Question

[Trâm Nguyễn]

Tìm x:

a) 3.(2x-\(\dfrac{1}{2}\))³ + \(\dfrac{1}{9}\)=1 

b) 2.(x-1/2)² + 1\(\dfrac{1}{3}\) = 2\(\dfrac{2}{9}\)

c) \(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+\(\dfrac{1}{15}\)+....+\(\dfrac{2}{x.\left(x+1\right)}\)=\(\dfrac{99}{101}\)

GIÚP EM VS Ạ

Answer 1

[Kiều Vũ Linh]

a) \(3.\left(2x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=1\)

\(3.\left(2x-\dfrac{1}{2}\right)^3=1-\dfrac{1}{9}\)

\(3.\left(2x-\dfrac{1}{2}\right)^3=\dfrac{8}{9}\)

\(\left(2x-\dfrac{1}{2}\right)^3=\dfrac{8}{9}:3\)

\(\left(2x-\dfrac{1}{2}\right)^3=\dfrac{8}{27}\)

\(2x-\dfrac{1}{2}=\dfrac{2}{3}\)

\(2x=\dfrac{2}{3}+\dfrac{1}{2}\)

\(2x=\dfrac{7}{6}\)

\(x=\dfrac{7}{6}:2\)

\(x=\dfrac{7}{12}\)

b) \(2.\left(x-\dfrac{1}{2}\right)^2+1\dfrac{1}{3}=2\dfrac{2}{9}\)

\(2\left(x-\dfrac{1}{2}\right)^2+\dfrac{4}{3}=\dfrac{20}{9}\)

\(2\left(x-\dfrac{1}{2}\right)^2=\dfrac{20}{9}-\dfrac{4}{3}\)

\(2\left(x-\dfrac{1}{2}\right)^2=\dfrac{8}{9}\)

\(\left(x-\dfrac{1}{2}\right)^2=\dfrac{8}{9}:2\)

\(\left(x-\dfrac{1}{2}\right)^2=\dfrac{4}{9}\)

\(x-\dfrac{1}{2}=-\dfrac{2}{3}\) hoặc \(x-\dfrac{1}{2}=\dfrac{2}{3}\)

*) \(x-\dfrac{1}{2}=-\dfrac{2}{3}\)

\(x=-\dfrac{2}{3}+\dfrac{1}{2}\)

\(x=-\dfrac{1}{6}\)

*) \(x-\dfrac{1}{2}=\dfrac{2}{3}\)

\(x=\dfrac{2}{3}+\dfrac{1}{2}\)

\(x=\dfrac{7}{6}\)

Vậy \(x=-\dfrac{1}{6};x=\dfrac{7}{6}\)

c) \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{99}{101}\)

\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{99}{101}\)

\(2.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{99}{101}\)

\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{99}{101}:2\)

\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{99}{202}\)

\(\dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{99}{202}\)

\(\dfrac{1}{x+1}=\dfrac{1}{101}\)

\(x+1=101\)

\(x=101-1\)

\(x=100\)