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\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+...+\frac{1}{9700}=\frac{0,33x}{2009}\)
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}=\frac{0,33x}{2009}\)
\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}=\frac{0,33x}{2009}\)
\(\frac{1}{1}-\frac{1}{100}=\frac{0,33x}{2009}\)
\(\frac{100}{100}-\frac{1}{100}=\frac{0,33x}{2009}\)
\(\frac{99}{100}=\frac{0,33x}{2009}\)
\(\Rightarrow2009.99=100.0,33x\)
\(\Rightarrow2009.99=33x\)
\(\Rightarrow2009.99:33=x\)
\(\Rightarrow2009.3=x\)
\(\Rightarrow6027=x\)
Vậy \(x=6027\)(MK KO CHẮC NÓ ĐÚNG NHÉ )
Ta có : \(\frac{1}{4}+\frac{1}{28}+....+\frac{1}{9700}=\frac{0,33x}{2009}\)
=> \(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{97.100}=\frac{0.99x}{2009}\)
=> \(\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)=\frac{0,33x}{2009}\)
=> \(\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)=\frac{0,33x}{2009}\)
=> \(\frac{1}{3}\left(1-\frac{1}{100}\right)=\frac{0,33x}{2009}\)
=> \(\frac{33}{100}=\frac{0,33x}{2009}\Rightarrow33.2009=100.0,33x\)
=> 33.2009 = 33x
=> x = 2009
Thanks bn nhìu nha, mình sẽ K cho bn ngay. Bn kb với mình nha.
A = 1/4 + 1/28 + 1/70 +...+ 1/9700
A = 1/1.4 + 1/4.7 + 1/7.10 +...+ 1/97.100
3A = 3/1.4 + 3/4.7 + 3/7.10 +...+ 3/97.100
3A = 1 - 1/100
3A = 99/100
A=99/100:3=33/100
\(=\frac{1}{1.4}+\frac{1}{4.7}+..+\frac{1}{97.100}\)
\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(=\frac{1}{3}.\frac{99}{100}=\frac{33}{100}\)
\(A=\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+...+\frac{1}{9700}\)
\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{97.100}\)
\(A=\frac{3}{3}\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{97.100}\right)\)
\(A=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{97.100}\right)\)
\(A=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=\frac{1}{3}\left(1-\frac{1}{100}\right)\)
\(A=\frac{1}{3}.\frac{99}{100}=\frac{33}{100}\)
\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+...+\frac{1}{9700}\)
\(=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}\)
\(=\frac{100}{100}-\frac{1}{100}\)
\(=\frac{99}{100}\)
ở câu 1 ở mỗi phẫn số chúng ta cộng thêm 1, tổng là ta cộng thêm 5. Lấy 5 + -5=0. Rồi ta được tất cả tử là x+200,đặt chung ra ngoài,từ đó tính x=-200
a) Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}\) ; \(\frac{1}{3^2}< \frac{1}{2.3}\) ; \(\frac{1}{4^2}< \frac{1}{3.4}\) ; ... ; \(\frac{1}{2010^2}< \frac{1}{2009.2010}\)
=> \(Vt< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(=1-\frac{1}{2010}< 1\)
\(\frac{8}{23}\cdot\frac{46}{24}-x=\frac{1}{3}\)
=> \(\frac{2}{3}-x=\frac{1}{3}\)
=> \(x=\frac{1}{3}\)
\(\frac{10}{12}\div x=\frac{28}{9}\cdot\frac{3}{56}\)
=> \(\frac{10}{12}\div x=\frac{1}{6}\)
=> \(x=\frac{60}{12}=5\)
\(\frac{x-12}{4}=\frac{1}{2}\)
=> \(\left(x-12\right)\cdot2=4\cdot1\)
=> \(2x-24=4\)
=> \(2x=28\)
=> \(x=14\)
\(\frac{3}{1.4}+\frac{3}{4.7}+..+\frac{3}{97.100}=\frac{0,33x}{2009}\)
\(1-\frac{1}{4}+\frac{1}{4}-...-\frac{1}{100}=\frac{0,33x}{2009}\)
\(1-\frac{1}{100}=\frac{0,33x}{2009}\)
\(\frac{99}{100}=\frac{0,33x}{20009}\Rightarrow2009.99=100.0,33x\)
x=6027