So sánh 2,4 và 2v3/5
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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 "=" \(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y-x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
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\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2013^2}\)
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
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