\(14.86+14^2\)

\(=14.\left(86+14\right)\)

\(=14.100=1400\)

= 14. 86 + 14.14 

= 14. ( 86 + 14) 

= 14. 100

=1400

= - 1/24

\(\dfrac{5}{8}-\dfrac{2}{3}\)

\(=\dfrac{15}{24}-\dfrac{16}{24}=-\dfrac{1}{24}\)

có:

(x-2)² ≥ 0 ∀ x

(y-x+1)² ≥ 0 ∀ x,y

⇒ (x-2)²+(y-x+1) ≥ 0 ∀ x,y

Dấu "=" xày ra

⇔(x-2)²+(y-x+1)²=0

⇔(x-2)²=0                                

⇔x - 2 = 0

⇔x=2

Có: x=2

⇒(y-x+1)²=0

⇔ (y-[2-1])² =0

⇔ (y-1)²=0

⇔y=1

vậy số nguyên x=2;y=1(đpcm)

Vì: 

\(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\forall x\\\left(y-x+1\right)^2\ge0\forall x,y\end{matrix}\right.\)

Dấu "=" \(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y-x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2013^2}\)

Ta có ; 

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

...

\(\dfrac{1}{2013^2}< \dfrac{1}{2012.2013}\)

\(\Rightarrow B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2013^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2012.2013}\)

\(\Rightarrow B< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2012}-\dfrac{1}{2013}\)

\(\Leftrightarrow B< 1-\dfrac{1}{2013}\)

\(\Rightarrow B< \dfrac{2012}{2013}\)

Lại có : \(\dfrac{2012}{2013}< \dfrac{3}{4}\)

\(\Rightarrow B< \dfrac{3}{4}\)

* Chắc vậy, sai thì thôg cảm ^^ * 

Còn j k hiểu thì ib nha