4/1.4+4/4.7+4/7.10+4/10.13
= 4/3(3/1.4+3/4.7+3/7.10+3/10.13)
=4/3(1/1-1/4+1/4-1/7+1/7-1/10+1/10-1/13)
=4/3(1/1-1/13)
=4/3.12/13
=16/13
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Các câu hỏi tương tự
Tính nhanh
\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{97.100}\)
Tính :
\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\frac{3^2}{10.13}+\frac{3^2}{13.16}+...+\frac{3^2}{97.100}\)
Tìm A :
A = \(\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\frac{3^2}{10.13}+...+\frac{3^2}{97.100}\)
Tính nhanh: \(\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\frac{3^2}{10.13}+\frac{3^2}{13.16}+...+\frac{3^2}{97.100}\)
Bài 3 Tính giá trị biểu thức\(\left(1_{ },5\right).\frac{-2}{3}+\left(2,5-\frac{3}{4}\right):1\frac{3}{4}\)
B=\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{100.103}\)
tinh A
A=\(\frac{3^2}{1.4}\)+\(\frac{3^2}{4.7}\)+\(\frac{3^2}{7.10}\)+\(\frac{3^2}{10.13}\)+.......+\(\frac{3^2}{97.100}\)
Cho mk hỏi nha ( Đề thi vào lớp 7 )
Thông cảm năm nay mk mới vào lớp 6
Tính :
\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\frac{3^2}{10.13}+...+\frac{3^2}{97.100}\)
\(\text{A=}\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\frac{3^2}{10.13}+\frac{3^2}{13.16}+...+\frac{3^2}{97.100}\)
Loi giai day du gium minh nha may ban
minh dang gap
Bài 2: a.Tínhfrac{1}{4}-frac{1}{7}frac{3}{4.7};_{ }_{ }frac{1}{7}-frac{1}{10}frac{3}{7.10};_{ }_{ }frac{1}{10}-frac{1}{13}frac{3}{10.13};_{ }_{ }frac{1}{13}-frac{1}{16}frac{3}{13.16};_{ }....._{ }frac{1}{x}-frac{1}{x+3}frac{3}{xleft(x+3right)}Qui luật: frac{m}{aleft(a+mright)}frac{1}{a}-frac{1}{a+m}b. A frac{1}{4.7}+frac{1}{7.10}+frac{1}{10.13}+...+frac{1}{xleft(x+3right)} 3A frac{3}{4.7}+frac{3}{7.10}+frac{3}{10.13}+...+frac{3}{xleft(x+3right)} 3A frac{1}{4}-frac{1}{7}+frac{1}{7}-frac{1}{...
Đọc tiếp
Bài 2: a.Tính
\(\frac{1}{4}-\frac{1}{7}=\frac{3}{4.7};_{ }_{ }\frac{1}{7}-\frac{1}{10}=\frac{3}{7.10};_{ }_{ }\frac{1}{10}-\frac{1}{13}=\frac{3}{10.13};_{ }_{ }\frac{1}{13}-\frac{1}{16}=\frac{3}{13.16};_{ }....._{ }\frac{1}{x}-\frac{1}{x+3}=\frac{3}{x\left(x+3\right)}\)Qui luật: \(\frac{m}{a\left(a+m\right)}=\frac{1}{a}-\frac{1}{a+m}\)
b. A = \(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{x\left(x+3\right)}\)
3A = \(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{x\left(x+3\right)}\)
3A = \(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{x}-\frac{1}{x+3}\)
3 A = \(\frac{1}{4}-\frac{1}{x+3}\)
\(\frac{63}{764}.3=\frac{1}{4}-\frac{1}{x+3}\)
\(\frac{189}{764}=\frac{1}{4}-\frac{1}{x+3}\)
\(\frac{1}{382}=\frac{1}{x+3}_{ }_{ }_{ }=>x=382-3=379\)