Tìm x
a. \(x-\dfrac{20}{11.13}-\dfrac{20}{13.15}-\dfrac{20}{15.17}-...-\dfrac{20}{53.55}\)
b. \(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
Tìm x, biết:
\(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=\dfrac{-1}{4}\)
Rút gọn
a) B=(\(1-\dfrac{1}{2}\))x\(\left(1-\dfrac{1}{3}\right)x\left(1-\dfrac{1}{4}\right)x....\left(1-\dfrac{1}{20}\right)\)
b) tính nhanh
P=\(\dfrac{\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}}{\dfrac{5}{12}+1-\dfrac{6}{11}}\)
\(\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)-x=\dfrac{-19}{24}\)
1)tìm x\(\in\)Q biết:
a)\(\left(5\cdot x+1\right)^2=\dfrac{36}{49}\)
2)tìm x,y biết:
a)\(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)
b)\(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{40}\le0\)
Bài 1. Tính giá trị biểu thức
125%.\((\dfrac{-1}{2})^2\):\((1\dfrac{5}{16}-1,5)+2008^0\)
\(\dfrac{2}{3.5}+\dfrac{2}{5.3}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)
Bài 2. Tìm x, biết
x:\(3\dfrac{1}{15}=1\dfrac{1}{12}\)
\(\dfrac{3}{4}.x=27\)
\(x.\dfrac{1}{4}=\dfrac{5}{8}.\dfrac{2}{3}\)
\(\left(x+\dfrac{1}{4}\right):\dfrac{3}{7}=\dfrac{1}{6}\)
\(\dfrac{3}{4}-2.|2x-\dfrac{2}{3}|=2\)
\(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{2}\)
Chứng minh rằng :
a)\(\dfrac{1}{x}\)-\(\dfrac{1}{x+a}=\dfrac{a}{x\left(x+a\right)}\)
b)\(\dfrac{1}{x\left(x+1\right)}-\dfrac{1}{\left(x+1\right)\left(x+2\right)}=\dfrac{2}{x\left(x+1\right)\left(x+2\right)}\)
c)\(\dfrac{1}{x\left(x+1\right)\left(x+2\right)}-\dfrac{1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\dfrac{3}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
Chứng minh rằng :
\(\dfrac{7}{12}< \dfrac{1}{21}+\dfrac{1}{20}+...+\dfrac{1}{40}< \dfrac{5}{6}\)
( Chú ý : \(\dfrac{1}{21}+\dfrac{1}{20}\)chứ k phải \(\dfrac{1}{21}+\dfrac{1}{22}\) nha )
Câu1 : Tìm x biết :
a) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
b)\(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)
c)\(\dfrac{x-1}{x+2}=\dfrac{x-2}{x+3}\)