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Ghép (11;9) ; (12;8) ; ....;(19;1) ta có giá trị mỗi cập là 20
Mà có tất cả: 18/2 = 9 cặp như thế ( do tổng trên có 18 số hạng , 2 số hạng ghép thành một cặp)
===> Tổng trên bằng 20 x 9 =180
11+12+13+.....+18+19+1+2+3+4+.....+8+9
= (11+9)+(12+8)+13+7)+....+(18+2)+(19+1)
= [(19-1)+1.(11+9)
= 19.20
=19.10+19.10
= 380
em mới lớp 6 :D
\(G=\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{2\left(\sqrt{x}-3\right)+7}{\sqrt{x}-3}=2+\dfrac{7}{\sqrt{x}-3}\)
\(G\in Z\Leftrightarrow\dfrac{7}{\sqrt{x}-3}\in Z\)
Tại \(x\in N\Rightarrow\left[{}\begin{matrix}\sqrt{x}\in N\\\sqrt{x}\in I\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-3\in Z\\\sqrt{x}-3\in I\end{matrix}\right.\)
TH1: \(\sqrt{x}-3\in I\) \(\Rightarrow\dfrac{7}{\sqrt{x}-3}\notin Z\forall x\) thỏa mãn đk
\(TH2:\sqrt{x}-3\in Z\).Để \(\dfrac{7}{\sqrt{x}-3}\in Z\) \(\Leftrightarrow\sqrt{x}-3\inƯ\left(7\right)=\left\{-1;1;-7;7\right\}\)
\(\Leftrightarrow x\in\left\{4;16;100\right\}\)
Tại x=4 =>G=-5
Tại x=16=>G=9
Tại x=100=>G=3
Vậy tại x=6 thì \(G_{max}\)=9
(I là số vô tỉ)
\(G=\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{2\left(\sqrt{x}-3\right)+7}{\sqrt{x}-3}=2+\dfrac{7}{\sqrt{x}-3}\)
Để \(G\in Z\Rightarrow7⋮\sqrt{x}-3\Rightarrow\sqrt{x}-3\in\left\{1;7;-1;-7\right\}\)
mà \(\sqrt{x}-3\ge-3\Rightarrow\sqrt{x}-3\in\left\{1;7;-1\right\}\)
Để \(G_{max}\Rightarrow\dfrac{7}{\sqrt{x}-3}_{max}\Rightarrow\left\{{}\begin{matrix}\sqrt{x}-3>0\\\sqrt{x}-3_{min}\end{matrix}\right.\Rightarrow\sqrt{x}-3=1\Rightarrow x=4\)
\(\Rightarrow G_{max}=5\)
a: căn 14<4
=>7+căn 14<4+7=11
b: -căn 5<-2
=>-căn 5+9<-2+9=7
d: \(\sqrt{145}< 13\)
=>-11+căn 145<-11+13=2
e: \(7-4\sqrt{5}+2=9-4\sqrt{5}>0\)
=>7-4căn 5>-2
f: -4căn 5>-9
=>-9-4căn 5>-9-9=-18
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