Sao chửi bn ấy

Bạn Thư kì ghê.

\(a,=x\left(\frac{3}{7}x+6+xy\right)\)

\(b,=\left(x+3y\right)\left(3x-6xy\right)=\left(x+3y\right).3x\left(1-2y\right)\)

\(c,=x\left(x+y\right).\left(-5\right)\left(x+y\right)=\left(x+y\right)\left[x.\left(-5\right)\right]\)

\(d,=3\left(x-y\right)+5x\left(x-y\right)=\left(x-y\right)\left(3+5x\right)\)

\(B3.a,x\left(1+6x\right)=0\)

\(Th1:x=0\)

\(Th2:1+6x=0=>x=-\frac{1}{6}\)

Vậy \(x\in\left\{0;-\frac{1}{6}\right\}\)

\(b,\left(x+3\right)\left(2-x\right)=0\)

\(Th1:x+3=0=>x=-3\)

\(Th2:2-x=0=>x=2\)

Vậy \(x\in\left\{-3;2\right\}\)

\(c,5x\left(x-2\right)+\left(x-2\right)=0\)

\(\left(x-2\right)\left(5x+1\right)=0\)

\(Th1:x-2=0=>x=2\)

\(5x +1=0=>x=-\frac{1}{5}\)

Vậy \(x\in\left\{-\frac{1}{5};2\right\}\)

Ta có: a^3+b^3 = (a+b)(a^2-ab+b^2)

                   = a^3-a^2b+ab^2+a^2b-ab^2+b^3

                   = a^3-3a^2b+2a^2b+3ab^2-2ab^2+3a^2b-2a^2b-3ab^2+2ab^2+b^3

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

                   = (a+b)^3-3ab(a+b) (đpcm)

 a³ + b³ = ( a + b ) ³ - 3ab( a + b )

 a³ + b³ =a^3+3a^2b+3ab^2-3a^b-3ab^2

 a³ + b³ =a^2+b^2(đpcm)

`a,`

`(1/3 a + 4y)^2`

`= (1/3 a)^2 + 2 . 1/3a . 4y + (4y)^2`

`=1/9 a^2 + 8/3 ay + 16y^2`

`b,`

`(1/x - 3/y)^2`

`= (1/x)^2 - 2 . 1/x . 3/y + (3/y)^2`

`= 1/x^2 - 6/(xy) + 9/y^2`