-5x chứ nhỉ?

\(n^5-n=n\left(n^4-1\right)=n\left(n^2+1\right)\left(n^2-1\right)=n\left(n^2+1\right)\left(n+1\right)\left(n-1\right)\)

\(=\left(n-1\right)n\left(n+1\right)\left(n^2-4+5\right)=\left(n-1\right)n\left(n+1\right)\left(n+2\right)\left(n-2\right)+5n\left(n+1\right)\left(n-1\right)⋮5\)

-có người nhờ t làm

\(\left\{{}\begin{matrix}x-y=3\\3x-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-3y=9\left(1\right)\\3x-4y=2\left(2\right)\end{matrix}\right.\) lấy (1)-(2) tìm được x;sau đó dễ dàng có y
\(\left\{{}\begin{matrix}7x-3y=5\\4x+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}28x-12y=20\left(1\right)\\28x+7y=14\left(2\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x+3y=-2\\5x-4y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+15y=-10\left(1\right)\\5x-4y=11\left(2\right)\end{matrix}\right.\)

Gt: Nhân sao cho cả 2 pt xuất hiện chung 1 thừa số,trừ đi chỉ còn 1 x or y

-Hoặc có thể đề bạn sai

hiểu đơn giản thì có thể hiểu như sau:

\(a^3+b^3+c^3\ge3\sqrt[3]{a^3b^3c^3}=3abc>abc\)

-Không có a;b;c thỏa mãn

-Thiếu số 3 ở gt kìa

a) Tính chất dãy tỉ số bằng nhau: \(\dfrac{x+y}{2014}=\dfrac{x-y}{2016}=\dfrac{x+y+x-y}{2014+2016}=\dfrac{2x}{4030}=\dfrac{x}{2015}\)

\(\dfrac{x+y}{2014}=\dfrac{x-y}{2016}=\dfrac{x+y-x+y}{2014-2016}=\dfrac{2y}{-2}=\dfrac{y}{-1}\)

Nên: \(\dfrac{x}{2015}=\dfrac{y}{-1}=\dfrac{xy}{2015}\)

Xét: \(\left\{{}\begin{matrix}\dfrac{x}{2015}=\dfrac{xy}{2015}\Leftrightarrow2015x=2015xy\Leftrightarrow y=1\\\dfrac{y}{-1}=\dfrac{xy}{2015}\Leftrightarrow2015y=-1xy\Leftrightarrow2015=-1x\Leftrightarrow x=-2015\end{matrix}\right.\)

2) \(VT=\left|x-6\right|+\left|x-10\right|+\left|x-2022\right|+\left|y-2014\right|+\left|z-2015\right|\)

\(VT=\left|x-6\right|+\left|2022-x\right|+\left|x-10\right|+\left|y-2014\right|+\left|z-2015\right|\)

\(VT\ge\left|x-6+2022-x\right|+\left|x-10\right|+\left|y-2014\right|+\left|z-2015\right|\)

\(VT\ge2016+\left|x-10\right|+\left|y-2014\right|+\left|z-2015\right|\ge2016=VP\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}6\le x\le2022\\x=10\\y=2014\\z=2015\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=2014\\z=2015\end{matrix}\right.\)

\(\dfrac{x^2+2x+5}{x+2}=\dfrac{x^2+2x}{x+2}+\dfrac{5}{x+2}=\dfrac{x\left(x+2\right)}{x+2}+\dfrac{5}{x+2}=x+\dfrac{5}{x+2}\)

\(\dfrac{x^2+4x+9}{x+2}=\dfrac{\left(x^2+4x+4\right)+5}{x+2}=\dfrac{\left(x+2\right)^2+5}{x+2}=x+2+\dfrac{5}{x+2}\)

\(pt\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+10\right)}+\dfrac{1}{\left(x+10\right)\left(x+13\right)}=\dfrac{4}{13}\)

\(a=1^2+2^2+3^2+4^2+...+100^2\)

\(a=1.\left(2-1\right)+2.\left(3-1\right)+3.\left(4-1\right)+...+100.\left(101-1\right)\)

\(a=1.2-1+2.3-2+3.4-3+...+100.101-100\)

\(a=\left(1.2+2.3+3.4+...+100.101\right)-\left(1+2+3+...+100\right)\)

Đặt: \(\left\{{}\begin{matrix}l=1.2+2.3+3.4+...+100.101\\n=1+2+3+...+100\end{matrix}\right.\)

Áp dụng tính ta được: \(l=1.2+2.3+3.4+...+100.101\)

\(3l=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+100.101.\left(102-99\right)\)

\(3l=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+100.101.102-99.100.101\)

\(3l=100.101.102\Leftrightarrow l=\dfrac{100.101.102}{3}=343400\)

\(n=1+2+3+...+100\)

\(n=\left[\left(100-1\right):1+1\right]:2.\left(100+1\right)=50.101=5050\)

\(a=l-n=343400-5050=338350\)

t ngửi thấy mùi đề sai