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)

1+1=mấy

1+1=2 chứ bao nhiêu

Ta có \(\frac{n^5}{30}+\frac{n^3}{6}+\frac{4n}{5}=\frac{n^5+5n^3+24n}{30}\)

Khi đó n⁵ + 5n³ + 24n 

= n(n⁴ + 5n² + 24)

=  n(n⁴ + 5n² - 6 + 30) 

= n(n⁴ - n² + 6n² - 6) + 30n 

= n[n²(n² - 1) + 6(n² - 1)] + 30n 

= n(n² + 6)(n² - 1) + 30n 

= n(n² - 4 + 10)(n² - 1) + 30n 

= n(n² - 4)(n² - 1) + 10n(n² - 1) + 30n 

= (n - 2)(n - 1)n(n + 1)(n + 2) + 10(n - 1)n(n + 1) + 30n

Nhận thây (n - 2)(n - 1)n(n + 1)(n + 2) \(⋮\)30 (tích 5 số nguyên liên tiếp) (1)

10(n - 1)n(n + 1) \(⋮\)30 (2)

30n \(⋮\)30 (3)

Từ (1) ; (2) ; (3) =>  (n - 2)(n - 1)n(n + 1)(n + 2) + 10(n - 1)n(n + 1) + 30n \(⋮\)30 

=> n⁵ + 5n³ + 24n \(⋮\)30 

=> P \(\inℤ\)(ĐPCM) 

5 phút nữa thì mk sẽ tự giải

#)Giải :

\(\left(-2a-4\right)\left(2a+1\right)\)

\(=\left(-2a\right)\left(2a+1\right)+4\left(2a+1\right)\)

\(=\left(-2a\right).2a+\left(-2a\right).1+4.2a+4.1\)

\(=-4a^2-2a+8a+4\)

\(=-4a^2+6a+4\)