\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{100}\right)=\frac{1}{3}\left(\frac{99}{100}\right)=\frac{33}{100}\)
\(\left(3\cdot x-0,8\right):x+14,5=15\)
1,2\(\cdot\)(\(\frac{2,4\cdot x-0,23}{x}\)\(-0,05=1,44\)
\(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{97\cdot100}=\frac{0,33\cdot x}{2009}\)
\(x\)-\(\frac{20}{11\cdot13}-\frac{20}{13\cdot15}-...-\frac{20}{53\cdot55}=\frac{3}{11}\)
\(\left(\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\right)\)\(\cdot x=\frac{5}{14}\)
Các bạn ơi trả lời giùm mình với nhé, cần gấp.
Tính tổng A=\(\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+...+\frac{2}{97\cdot100}\)
\(A=\frac{^{3^2}}{1\cdot4}+\frac{3^2}{4\cdot7}+\frac{3^2}{7\cdot10}+...+\frac{3^2}{97\cdot100}\) Tính Nhanh
Bài 2 : Tính tổng A :
\(A=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+...+\frac{2}{97\cdot100}\)
A=\(\frac{^{3^2}}{1\cdot4}\)+ \(\frac{3^2}{4\cdot7}\)+ \(\frac{3^2}{7\cdot10}\)+ \(\frac{3^2}{10\cdot13}\)+\(\frac{3^2}{13\cdot16}\)+......+ \(\frac{3^2}{97\cdot100}\)
CHỨNG TỎ RẰNG :
\(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{67\cdot70}< 1\)
\(20\%x+\frac{2}{5}x=x-4\)
\(\frac{\frac{2}{3}+\frac{2}{7}\cdot\frac{1}{14}}{-1-\frac{3}{7}+\frac{3}{28}}\)
\(\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10}{3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20}\)
giải hội mik mới cách làm lun nha
chứng minh rằng \(\frac{91}{1\cdot4}+\frac{91}{4\cdot7}+\frac{91}{7\cdot10}+......+\frac{91}{88\cdot91}\)=30
Chứng tỏ rằng :
\(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+...\frac{1}{2010\cdot2013}<\frac{1}{3}\)
