Các câu hỏi tương tự
\(\left(\frac{1}{25.26}+\frac{1}{26.27}+........+\frac{1}{29.30}\right).150+103:\left[1,03+\left(x+1\right)\right]=22\)
\(\left(\frac{1}{25.21}+\frac{1}{26.27}+........+\frac{1}{29.30}\right).150+103\left[1,03+\left(x+1\right)\right]=22\)
\(\left(\frac{1}{25.21}+\frac{1}{26.27}+...........+\frac{1}{29.30}\right).150+103\div\left[1.03+\left(x+1\right)\right]=22\)
Tĩm x:
1,\(\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+5}\)
2,\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}+......+x+\frac{1}{23.25}\right)=11.x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
Giải chi tiết hộ mình nha mình tik cho
\(A=\left(1+\frac{1}{2}\right)x\left(1+\frac{1}{3}\right)x\left(1+\frac{1}{4}\right)x...x\left(1+\frac{1}{98}\right)x\left(1+\frac{1}{99}\right)\)
\(A=\)
Tìm \(x\)sao cho:
\(\left(x+\frac{1}{1\cdot3}\right)+\left(x+\frac{1}{3\cdot5}\right)+\left(x+\frac{1}{5\cdot7}\right)+...+\left(x+\frac{1}{23\cdot25}\right)=11\cdot x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
x-left(frac{20}{11.13}+frac{20}{13.15}+frac{20}{15.17}+...+frac{20}{53.55}right)frac{3}{11}frac{7}{4}x.left(frac{33}{12}+frac{3333}{2020}+frac{333333}{303030}+frac{33333333}{42424242}right)22frac{1}{3}+frac{1}{6}+frac{1}{10}+frac{1}{15}+...+frac{2}{xleft(x-1right)}frac{2007}{2009}left(frac{1}{4}+frac{1}{8}+frac{1}{16}+frac{1}{32}+frac{1}{64}+frac{1}{128}+frac{1}{256}right).x1
Đọc tiếp
x-\(\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(\frac{7}{4}x.\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)=22\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{2}{x\left(x-1\right)}=\frac{2007}{2009}\)
\(\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\right).x=1\)
C=\(\left(1+\frac{1}{2}\right)x\left(1+\frac{1}{3}\right)x\left(1+\frac{1}{4}\right)x......x\left(1+\frac{1}{2015}\right)\)
\(C=\left(\frac{1}{2}+1\right)x\left(\frac{1}{3}+1\right)x\left(\frac{1}{4}+1\right)x...x\left(\frac{1}{99}+1\right)\)