Tìm x , biết (\(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110}\)).x = \(\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)
Tìm x biết: \(\left(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110}\right).x=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)
Tìm x bít
\(\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}.x=\frac{1}{1.11}+\frac{1}{2.12}+\frac{1}{3.13}+...+\frac{1}{100.110}\)
\(A=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103+...}+\frac{1}{10.110}\)
\(A=\frac{1}{100}(\frac{100}{1.101}+\frac{100}{2.102}+\frac{100}{3.103}+...+\frac{100}{10.110})\)
\(A=\frac{1}{100}(\frac{1}{1}-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110})\)
\(A=\frac{1}{100}((\frac{1}{1}+\frac{1}{2}+...+\frac{1}{10})-(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}))\) ok?
\(B=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)
\(B=\frac{1}{10}(\frac{10}{1.11}+\frac{10}{2.12}+...+\frac{10}{100.110})\)
\(B=\frac{1}{10}(\frac{1}{1}-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110})\)
\(B=\frac{1}{10}((\frac{1}{1}+\frac{1}{2}+...+\frac{1}{100})-(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{110}))\)=\(\frac{1}{10}((\frac{1}{1}+\frac{1}{2}+...+\frac{1}{10})-(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}))\)
B=10A
A.x=10A suy ra x=10
gõ xong mém xỉu. :)
tìm x
\(\left(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110}\right).x=\frac{1}{100.110}+\frac{1}{99.109}+...+\frac{1}{2.12}+1.11\)
Vì gõ trên Hoc24 khá lâu nên mình gửi hình ảnh cho lẹ
Tìm x , bíÊt:
\(\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}x=\frac{1}{1.11}+\frac{1}{2.12}+\frac{1}{3.13}+...+\frac{1}{100.110}\)
Tìm x:
( \(\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}\) ) . x = \(\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)
Bài 1 : Tìm x biết : ( Giải rõ ràng => like )
\(\left(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110}\right)x=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)
khi ko mún tích thì tích 1 tích
khi mún tích thì tích 50 tích
Giải phương trình :\(\left(\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}\right).x=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)
Câu hỏi của Huỳnh Ngọc Cẩm Tú - Toán lớp 6 - Học toán với OnlineMath
Tìm số nguyên x,biết:
\(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110}\left(x\right)=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.101}\)
Tìm x
:\(\left(\frac{2011}{1.11}+\frac{2011}{2.12}+...+\frac{2011}{100.110}\right)=\frac{2012}{1.101}+\frac{2012}{2.102}+...+\frac{2012}{10.110}\)