\(5\left(3x^{n+1}-y^{n-1}\right)+\left(x^{n+1}+5y^{n-1}\right)-5\left(3x^{n+1}+2y^{n-1}\right)-\left(3x^{n+1}-10\right)\)

\(=5\left(3x^{n+1}-y^{n-1}-3x^{n-1}-2y^{n-1}\right)+\left(x^{n+1}+5y^{n-1}\right)-3x^{n+1}+10\)

\(=5\left(-3y^{n-1}\right)+\left(x^{n+1}-3x^{n+1}\right)+5y^{n-1}+10\)

\(=-15y^{n-1}-2x^{n+1}+5y^{n-1}+10\)

\(=-10y^{n-1}-2x^{n+1}+10\)

P/s : Bạn xem lại đề đi , hình như đề sai : 

Tham khảo nha : https://olm.vn/hoi-dap/question/131605.html

https://olm.vn/hoi-dap/detail/9148079003.html

Tham khảo nhé

(๑￫ܫ￩)Hanna Maia(❍ᴥ❍ʋ)

\(\left(a+b\right)^3-\left(a-b\right)^3\)

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

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

\(=6a^2b+2b^3\)

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

a/\(\left(a+b\right)^3-\left(a-b\right)^3\)

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

\(=6ab^2+2b^3\)(rút gọn hết)

b/\(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3xy\left(x-y\right)+z^3-3xyz\)

\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x-y-z\right)\)

\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3z\left(x+y\right)-3xy\right]\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-2xz+2xz+2xy-3xz-3yz-3xy\right).\)

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

Hok tốt

\(a,\frac{1}{64}x^6-125y^3\)

\(=\left(\frac{1}{2}x\right)^6-\left(5y\right)^3\)

\(=\left(\frac{1}{4}x^2\right)^3-\left(5y\right)^3\)

\(\left(\frac{1}{4}x^2-5y\right)\left[\left(\frac{1}{4}x^2\right)^2+\left(\frac{1}{4}x^2\right).5y+25y^2\right]\)

\(b,27a^3-54a^2b+36ab^2-8b^3\)

\(=\left(3a\right)^3-3.2.\left(3a\right)^2b+3.3a.\left(2b\right)^2-\left(2b\right)^3\)

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

\(c,x^6-x^6\)

\(=0\)

\(d,10x-25-x^2\)

\(=-x^2+10x-25\)

\(=-\left(x^2-10x+25\right)\)

\(=-\left(x-5\right)^2\)

Vì a>2=>a=2+m, b>2=>b=2+n (m,n thuộc N*)

=>a.b=(2+m).(2+n)=2.(2+n)+m.(2+n)=4+2n+2m+mn=4+m+m+n+n+mn=(4+m+n)+(m+n+mn)=(2+m)+(2+n)+(m+n+mn)>(2+m)+(2+m)=a.b

=>ĐPCM

Vì \(a>2\)

và \(b>2\)

\(\Rightarrow a>0\)và \(b>0\)

Vì \(a>2\)và \(b>0\)

\(\Rightarrow ab>2b\)(1)

Vì \(b>2\)và \(a>0\)

\(\Rightarrow ab>2a\) (2)

Cộng vế tương ứng (1) và (2) ta có :

\(2ab>2\left(a+b\right)\)

\(\Rightarrow ab>a+b\)(đpcm)

Ta có : 

\(\left(a+b\right)^2\ge4ab\)

\(\Leftrightarrow a^2+2ab+b^2\ge4ab\)

\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\)( luôn đúng ) 

\(\left(đpcm\right)\)

Chứng minh phản chứng :

Giả sử : \(\left(a+b\right)^2< 4ab\)

\(\Rightarrow a^2+2ab+b^2< 4ab\)

\(\Rightarrow a^2-2ab+b^2< 0\)

\(\Rightarrow\left(a-b\right)^2< 0\) (vô lí )

Vậy cần có :

\(\left(a+b\right)^2\ge4ab\)

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

\(x\left(x^2-4^2\right)-\left[\left(x^2\right)^2-1^2\right]\)

\(x\left(x^2-16\right)-\left[x^4-1\right]\)

HOK tốt 

Bạn tự vẽ hình nhé.

a) Ta có: EF, FG; GN; NE lần lượt là đường trung bình của \(\Delta ABC;\Delta BCD;\Delta CDA;\Delta DAB\)

\(\Rightarrow\hept{\begin{cases}EF=\frac{1}{2}AB;EF//AC\\GN=\frac{1}{2}AB;GN//AC\\FC//BC\end{cases}}\Rightarrow AC\perp BD\)

\(\Rightarrow\hept{\begin{cases}EFGH\text{ là HBH}\\AC\perp BD\\FG//BD;EF//AC\end{cases}}\Rightarrow EF\perp FG\)

=> EFGH là HCN

b) Dựa câu a) để làm nhé

bạn bấm vào câu hỏi tương tự nhé.

Ta có :

\(a+b+c+d=0\)

\(\Rightarrow c=-\left(a+b\right)\)

Do đó :

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

\(\Rightarrow a^3+b^3+3ab\left(a+b\right)\)

\(=-c^3-d^3-3cd\left(c+d\right)\)

\(\Rightarrow a^3+b^3+c^3=-3ab\left(a+b\right)-3cd\left(c+d\right)\)

Vì \(a+b=-\left(c+d\right)\)

\(\Rightarrow3ab\left(c+d\right)-3cd\left(c+d\right)=3\left(c+d\right)\left(ab-cd\right)\)