\(\left(1-\frac{4}{7}\right).\left(1-\frac{1}{8}\right).\left(1-\frac{1}{9}\right)..........\left(1-\frac{1}{2011}\right)\)
Giúp mình với d Mình vote cho gấp lắm rồi
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(1-1/3).(1-1/5).(1-1/7).(1-1/9).(1-1/11).(1-1/13).(1-1/2).(1-1/4).(1-1/6).(1-1/8).(1-1/10)
=2/3.4/5.6/7.8/9.10/11.12/13.1/2.3/4.5/6.7/8.9/10
=8/15.48/63.120/143.3/8.35/48.9/10
=384/945.360/1144.315/480
=138240/1081080.315/480
=43545600/518918400=84/1001
=\(-\frac{6}{5}\).\(\frac{-7}{6}\).\(\frac{-8}{7}\).\(\frac{-9}{8}\).\(\frac{-10}{9}\).\(\frac{-11}{10}\)
=\(\frac{7}{5}\).\(\frac{9}{7}\).\(\frac{11}{9}\)
=\(\frac{11}{5}\)
\(=\frac{-6}{5}\times\frac{-7}{6}\times\frac{-8}{7}\times\frac{-9}{8}\times\frac{-10}{9}\times\frac{-11}{10}\)
\(=\frac{\left(-6\right).\left(-7\right).\left(-8\right).\left(-9\right).\left(-10\right).\left(-11\right)}{5.6.7.8.9.10}\)
\(=\frac{6\times7\times8\times9\times10\times11}{5\times6\times7\times8\times9\times10}\)
Triệt tiêu các thừa số bằng nhau ở tử và mẫu, ta có kết quả là \(\frac{11}{5}\)
c) Ta có: \(\left\{{}\begin{matrix}\dfrac{x+2}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x+1}+\dfrac{10}{y-2}=25\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y-2}=22\\\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=\dfrac{1}{2}\\\dfrac{1}{x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=1\\y-2=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{5}{2}\end{matrix}\right.\)
Câu b: Đặt \(B=\left(\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}-1\right)\cdot\left(\frac{1}{4}-1\right)\cdot...\cdot\left(\frac{1}{2004}-1\right)\)
Ta có: \(\frac{1}{2}-1=\left(-\frac{1}{2}\right);\frac{1}{3}-1=\left(-\frac{2}{3}\right);...;\frac{1}{2004}-1=\left(-\frac{2003}{2004}\right)\)
\(\Rightarrow B=\left(-\frac{1}{2}\right)\cdot\left(-\frac{2}{3}\right)\cdot...\cdot\left(-\frac{2003}{2004}\right)\)
Vì B là 2003 thừa số âm nhân lại với nhau nên B là số âm
\(\Rightarrow B=-\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2003}{2004}\right)=-\frac{1}{2004}\)
Câu a: Đặt \(A=1+2^4+2^8;B=1+2+2^2+...+2^{11}\)
\(\Rightarrow16A=2^4+2^8+2^{12}\) \(\Rightarrow15A=2^{12}-1\) \(\Rightarrow A=\frac{2^{12}-1}{15}\) \(\left(1\right)\)
\(\Rightarrow2B=2+2^2+2^3+...+2^{12}\) \(\Rightarrow B=2^{12}-1\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow A:B=\frac{2^{12}-1}{15}:\left(2^{12}-1\right)=\frac{1}{15}\)
\(\left(1-\frac{1}{6}\right)x\left(1-\frac{1}{7}\right)x\left(1-\frac{1}{8}\right)x\left(1-\frac{1}{9}\right)x\left(1-\frac{1}{10}\right)\)
\(=\frac{5}{6}x\frac{6}{7}x\frac{7}{8}x\frac{8}{9}x\frac{9}{10}\)
\(=\frac{1}{2}\)
=(1/1-1/6)x(1/1-1/7)x(1/1-1/8)x(1/1-19)x(1/1-1/10)
=5/6x6/6x7/8x8/9x9/10
A) \(\frac{1}{2}\cdot\left(\frac{2}{9}+\frac{3}{7}-\frac{5}{27}\right)\)
\(=\frac{1}{2}\cdot\frac{1}{2}\)
\(=\frac{1}{4}\)
B) \(\left(\frac{-5}{28}+1.75+\frac{8}{35}\right):\left(-3\frac{9}{20}\right)\)
\(=\left(\frac{-5}{28}+\frac{7}{4}+\frac{8}{35}\right):\frac{-69}{20}\)
\(=\frac{14}{5}:\frac{-69}{20}\)
\(=\frac{-56}{69}\)
Đặt A bằng biểu thức trên
Ta có:
\(A=\left(1-\frac{4}{7}\right).\left(1-\frac{1}{8}\right).\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{2011}\right)\)
\(A=\frac{3}{7}.\left(\frac{7}{8}\right).\left(\frac{8}{9}\right)...\left(\frac{2009}{2010}\right).\left(\frac{2010}{2011}\right)\)
\(A=\frac{3}{7}.\left(\frac{\left(7.8...2009.2010\right)}{\left(8.9...2010.2011\right)}\right)\)
\(A=\frac{3}{7}.\frac{7}{2011}\)\(=\frac{3}{2011}\)
Ok nhé, bài này khá dễ !
đề đúng không bạn phân số đầu đó