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Bài 1:
a) \(\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{10}-1\right)......\left(\dfrac{1}{2004}-1\right)\left(\dfrac{1}{2005}-1\right)\)
= \(\dfrac{-8}{9}.\dfrac{-9}{10}.......\dfrac{-2003}{2004}.\dfrac{-2004}{2005}\) = \(\dfrac{-8}{2005}\)
b) \(-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{-2+3}}}\) = \(-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{1}}}\)
= \(-2+\dfrac{1}{-2+\dfrac{1}{-1}}\) = \(-2+\dfrac{1}{-3}\) = \(\dfrac{-7}{3}\)
\(\text{Câu 1 : }\) Tính
\(\text{a) }\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{10}-1\right)...\left(\dfrac{1}{2004}-1\right)\left(\dfrac{1}{2005}-1\right)\\ =\left(1-\dfrac{9}{9}\right)\left(\dfrac{1}{10}-\dfrac{10}{10}\right)...\left(\dfrac{1}{2004}-1\right)\left(\dfrac{1}{2005}-\dfrac{2005}{2005}\right)\\ =\dfrac{-8}{9}\cdot\dfrac{-9}{10}\cdot...\cdot\dfrac{-2003}{2004}\cdot\dfrac{-2004}{2005}\\ =\dfrac{\left(-8\right)\cdot\left(-9\right)\cdot..\cdot\left(-2003\right)\cdot\left(-2004\right)}{9\cdot10\cdot...\cdot2004\cdot2005}\\ =-\dfrac{8\cdot9\cdot...\cdot2003\cdot2004}{9\cdot10\cdot...\cdot2004\cdot2005}\\ =-\dfrac{8}{2005}\)
\(-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{-2+3}}}\\ =-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{1}}}\\ =-2+\dfrac{1}{-2+\dfrac{1}{-1}}\\ =-2+\dfrac{1}{-3}\\ =-2+\dfrac{-1}{3}=-\dfrac{7}{3}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a_1}{a_2}=\dfrac{a_2}{a_3}=\dfrac{a_3}{a_4}=....=\dfrac{a_{2000}}{a_{2001}}=\dfrac{a_1+a_2+a_3+....+a_{2000}}{a_2+a_3+a_4+....+a_{2001}}\)
\(\Rightarrow\dfrac{a_1}{a_2}.\dfrac{a_2}{a_3}.\dfrac{a_3}{a_4}......\dfrac{a_{2000}}{a_{2001}}=\left(\dfrac{a_1+a_2+a_3+....+a_{2000}}{a_2+a_3+a_4+....+a_{2001}}\right)^{2000}\)
\(\Rightarrow\dfrac{a_1}{a_{2001}}=\left(\dfrac{a_1+a_2+a_3+....+a_{2000}}{a_2+a_3+a_4+....+a_{2001}}\right)^{2000}\)(đpcm)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a_1}{a_2}=\dfrac{a_2}{a_3}=\dfrac{a_3}{a_4}=...=\dfrac{a_{2017}}{a_{2018}}=\dfrac{a_1+a_2+a_3+...+a_{2017}}{a_2+a_3+a_4+...+a_{2018}}\)
Đặt:
\(\dfrac{a_1}{a_2}=\dfrac{a_2}{a_3}=\dfrac{a_3}{a_4}=...=\dfrac{a_{2017}}{a_{2018}}=\dfrac{a_1+a_2+a_3+...+a_{2017}}{a_2+a_3+a_4+....+a_{2018}}=k\)
\(\circledast\)\(\left(\dfrac{a_1+a_2+a_3+...+a_{2017}}{a_2+a_3+a_4+...+a_{2018}}\right)^{2017}=k^{2017}\)
\(\circledast\) \(\dfrac{a_1}{a_2}.\dfrac{a_2}{a_3}.\dfrac{a_3}{a_4}....\dfrac{a_{2017}}{a_{2018}}=\dfrac{a_1}{a_{2018}}=k^{2017}\)
Ta có đpcm
5a
Ta có \(\dfrac{a}{b}=\dfrac{a^2}{b^2}\) ; \(\dfrac{c}{d}=\dfrac{c^2}{d^2}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\)=> \(\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}\)=>\(\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}\)=\(\dfrac{a^2+c^2}{b^2+d^2}\)(T/c cuả dãy tỉ số bằng nhau)
=> ĐPCM
Xin lỗi nha mình nhầm đề. Nhưng bạn chỉ cần thay d bằng c là được.