x\left(x-y\right)+4y\left(x-y\right)\)
\(=2\left(x+2y\right)\left(x-y\right)\)
\(1)2x^2+5xy+2y^2=2x^2+4xy+xy+2y^2=\left(2x^2+4xy\right)+\left(xy+2y^2\right)=2x\left(x+2y\right)+y\left(x+2y\right)=\left(2x+y\right)\left(x+2y\right)\)\(2)2x^2+2xy-4y^2=2x^2+4xy-2xy-4y^2=\left(2x^2-2xy\right)+\left(4xy-4y^2\right)=2x\left(x-y\right)+4y\left(x-y\right)=\left(2x+4y\right)\left(x-y\right)\)
Bài giải:
a) x3 + 2x2y + xy2– 9x = x(x2 +2xy + y2 – 9)
= x[(x2 + 2xy + y2) – 9]
= x[(x + y)2 – 32]
= x(x + y – 3)(x + y + 3)
b) 2x – 2y – x2 + 2xy – y2 = (2x – 2y) – (x2 – 2xy + y2)
= 2(x – y) – (x – y)2
= (x – y)[2 – (x – y)]
= (x – y)(2 – x + y)
c) x4 – 2x2 = x2(x2 – (√2)2) = x2(x - √2)(x + √2).
a) x3 + 2x2y + xy2– 9x = x(x2 +2xy + y2 – 9)
= x[(x2 + 2xy + y2) – 9]
= x[(x + y)2 – 32]
= x(x + y – 3)(x + y + 3)
b) 2x – 2y – x2 + 2xy – y2 = (2x – 2y) – (x2 – 2xy + y2)
= 2(x – y) – (x – y)2
= (x – y)[2 – (x – y)]
= (x – y)(2 – x + y)
c) x4 – 2x2 = x2(x2 – (√2)2) = x2(x - √2)(x + √2).
a)\(x^2-y^2-2x+2y=\left(x^2-2x+1\right)-\left(y^2-2y+1\right)\)
\(=\left(x-1\right)^2-\left(y-1\right)^2=\left(x-1+y-1\right)\left(x-1-y+1\right)\)
\(=\left(x+y-2\right)\left(x-y\right)\)
b)\(3a^2-6ab+3b^2-12c^2=3\left(a^2-2ab+b^2-4c^2\right)\)\(=3\left[\left(a-b\right)^2-\left(2c\right)^2\right]\)
\(=3\left(a-b+2c\right)\left(a-b-2c\right)\)
Bài giải:
a) x3 – 2x2 + x = x(x2 – 2x + 1) = x(x – 1)2
b) 2x2 + 4x + 2 – 2y2 = 2[(x2 + 2x + 1) – y2]
= 2[(x + 1)2 – y2]
= 2(x + 1 – y)(x + 1 + y)
c) 2xy – x2 – y2 + 16 = 16 – (x2 – 2xy + y2) = 42 – (x – y)2
= (4 – x + y)(4 + x – y)
a) \(x^3 - 2x^2 + x\) \(= x(x^2 - 2x + 1)\)
\(= x (x - 1 )^2\)
b) \(2x^2 + 4x + 2 - 2y^2\) \(= 2(x^2 + 2x + 1 - y^2)\)
\(=2\left[\left(x^2+2x+1\right)-y^2\right]\)
\(=2\left[\left(x+1^2\right)-y^2\right]\)
\(= 2 (x+1-y) (x+1+y)\)
c) \(2xy - x^2 - y^2 + 16\) \(= - (x^2 - 2xy + y^2 - 4^2)\)
\(= - [(x^2 - 2xy + y^2) - 4^2]\)
\(= - [(x-y)^2 - 4^2 ]\)
\(= - (x - y - 4) (x- y + 4)\)
\(g,8x^3-27y^3=\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)\)
\(h,x^3+y^3+2x^2-2xy+2y^2\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+2\left(x^2-xy+y^2\right)\)
\(=\left(x^2-xy+y^2\right)\left(x+y+2\right)\)
Câu 2 nha
\(a,x^4+2x^3+x^2\)
\(=x^2\left(x^2+2x+1\right)\)
\(=x^2\left(x+1\right)^2\)
\(c,x^2-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
A = x^2 + y^2 + 2xy - 2x -2y +1
= (x+y)^2 -2.(x+y) + 1
=(x+y -1 )^2
\(x^4-2x^2y^2+y^4-1=0\Leftrightarrow\left(x^2-y^2\right)^2-1=0\Leftrightarrow\left(x^2-y^2-1\right).\left(x^2-y^2+1\right)=0\\ \)
\(x^2+2xy+2x+2y+y^2+1=0\Leftrightarrow\left(x+y+1\right)^2=0\)
A = (x^2 + 2xy + y^2) + 2.(x+y) + 1
=(x+y)^2 + 2.(x+y).1 + 1
=(x+y+1)^2
\(x^3+8y^3+2xy^2+x^2y\)
\(=x^3+2x^2y-x^2y-2xy^2+4xy^2+8y^3\)
\(=x^2\left(x+2y\right)-xy\left(x+2y\right)+4y^2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x^2-xy+4y^2\right)\)
2x – 2y – x2 + 2xy – y2
(Có x2 ; 2xy ; y2 ta liên tưởng đến HĐT (1) hoặc (2))
= (2x – 2y) – (x2 – 2xy + y2)
= 2(x – y) – (x – y)2
(Có x – y là nhân tử chung)
= (x – y)[2 – (x – y)]
= (x – y)(2 – x + y)