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Ta có:
\(\left(\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{10}{10.110}\right)x=\)\(\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)
\(\Rightarrow\left(\frac{100}{1.101}+\frac{100}{2.202}+\frac{100}{3.303}+...+\frac{100}{10.101}\right)x=\)\(10.\left(\frac{10}{1.11}+\frac{10}{2.12}+...+\frac{10}{100.110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=\)\(10.\left(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)
\(\Rightarrow\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{110}\right)\right]x\)\(=\)\(10.\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{110}\right)\right]\)
\(\Rightarrow\left[ \left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right]x=\)\(10.\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right]\)
\(\Rightarrow x=10\)
Ta có:
$(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110}).x=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}$
$\Leftrightarrow \frac{1}{100}\left ( \frac{1}{1}-\frac{1}{100}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110} \right )x=\frac{1}{10}\left ( \frac{1}{1}-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110} \right )$
$\Leftrightarrow \left ( \frac{1}{1}-\frac{1}{100}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110} \right )x=10\left ( \frac{1}{1}-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110} \right )$
Đặt $A=\frac{1}{1}-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}$
$\Rightarrow A=\left ( 1+\frac{1}{2}+...+\frac{1}{10} \right )+\left ( \frac{1}{11}+\frac{1}{12}+...+\frac{1}{100} \right )-\left ( \frac{1}{11}+\frac{1}{12}+...+\frac{1}{100} \right )-\left (\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110} \right )$
$\Rightarrow A=\left ( 1+\frac{1}{2}+...+\frac{1}{10} \right )-\left (\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110} \right )$
$\Rightarrow A=\frac{1}{1}-\frac{1}{100}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}$
Thay vào phương trình, ta có:
$\left ( \frac{1}{1}-\frac{1}{100}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110} \right )x=10\left ( \frac{1}{1}-\frac{1}{100}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110} \right )$
$\Leftrightarrow x=10$
Tìm x, biết: \(\left ( \frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110} \right ).x = \frac{1}{1.11}+\frac{1}{1.12}+...+\frac{1}{100.110}\)- Trường Toán Trực tuyến Pitago – Giải pháp giúp em học toán vững vàng!
a) Vì (2x - 5)2000 và (3y + 4)2002 đều có số mũ là chẵn => (2x - 5)2000 \(\ge\) 0; (3y + 4)2002 \(\ge\) 0
Mà tổng trên lại \(\le\) 0
=> (2x - 5)2000 = (3y + 4)2002 = 0
=> 2x - 5 = 3y + 4 = 0
=> x = 2,5; y = \(\frac{-4}{3}\)
b) x = 18 - 0,8 : \(\frac{1,5}{\frac{3}{2}.\frac{4}{10}.\frac{50}{2}}\)+ \(\frac{1}{4}\)+ \(\frac{1+0,5.4}{6-\frac{46}{23}}\)
= 18 - \(\frac{8}{10}:\frac{1,5}{15}+\frac{1}{4}+\frac{3}{4}\)
= \(18-8+1=11\)