\(\left(x+y+z\right)\left(x-y-z\right)\)

\(=\left(x.x\right)-\left(x.y\right)-\left(x.z\right)+\left(y.x\right)-\left(y.y\right)-\left(y.z\right)+\left(z.x\right)-\left(z.y\right)-\left(z.z\right)\)

\(=x^2-xy-xz+xy-y^2-yz+xz-yz-z^2\)

\(=x^2-y^2-z^2\)

n(2n-3) - 2n(n+1)

= 2n² - 3n - 2n² - 2n

= (2n² - 2n²) - (3n + 2n)

= 0 - (-5)n

= (-5)n

Vì tích có chứa thừa số -5\(⋮\)5 nên chia hết cho 5

Vậy n(2n-3) - 2n(n+1)\(⋮\)5 với \(\forall\)n\(\in\)Z

\(1)\)

\(a)\)\(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)

\(A=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)

\(A=100+99+98+97+...+2+1\)

\(A=\frac{100\left(100+1\right)}{2}\)

\(A=5050\)

\(b)\)\(B=3\left(2^2+1\right)\left(2^4+1\right).....\left(2^{64}+1\right)+1\)

\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right).....\left(2^{64}+1\right)+1\)

\(B=\left(2^4-1\right)\left(2^4+1\right).....\left(2^{64}+1\right)+1\)

\(B=\left(2^8+1\right).....\left(2^{64}+1\right)+1\)

\(............\)

\(B=\left(2^{64}-1\right)\left(2^{64}+1\right)+1\)

\(B=2^{128}-1+1\)

\(B=2^{128}\)

Chúc bạn học tốt ~ 

\(1)\)

\(c)\)\(C=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)

\(C=\left(a+b\right)^2+2\left(a+b\right)c+c^2+\left(a+b\right)^2-2\left(a+b\right)c+c^2-2\left(a+b\right)^2\)

\(C=2\left(a+b\right)^2+2c^2-2\left(a+b\right)^2\)

\(C=2c^2\)

\(2)\)

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

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

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

\(VP=a^3+b^3=VT\) ( đpcm ) 

\(b)\)\(VT=a^3+b^3+c^3-3abc\)

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

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

\(VT=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

\(VT=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=VP\) ( đpcm ) 

Từ đó suy ra : 

\(i)\)\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow\)\(a^3+b^3+c^3-3abc=0\)\(\Rightarrow\)\(a+b+c=0\)

Hoặc \(a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow\)\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Leftrightarrow\)\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Leftrightarrow\)\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}a=b=c}\)

Chúc bạn học tốt ~ 

\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)

Do \(a+b+c=0\)

\(\Rightarrow a^3+b^3+c^3-3abc=0\Rightarrow a^3+b^3+c^3=3abc\)

a+b+c=0\(\Rightarrow\)a+b= -c

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

\(\Rightarrow\)a^3+b^3+c^3= -3ab(a+b)

\(\Rightarrow\)a^3+b^3+c^3=3abc

(3x+5)(4-3x)=0

3x+5 =0 hoặc 4-3x=0

3x=-5 hoặc 3x=-4

x=-5/3 hoặc x=-4/3

9(3x-2)=x(2-3x)

9(3x-2)-x(3x-2)=0

(3x-2)(9-x)=0

3x-2=0 hoặc 9-x=0

3x=2 hoặc x= -9

x =2/3 hoặc x=-9 

vậy x =2/3 ; x= -9