Tính nhanh: \(\left(2012\times2010+2010\times2008\right)\times\left(1+\dfrac{1}{2}\div1\dfrac{1}{2}-1\dfrac{1}{3}\right)\)
Những câu hỏi liên quan
cho \(f\left(x\right)=\dfrac{x^3}{1-3x-3x^2}\). hãy tính giá trị biểu thức sau: \(A=f\left(\dfrac{1}{2012}\right)+f\left(\dfrac{2}{2012}\right)+...+f\left(\dfrac{2010}{2012}\right)+f\left(\dfrac{2011}{2012}\right)\)
Bạn kiểm tra lại đề, \(f\left(x\right)=\dfrac{x^3}{1-3x-3x^2}\) hay \(f\left(x\right)=\dfrac{x^3}{1-3x+3x^2}\)
Đúng 1
Bình luận (0)
Tính:
A = \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+99\right)}{1\times99+2\times98+3\times97+...+99\times1}\)
B = \(\frac{1\times2010+2\times2009+3\times2008+...+2010\times1}{\left(1+2+3+...+2010\right)+\left(1+2+3+...+2009\right)+...+\left(1+2\right)+1}\)
cho \(f\left(x\right)=\dfrac{x^3}{1-3x+3x^2}\) hãy tính giá trị của biểu thức sau:
\(A=f\left(\dfrac{1}{2012}\right)+f\left(\dfrac{2}{2012}\right)+...+f\left(\dfrac{2010}{2012}\right)+f\left(\dfrac{2011}{2012}\right)\)
1/ Chứng tỏ rằng : Bdfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+...+dfrac{1}{8^2} 1
2/ Rút gọn: Bleft(1-dfrac{1}{2}right)cdotleft(1-dfrac{1}{3}right)cdotleft(1-dfrac{1}{4}right)cdot...cdotleft(1-dfrac{1}{20}right)
3/ Tính giá trị của biểu thức: A dfrac{7}{4}cdotleft(dfrac{3333}{1212}+dfrac{3333}{2020}+dfrac{3333}{3030}+dfrac{3333}{4242}right)
4/ So sánh : A dfrac{2011+2012}{2010+2013} và B dfrac{2011}{2012}+dfrac{2012}{2013}
Đọc tiếp
1/ Chứng tỏ rằng : B=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{8^2}< 1\)
2/ Rút gọn: B=\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{20}\right)\)
3/ Tính giá trị của biểu thức: A= \(\dfrac{7}{4}\cdot\left(\dfrac{3333}{1212}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{3333}{4242}\right)\)
4/ So sánh : A= \(\dfrac{2011+2012}{2010+2013}\) và B= \(\dfrac{2011}{2012}+\dfrac{2012}{2013}\)
1/ \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}\)
\(B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)
\(B< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(B< \dfrac{1}{1}-\dfrac{1}{8}< 1\)
\(B< 1\)
2/ \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{20}\right)\)
\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{19}{20}\)
\(B=\dfrac{1\times2\times3\times...\times19}{2\times3\times4\times...\times20}\)
\(B=\dfrac{1}{20}\)
3/ \(A=\dfrac{7}{4}\cdot\left(\dfrac{3333}{1212}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{3333}{4242}\right)\)
\(A=\dfrac{7}{4}\cdot\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(A=\dfrac{7}{4}\cdot\left(\dfrac{33}{3.4}+\dfrac{33}{4.5}+\dfrac{33}{5.6}+\dfrac{33}{6.7}\right)\)
\(A=\dfrac{7}{4}.33.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(A=\dfrac{231}{4}.\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)
\(A=\dfrac{231}{4}\cdot\dfrac{4}{21}\)
\(A=11\)
4/ A phải là \(\dfrac{2011+2012}{2012+2013}\)
Ta có : \(B=\dfrac{2011}{2012}+\dfrac{2012}{2013}>\dfrac{2011}{2013}+\dfrac{2012}{2013}=\dfrac{2011+2012}{2013}>\dfrac{2011+2012}{2012+2013}=A\)
\(\Rightarrow B>A\)
Đúng 0
Bình luận (0)
25 tháng 7 2018 lúc 11:34
Bài 1:
a. \(\left(5^{2010}+5^{2012}+5^{2014}\right):\left(5^{2011}+5^{2009}+5^{2007}\right)\)
b. \(\left(-\dfrac{7}{45}\right)-\left(-\dfrac{1}{4}\right)-\left(-\dfrac{3}{5}\right)+\dfrac{1}{12}+\dfrac{2}{3}+\dfrac{1}{39}-\left(-\dfrac{5}{9}\right)\)
\(a)\left(5^{2010}+5^{2012}+5^{2014}\right):\left(5^{2011}+5^{2009}+5^{2007}\right)\)
\(=\dfrac{5^{2007}\left(5^3+5^5+5^7\right)}{5^{2007}\left(5^4+5^2+1\right)}=\dfrac{5^3+5^5+5^7}{5^4+5^2+1}\)
\(=\dfrac{125+3125+78125}{625+25+1}=\dfrac{81375}{651}=125\)
\(b)-\dfrac{7}{45}+\dfrac{1}{4}+\dfrac{3}{5}+\dfrac{1}{12}+\dfrac{2}{3}+\dfrac{1}{39}+\dfrac{5}{9}\)
\(=\dfrac{-7.52+1.585+3.468+1.195+2.780+1.60-5.260}{2340}\)
\(=\dfrac{-364+585+1404+195+1560+60-1300}{2340}\)
\(=\dfrac{2140}{2340}=\dfrac{107}{117}\)
Đúng 0
Bình luận (3)
Tính \(\left(1+\dfrac{1}{1+2}\right)\times\left(1+\dfrac{1}{1+2+3}\right)\times\left(1+\dfrac{1}{1+2+3+4}\right)\times...\times\left(1+\dfrac{1}{1+2+3+...+997}\right)\)
\(\left(\dfrac{2\sqrt{x}-x}{x\sqrt{x-1}}-\dfrac{1}{\sqrt{x}-1}\right)\div1\left(-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\left(x>0,x\ne1\right)\)
\(\left(\dfrac{2\sqrt{x}-x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right)\div\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\left[\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}-1}\right]\div\left[\dfrac{\left(x+\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)}{x+\sqrt{x}+1}\right]\)
\(=\dfrac{\left(2\sqrt{x}-x\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\times\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x-1\right)}=\dfrac{1}{x-1}\)
Đúng 0
Bình luận (0)
Tính
P=\(\left(1-\dfrac{28}{10}\right)\left(1-\dfrac{52}{22}\right)\left(1-\dfrac{80}{36}\right)...\left(1-\dfrac{21808}{10900}\right)\)
Q=\(\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}}{\dfrac{2013}{1}+\dfrac{2014}{2}+\dfrac{2015}{3}+...+\dfrac{4023}{2011}+\dfrac{4024}{2012}-2012}\)
\(F=\dfrac{1}{2\times\left(1+2\right)}+\dfrac{1}{2\times\left(1+2+3\right)}+....+\dfrac{1}{2\times\left(1+2+3+...+9\right)}\)