Vì \(\left|x+23\right|^{2017}\ge0;\left|y-23\right|^{2015}\ge0\)

\(\Rightarrow\left|x+23\right|^{2017}+\left|y-23\right|^{2015}\ge0\)

Dấu "=" xảy ra <=> \(\orbr{\begin{cases}\left|x+23\right|^{2017}=0\\\left|y-23\right|^{2015}=0\end{cases}\Rightarrow\orbr{\begin{cases}x+23=0\\y-23=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-23\\y=23\end{cases}}}\)

Đặt A = 1.2 + 2.3 + 3.4 + ..... + n(n + 1)

3A = 1.2.3 + 2.3.3 + 3.4.3 + ..... + n(n + 1)3

= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + n(n + 1)[(n + 2) - (n - 1)]

= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + n(n + 1)(n + 2) - (n - 1)n(n + 1)

= n(n + 1)(n + 2)

\(\Rightarrow A=\frac{N\left(N+1\right)\left(N+2\right)}{3}\)

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{99.100.101}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)=\frac{5049}{20200}\)

1/1.2.3+1/2.3.4+...+1/99.100.101

= 1/2 ( 1/1.2-1/2.3+1/2.3-1/3.4+...+1/99.100-1/100.101)

=1/2(1/1.2-1/.100.101)=5049/20200

Ta có:1/1.2+1/2.3+...+1/99.100

       =1-1/2+1/2-1/3+...+1/99-1/100

       =1-1/100

       =100-1/100

       =99/100

d/s=99/100 

đây là câu lớp 5,đáp số là 99/100

N=-y⁵​+(15-4)y³-2y= -y⁵+11y³​-2y

​M=8y⁵​-3y+1

​N+M=7y⁵+11y³​-5y+1

​N-M=-9y⁵+11y³+y-1

​Không biết đúng không nữa :)

Vì \(\left(x+y\right)^{2016}\ge0;\left|y-34\right|^{2018}\ge0\)

\(\Rightarrow\left(x+y\right)^{2016}+\left|y-34\right|^{2018}\ge0\)

Dấu "=" xảy ra <=> \(\orbr{\begin{cases}\left(x+y\right)^{2016}=0\\\left|y-34\right|^{2018}=0\end{cases}\Rightarrow\orbr{\begin{cases}x+y=0\\y-34=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-34\\y=34\end{cases}}}\)

Đặt A = 1.2 + 2.3 + 3.4 + .... + n(n + 1)

3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + n(n + 1).3

= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ..... + n(n + 1)[(n + 2) - (n - 1)]

= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + n(n + 1)(n + 2) - (n - 1)n(n + 1)

= n(n + 1)(n + 2)

\(\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

A=1.2.3+2.3.3+3.4.3+.....+N(N+1).3

3A=1.2(3-0)+2.3(4-1)+3.4(5-2)+........+N(N+1)-(N-2)(N-1)

3A=1.2.3-1.2.0-2.3.4-2.3.1+......+N(N-1)+(N+2)-N(N-1)-N-1

3A=N(N-1)+(N+2)/3