\(a^3-b^3+3a^2+3ab+b^2\)

\(=\left(a-b\right)\left(a^2+ab+b^2\right)+3\left(a^2+ab+b^2\right)\)

\(=\left(a-b+3\right)\left(a^2+ab+b^2\right)\)

\(a^3+3a^2b+3ab^2+b^3-8c^3\)

\(=\left(a+b\right)^3-8c^3\)

\(=\left(a+b-2c\right)\left[\left(a+b\right)^2+2\left(a+b\right)c+4c^2\right]\)

\(=\left(a+b-2c\right)\left(a^2+b^2+2ab+2ac+2bc+4c^2\right)\)

thank bạn

a) Ta có: \(a^2-b^2-5a+5b\)

\(=\left(a-b\right)\left(a+b\right)-5\left(a-b\right)\)

\(=\left(a-b\right)\left(a+b-5\right)\)

b) Ta có: \(a^2-b^2-3ab^2-3a^2b\)

\(=\left(a-b\right)\left(a+b\right)-3ab\left(a+b\right)\)

\(=\left(a+b\right)\left(a-b-3ab\right)\)

c) Ta có: \(x^2-xy+2y-4\)

\(=\left(x^2-4\right)-\left(xy-2y\right)\)

\(=\left(x-2\right)\left(x+2\right)-y\left(x-2\right)\)

\(=\left(x-2\right)\left(x+2-y\right)\)

d) Ta có: \(4a^2-10ax+15x-9\)

\(=\left(4a^2-9\right)-\left(10ax-15x\right)\)

\(=\left(2a-3\right)\left(2a+3\right)-5x\left(2a-3\right)\)

\(=\left(2a-3\right)\left(2a+3-5x\right)\)

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

\(\left(a+b\right)^3-3ab.\left(a+b\right)=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]=\left(a+b\right)\left(a^2+b^2-ab\right)\)

`(a+b)^3-3ab(a+b)`

`=(a+b)(a+b)^2-3ab(a+b)`

`=(a+b)[(a+b)^2-3ab]`

`=(a+b)(a^2+2ab+b^2-3ab)`

`=(a+b)(a^2-ab+b^2)`

\(\left(a+b\right)^3-3ab\left(a+b\right)=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

(2 - 3a)² - (3 - a)²

= (2 - 3a - 3 + a)(2 - 3a + 3 - a)

= (-2a - 1)(1 - 4a)

2, a^3-3ab^2 = 5

<=> (a^3-3ab^2)^2 = 25

<=> a^6-6a^4b^2+9a^2b^4 = 25

b^3-3a^2b=10

<=> (b^3-3a^2b)^2 = 100

<=> b^6-6a^2b^4+9a^4b^2 = 100

=> 100+25 = a^6-6a^4b^2+9a^2b^4+b^6+6a^2b^4+9a^4b^2

<=> 125 = a^6+3a^4b^2+3a^3b^4+b^6 = (a^2+b^2)^3

<=> a^2+b^2 = 5

Khi đó : S = 2016.(a^2+b^2) = 2016.5 = 10080

Tk mk nha

1) \(x^2+6xy+5y^2-5y-x=\left(x^2+xy-x\right)+\left(5xy+5y^2-5y\right)\)

\(=x\left(x+y-1\right)+5y\left(x+y-1\right)\)

\(=\left(x+5y\right)\left(x+y-1\right)\)

2) Ta có : \(a^3-3ab^2-5\Rightarrow\left(a^3-3ab^2\right)^2=25\Rightarrow a^6-6a^4b^2+9a^2b^4=25\)

và \(b^3-3a^2b=10\Rightarrow\left(b^3-3a^2b\right)^2=100\Rightarrow b^6-6b^4a^2+9a^4b^2=100\)

\(\Rightarrow\)\(125=a^6+b^6+3a^2b^4+3a^4b^2\)

Hay \(125=\left(a^2+b^2\right)^2\Rightarrow a^2+b^2=5\)

Nên \(S=2016\left(a^2+b^2\right)=2016.5=10080\)

1,\(x^2+6xy+5y^2-5y-x=x^2+xy-x+5y^2-5y+5xy\)

\(=x\left(x+y-1\right)+5y\left(x+y-1\right)\)

\(=\left(x+y-1\right)\left(x+5y\right)\)

2,

\(a^3-3ab^2=5\Leftrightarrow\left(a^3-3ab^2\right)^2=25\Leftrightarrow a^6-6a^4b^2+9a^2b^4=25\)

\(b^3-3a^2b=10\Leftrightarrow\left(b^3-3a^2b\right)^2=100\Leftrightarrow b^6-6a^2b^4+9a^4b^2=100\)

\(\Rightarrow a^6-6a^4b^2+9a^2b^4+b^6-6a^2b^4+9a^4b^2=125\)

\(\Leftrightarrow a^6+3a^4b^2+3a^2b^4+b^6=125\)

\(\Leftrightarrow\left(a^2+b^2\right)^3=125\)

\(\Rightarrow a^2+b^2=5\Rightarrow2016a^2+2016b^2=2016.5=10080\)

Ta có:A= 3a+3b-a^2-ab

=>A= (3a-a^2)+(3b-ab)

=>A= a(3-a)+b(3-a)

=>A= (a+b)(3-a)

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