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B=\(\left(\dfrac{2012}{x^2}\right)+4x+2013\) hay B=\(\left(\dfrac{2012}{x}\right)^2+4x+2013\) vậy bn
A=\(\left(-x\right)^2+3x+4=x^2+3x+4\)
=\(x^2+3x+2,25+1,75=x^2+1,5x+1,5^2+1,75\)
=\(\left(x+1,5\right)^2+1,75\)
Thấy\(\left(x+1,5\right)^2\ge0\Rightarrow\left(x+1,5\right)^2+1,75\ge1,75\) hay \(A\ge1,75\)
Vậy GTNN của A là 1,75 khi x=-1,5
Ta có: \(\left|x-1\right|+\left|x-5\right|=\left|x-1\right|+\left|5-x\right|\)
Nhận thấy: \(\left[{}\begin{matrix}\left|x-1\right|\ge x-1\\\left|5-x\right|\ge5-x\end{matrix}\right.\)
\(\Rightarrow\left|x-1\right|+\left|5-x\right|\ge x-1+5-x\)
\(\Rightarrow\left|x-1\right|+\left|5-x\right|\ge4\)
Dấu \("="\) xảy ra khi:
\(\left[{}\begin{matrix}x-1\ge0\\5-x\ge0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x\ge1\\x\le5\end{matrix}\right.\) \(\Rightarrow1\le x\le5\)
Vậy \(1\le x\le5.\)
Cho mk thêm cái ạ:
\(x\in\left\{1;2;3;4;5\right\}\)
Vậy \(x\in\left\{1;2;3;4;5\right\}\)
Theo bài ra ta có :
\(A=\frac{2011}{1.2}+\frac{2011}{3.4}+\frac{2011}{4.5}+...+\frac{2011}{1999.2000}\)
\(\Rightarrow\frac{A}{2011}=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{1999.2000}\)
\(\Rightarrow\frac{A}{2011}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1999}-\frac{1}{2000}\)
\(\Rightarrow\frac{A}{2011}=\left(\frac{1}{1}+\frac{1}{3}+...+\frac{1}{1999}\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2000}\right)\)
\(\Rightarrow\frac{A}{2011}=\left(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{2000}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2000}\right)\) \(-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2000}\right)\)
\(\Rightarrow\frac{A}{2011}=\left(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{2000}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2000}\right)\)
\(\Rightarrow\frac{A}{2011}=\left(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{2000}\right)-\left(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{1000}\right)\)
\(\Rightarrow\frac{A}{2011}=\frac{1}{1001}+\frac{1}{1002}+...+\frac{1}{2000}\)
\(\Rightarrow A=2011\left(\frac{1}{1001}+\frac{1}{1002}+...+\frac{1}{2000}\right)\left(1\right)\)
Ta lại có :
\(B=\frac{2012}{1001}+\frac{2012}{1002}+...+\frac{2012}{2000}\)
\(\Rightarrow B=2012\left(\frac{1}{1001}+\frac{1}{1002}+...+\frac{1}{2000}\right)\)\(\left(2\right)\)
Từ (1) và (2) => A < B
Vậy A < B
2.
a) +) ta co: tam giác GLO
GL = 6, LO = 8, OG = 10
=> GL < LO < GO ( 6<8<10)
=> góc O < góc G < góc L ( quan hệ giữa góc và cạnh đối diện trong tam giác LOG )
+) ta co: tam giac UVW
góc V = 40, góc U = 50
=> góc W = 180 - ( góc V + goc Ư )
= 180 - ( 50 + 40)
= 90
=> góc V < góc U < góc W
=> UW < VW < VU ( quan hệ giữa cạnh và góc trong tam giác ACB )
xin lỗi nhưng bài này mik cũng ko bt giải
theo bài ra ta có:
\(A=\dfrac{2011}{1.2}+\dfrac{2011}{3.4}+...+\dfrac{2011}{1999.2000}\)
\(\Rightarrow\dfrac{A}{2011}=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{1999.2000}\)
\(\Rightarrow\dfrac{A}{2011}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1999}-\dfrac{1}{2000}\)
\(\Rightarrow\dfrac{A}{2011}=\left(\dfrac{1}{1}+\dfrac{1}{3}+...+\dfrac{1}{1999}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2000}\right)\)
\(\Rightarrow\dfrac{A}{2011}=\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{2000}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2000}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2000}\right)\) \(\Rightarrow\dfrac{A}{2011}=\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{2000}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2000}\right)\) \(\Rightarrow\dfrac{A}{2011}=\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{2000}\right)-\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{1000}\right)\) \(\Rightarrow\dfrac{A}{2011}=\dfrac{1}{1001}+\dfrac{1}{1002}+...+\dfrac{1}{2000}\)
\(\Rightarrow A=2011\left(\dfrac{1}{1001}+\dfrac{1}{1002}+...+\dfrac{1}{2000}\right)\left(1\right)\)
ta lại có:
\(B=\dfrac{2012}{1001}+\dfrac{2012}{1002}+...+\dfrac{2012}{2000}\\ \Rightarrow B=2012\left(\dfrac{1}{1001}+\dfrac{1}{1002}+...+\dfrac{1}{2000}\right)\left(2\right)\)
Từ 1 và 2 => A < B\
vậy A < B