Từ giả thiết: \(\frac{a}{b}=\frac{b}{c}\Rightarrow ac=b^2\Rightarrow abc=b^3\)

Ta có: \(\frac{a^3-2b^3+c^3}{a+b+c}=\frac{a^3+b^3+c^3-3c^3}{a+b+c}=\frac{a^3+b^3+c^3-3abc}{a+b+c}\)

Xét: \(a^3+b^3+c^3=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

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

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

\(\Rightarrow\frac{a^3-2b^3+c^3}{a+b+c}=a^2+b^2+c^2-ab-bc-ac\) là 1 số nguyên (đpcm)

Sai r bạn ơi

Đặt x=a + b - 2c
       y=b+c-2a
       z=c+a-2b
=>x+y+z=(a + b - 2c)+(b+c-2a)+(c+a-2b)
=>x+y+z=0
=>x+y= - z                         (1)
=>(x+y)^3=(-z)^3
=>x^3+y^3+3xy(x+y)=(-z)^3
=>x^3+y^3+z^3 +3xy(-z)=0        {vì x+y=-z [theo (1)]}
=>x^3+y^3+z^3 -3xyz=0
=>x^3+y^3+z^3 =3xyz
Vậy (a + b - 2c)^3 + (b + c - 2a)^3 + (c + a - 2b)^3=3(a + b - 2c) (b + c - 2a)(c + a - 2b)

\(\left(x-1\right)^3+\left(x-3\right)^3+8\left(2-x\right)^3=0\)

\(\left(x-1+x-3\right)\left[\left(x-1\right)^2-\left(x-1\right)\left(x-3\right)+\left(x-3\right)^2\right]+\left[2\left(2-x\right)\right]^3=0\)

\(\left(2x-4\right)\left(x^2-2x+1-x^2+4x-3+x^2-4x+4\right)+\left(4-2x\right)^3=0\)

\(\left(2x-4\right)\left(x^2-4x+7\right)-\left(2x-4\right)^3=0\)

\(\left(2x-4\right)\left[x^2-4x+7-\left(2x-4\right)^2\right]=0\)

\(2\left(x-2\right)\left(x^2-4x+7-4x^2+16x-16\right)=0\)

\(2\left(x-2\right)\left(12x-3x^2-9\right)=0\)

\(6\left(x-2\right)\left(4x-x^2-3\right)=0\)

\(6\left(x-2\right)\left(3x-x^2+x-3\right)=0\)

\(6\left(x-2\right)\left[x\left(3-x\right)-\left(3-x\right)\right]=0\)

\(6\left(x-2\right)\left(3-x\right)\left(x-1\right)=0\)

\(\Rightarrow x=\left\{1;2;3\right\}\)

\(\left(x-1\right)^3+\left(x-3\right)^3+8\left(2-x\right)^3=0\)

\(\Rightarrow x^3-2x^2+x-x^2+2x+1+x^3-6x^2+9x-3x^2+18x-27+64-64x+16x^2-32x+32x^2-8x^3=0\)

\(\Rightarrow-6x^3+36x^2-66x+36=0\)

\(\Rightarrow-6\left(x^3-6x^2+11x-6\right)=0\)

\(\Rightarrow\left(x^2-5x+6\right)\left(x-1\right)=0\)

\(\Rightarrow\left(x-3\right)\left(x-2\right)\left(x-1\right)=0\)

=> x - 3 = 0 ; x - 2 = 0 hoặc x - 1 = 0

=> x = 3 ; x = 2 hoặc x = 1

a) 16(4x+5)² - 25(2x+2)²

^{\(=\left[4\left(4x+5\right)\right]^2-\left[5\left(2x+2\right)\right]^2\)}

^{\(=\left[4\left(4x+5\right)+5\left(2x+2\right)\right]\left[4\left(4x+5\right)-5\left(2x+2\right)\right]\)}

^{\(=\left(16x+20+10x+10\right)\left(16x+20-10x-10\right)\)}

^{\(=\left(26x+30\right)\left(6x+10\right)\)}

\(b,\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)

\(=\left(x-y+4+2x+3y-1\right)\left(x-y+4-2x-2y+1\right)\)

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

\(c,\left(x+1\right)^4-\left(x-1\right)^4\)

\(=\left(x+1\right)^{2^2}-\left(x-1\right)^{2^2}\)

\(=\left[\left(x+1\right)^2+\left(x-1\right)^2\right]\left[\left(x+1\right)^2-\left(x-1\right)^2\right]\)

\(=\left(x^2+2x+1+x^2-2x+1\right)\left[\left(x+1+x-1\right)\left(x+1-x+1\right)\right]\)

\(=\left(2x^2+2\right)2x.2\)

\(=4x.2\left(x^2+1\right)\)

\(=8x\left(x^2+1\right)\)

\(\left(2x-1\right)^2=\left(x+1\right)^2\)

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

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

\(\Rightarrow\left(x-2\right)3x=0\)

\(\Rightarrow\orbr{\begin{cases}3x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Câu a) 

Em tham khảo link: Câu hỏi của I have a crazy idea - Toán lớp 6 - Học toán với OnlineMath

Ta có bài toán

P_n-P_n-1=(n-1)P_n-1

Chứng minh

Ta có    P_n-P_n-1=n!-(n-1)!

                         =n(n-1)!-(n-1)!

                         =(n-1)(n-1)!=(n-1)P_n-1

=>P_n-P_n-1=(n-1)P_n-1

Từ kết quả trên ta có

P₂-P₁=(2-1)P₁

P₃-P₂=(3-1)P₂

...............

P_n=P_n-1=(n-1)P_n-1

-----------------------------

P_n-P₁=P₁+2P₂+3P₃+.........+(n-1)P₁

=>1+1.P₁+2P₂+3P₃+...+n.P_n=P_n+1

#)Giải :

\(2x^3-3x^2+x+m=\left(x+2\right)\left(2x^2-7x+15\right)\)

\(\Leftrightarrow2x^3-3x^2+x+m=2x^3-7x^2+15x+4x^2-14x+30\)

\(\Leftrightarrow-3x^2+x+m=-3x^2+x+30\)

\(\Leftrightarrow m=30\)