|x-2010| + |x+2011| = 4021
PL
Những câu hỏi liên quan
Cho góc nhọn có số đo là x. Chứng minh: (2010 - sin x)/2011 + 2011/(2011 - sin x) > 4021/2011
Ta có VT = 2010/2011 -sinx/2011 + 1 + sinx/(2011-sinx) = 4021/2011 +[(2011sinx - 2011sinx + sin2 x)/(2011-sinx) = 4021/2011 + sin2 x/(2011-sinx) > 4021/2011
Đúng 0
Bình luận (0)
Cho góc nhọn có số đo là x. Cm bđt:
\(\frac{2010-sinx}{2011}+\frac{2011}{2011-sĩn}>\frac{4021}{2011}\)
1^2-2^2+3^2-4^2+...+2009^2-2010^2+2011^21-left(2-3right)left(2+3right)-left(4-5right)left(4+5right)-....-left(2010-2011right)left(2010+2011right)1-left(-1right).5-left(-1right).9-...-left(-1right).40211+5+9+...+4021frac{left[left(4021-1right):4+1right]left(4021+1right)}{2}2023066
Đọc tiếp
\(1^2-2^2+3^2-4^2+...+2009^2-2010^2+2011^2\)
\(=1-\left(2-3\right)\left(2+3\right)-\left(4-5\right)\left(4+5\right)-....-\left(2010-2011\right)\left(2010+2011\right)\)
\(=1-\left(-1\right).5-\left(-1\right).9-...-\left(-1\right).4021\)
\(=1+5+9+...+4021\)
\(=\frac{\left[\left(4021-1\right):4+1\right]\left(4021+1\right)}{2}\)
\(=2023066\)
thì làm sao???Hỏi xong rồi tự trả lời thì có ích gì
(✿◠‿◠)(๛ČℌUƔÊŇ♥Ť❍Ą́Ňツ)
Ê nhóc đừng có nghĩ lung tung
Chứng minh rằng:
\(\dfrac{1}{3\left(\sqrt{2}+1\right)}+\dfrac{1}{5\left(\sqrt{3}+\sqrt{2}\right)}+\dfrac{1}{7\left(\sqrt{4}+\sqrt{3}\right)}+...+\dfrac{1}{4021\left(\sqrt{2011}+\sqrt{2010}\right)}< \dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{2011}}\right)\)
\(\dfrac{1}{\sqrt{k}+\sqrt{k+1}}=\dfrac{\sqrt{k}-\sqrt{k+1}}{k-k-1}=\sqrt{k+1}-\sqrt{k}\\ \Leftrightarrow\text{Đặt}\text{ }A=\dfrac{1}{3\left(\sqrt{2}+\sqrt{1}\right)}+\dfrac{1}{5\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{1}{4021\left(\sqrt{2011}+\sqrt{2010}\right)}< \dfrac{1}{2\left(\sqrt{2}+\sqrt{1}\right)}+\dfrac{1}{2\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{1}{2\left(\sqrt{2011}+\sqrt{2010}\right)}\\ \Leftrightarrow A< \dfrac{1}{2}\left(\dfrac{1}{\sqrt{2}+\sqrt{1}}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{2011}+\sqrt{2010}}\right)\)
\(\Leftrightarrow A< \dfrac{1}{2}\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2011}-\sqrt{2010}\right)\\ \Leftrightarrow A< \dfrac{1}{2}\left(\sqrt{2011}-1\right)< \dfrac{1}{2}\cdot\dfrac{\sqrt{2011}-1}{\sqrt{2011}}=\dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{2011}}\right)\)
Đúng 1
Bình luận (0)
Chứng minh : \(\frac{\sqrt{2}-\sqrt{1}}{3}+\frac{\sqrt{3}-\sqrt{2}}{5}+\frac{\sqrt{4}-\sqrt{3}}{7}+...+\frac{\sqrt{2011}-\sqrt{2010}}{4021}\)<\(\frac{1}{2}\)
\(\left(2n+1\right)^2=4n^2+4n+1\)
\(>4n^2+4n=4n\left(n+1\right)\)
\(\Rightarrow2n+1>\sqrt{4n\left(n+1\right)}=2\sqrt{n\left(n+1\right)}\)
\(\Rightarrow\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}< \frac{1}{2}\cdot\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\) \(=\frac{1}{2}\cdot\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Do đó : \(\frac{\sqrt{2}-\sqrt{1}}{3}+\frac{\sqrt{3}-\sqrt{2}}{5}+...+\frac{\sqrt{2011}-\sqrt{2010}}{4021}\)
\(< \frac{1}{2}\cdot\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2010}}-\frac{1}{\sqrt{2011}}\right)\)
\(< \frac{1}{2}\)
Đúng 0
Bình luận (0)
Chứng minh \(\dfrac{\sqrt{2}-\sqrt{1}}{3}+\dfrac{\sqrt{3}-\sqrt{2}}{5}+\dfrac{\sqrt{4}-\sqrt{3}}{7}+...+\dfrac{\sqrt{2011}-\sqrt{2010}}{4021}< \dfrac{1}{2}\)
giúp mk vs
https://hoc24.vn/hoi-dap/question/817465.html
Bn tham khảo ở đây nha!
Mk lm r, k muốn lm lại
Đúng 0
Bình luận (0)
Tính nhanh:
a, 2010 x 3+ 2010 x 6 + 2010
b, 2011 x 89 + 10 x 2011 + 2011
a, 2010 x 3+ 2010 x 6 + 2010
= 2010 x ( 3 + 6 + 1)
= 2010 x 10
= 20100
b, 2011 x 89 + 10 x 2011 + 2011
= 2011 x (89 + 10 + 1)
= 2011 x 100
= 201100
Đúng 0
Bình luận (0)
Tính nhanh:
a, 2011 x 3+ 2011 x 6 + 2011
b, 2010 x 89 + 10 x 2010 + 2010
a, 2011 x 3+ 2011 x 6 + 2011
= 2011 x ( 3+6+1)
= 2011 x 10
= 20110
b, 2010 x 89 + 10 x 2010 + 2010
= 2010 x (89+10+1)
= 2010 x 100
= 201000
Đúng 0
Bình luận (0)
(2009 x 2010 + 2011 x 12 + 1998)/(2011 x 2010 - 2010 x 2009)