bài này hỏi gì vậy?mình đoán là tìm Amin đúng không?Nếu vậy thì làm như sau:
Ta có:
\(A=|x-1010|+|x-1011|\)
\(\Rightarrow A=|x-1010|+|1011-x|\) \(\left(|a-b|=|b-a|\right)\)
\(\Rightarrow A\ge|x-1010+1011-x|\)
\(\Rightarrow A\ge1\)
\(\Rightarrow Amin=1\)KHI \(1010\le x\le1011\)
tim gia tri nho nhat
1)\(A=\left|x-1010\right|+\left|x-1011\right|\)\
2)\(B=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+2011\)
3)\(C=5x^2+20x+2010\)
tính hợp lí:
\(\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2017.2018}\right)-\left(\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}\right)\)
\(\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+.....+\frac{1}{2017.2018}\right)-\left(\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+.....+\frac{1}{2017}\right)\)
Cho A=\(\frac{1}{1011}\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2019}\right)\)và B=\(\frac{1}{1010}\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)So sánh A và B
â , tính M = \(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right).......\left(1+\frac{1}{2017}\right)\left(1+\frac{1}{2018}\right)\)
b , Cho A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2017}-\frac{1}{2018}\)
c , B = \(\frac{1}{1010}+\frac{1}{1011}+.....+\frac{1}{2017}+\frac{1}{2018}.tinh\left(\frac{A}{B}\right)^{2018}\)
A=1-\(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}+\dfrac{1}{2019}\)
B=\(\dfrac{1}{1010}+\dfrac{1}{1011}+\dfrac{1}{1012}+...+\dfrac{1}{2019}\)
Tính \(^{\left(A-B\right)^{2019}}\)
\(S=\left(-3\right)^0+\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{2017}\)
\(P=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(Q=\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2016}+\frac{1}{2017}\)
\(T\text{ÍNH}\)\(T\text{ỔNG}\)\(\left(S,P,Q\right)\)
cho x2+y2=1 và b.x2=a.y2CMR:\(\frac{x^{2020}}{a^{1010}}+\frac{y^{2020}}{b^{1010}}=\frac{2}{\left(a+b\right)^{1010}}\)
Cho x, y, z là các số khác 0 và x2 = yz, y2=xz, z2=xy
Tính \(\dfrac{\left(x+y+z\right)^{2022}}{x^{337}y^{674}z^{1011}}\)
