Tính nhanh: \(\left(2012\times2010+2010\times2008\right)\times\left(1+\dfrac{1}{2}\div1\dfrac{1}{2}-1\dfrac{1}{3}\right)\)
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NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 12 2020
Bạn kiểm tra lại đề, \(f\left(x\right)=\dfrac{x^3}{1-3x-3x^2}\) hay \(f\left(x\right)=\dfrac{x^3}{1-3x+3x^2}\)
7 tháng 5 2018
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 : \(B=\dfrac{2011}{2012}+\dfrac{2012}{2013}>\dfrac{2011}{2013}+\dfrac{2012}{2013}=\dfrac{2011+2012}{2013}>\dfrac{2011+2012}{2012+2013}=A\)
\(\Rightarrow B>A\)
T
1 tháng 8 2018
\(a)\left(5^{2010}+5^{2012}+5^{2014}\right):\left(5^{2011}+5^{2009}+5^{2007}\right)\)
\(=\dfrac{5^{2007}\left(5^3+5^5+5^7\right)}{5^{2007}\left(5^4+5^2+1\right)}=\dfrac{5^3+5^5+5^7}{5^4+5^2+1}\)
\(=\dfrac{125+3125+78125}{625+25+1}=\dfrac{81375}{651}=125\)
\(b)-\dfrac{7}{45}+\dfrac{1}{4}+\dfrac{3}{5}+\dfrac{1}{12}+\dfrac{2}{3}+\dfrac{1}{39}+\dfrac{5}{9}\)
\(=\dfrac{-7.52+1.585+3.468+1.195+2.780+1.60-5.260}{2340}\)
\(=\dfrac{-364+585+1404+195+1560+60-1300}{2340}\)
\(=\dfrac{2140}{2340}=\dfrac{107}{117}\)
PA
19 tháng 7 2017
\(\left(\dfrac{2\sqrt{x}-x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right)\div\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\left[\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}-1}\right]\div\left[\dfrac{\left(x+\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)}{x+\sqrt{x}+1}\right]\)
\(=\dfrac{\left(2\sqrt{x}-x\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\times\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x-1\right)}=\dfrac{1}{x-1}\)
Lời giải:
$(2012\times 2010+2010\times 2008)\times (1+\frac{1}{2}: 1\frac{1}{2}-1\frac{1}{3})$
$=2010\times (2012+2008)\times (1+\frac{1}{2}\times \frac{2}{3}-1\frac{1}{3})$
$=2010\times 4020\times (1+\frac{1}{3}-1\frac{1}{3})$
$=2010\times 4020\times 0=0$
=0 nha