tìm X phải không bạn!

vậy thì x sẽ bằng 0 hoặc 3

Tập hợp các chữ cái trong từ "TOÁN HỌC"là:

B=\(\left\{T;O;A;N;H;C\right\}\)

-40.(-8).7.5

=-40.(-8).5.7

=-40.(-40).7

=1600.7=11200

\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{43.46}\)

\(S=\dfrac{3}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{43.46}\right)\)

Ta thấy:

\(\dfrac{3}{1.4}=1-\dfrac{1}{4};\dfrac{3}{4.7}=\dfrac{1}{4}-\dfrac{1}{7};\dfrac{3}{7.10}=\dfrac{1}{7}-\dfrac{1}{10};\)

\(...;\dfrac{3}{43.46}=\dfrac{1}{43}-\dfrac{1}{46}\)

\(\Rightarrow S=1\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{43}-\dfrac{1}{46}\right)\)

\(\Rightarrow S=1\left(1-\dfrac{1}{46}\right)\)

\(\Rightarrow S=1.\dfrac{45}{46}=\dfrac{45}{46}\)

\(M=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{997.999}\)

Ta thấy:

\(\dfrac{1}{1.3}=1-\dfrac{1}{3};\dfrac{1}{3.5}=\dfrac{1}{3}-\dfrac{1}{5};\dfrac{1}{5.7}=\dfrac{1}{5}-\dfrac{1}{7};\)

\(...;\dfrac{1}{997.999}=\dfrac{1}{997}-\dfrac{1}{999}\)

\(\Rightarrow M=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\)

\(\dfrac{1}{997}-\dfrac{1}{999}\)

\(\Rightarrow M=1-\dfrac{1}{999}=\dfrac{998}{999}\)

Vậy M=\(\dfrac{998}{999}\)

đề bài đâu rồi Đỗ Thanh Nga???

\(B=\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{5.7}+...+\dfrac{5}{59.61}\)

\(B=\dfrac{5}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{59.61}\right)\)

Nhận xét :

\(\dfrac{2}{1.3}=1-\dfrac{1}{3};\dfrac{2}{3.5}=\dfrac{1}{3}-\dfrac{1}{5};\dfrac{2}{5.7}=\dfrac{1}{5}-\dfrac{1}{7};...;\)

\(\dfrac{2}{59.61}=\dfrac{1}{59}-\dfrac{1}{61}\)

\(\Rightarrow B=\dfrac{5}{2}\left(1-\dfrac{1}{61}\right)\)

\(\Rightarrow B=\dfrac{5}{2}.\dfrac{60}{61}=\dfrac{150}{61}\)

Vậy B=\(\dfrac{150}{61}\)

bn vô nick này nha:

https://hoc24vn/hoi-dap/question/272030.html