Sửa đề: 3x1⋅5+3x5⋅9+3x9⋅13+...+3x81⋅85=4153x1⋅5+3x5⋅9+3x9⋅13+...+3x81⋅85=415

⇔x⋅34(1−15+15−19+19−113+...+181−185)=415⇔x⋅34(1−15+15−19+19−113+...+181−185)=415

⇔x⋅6385=415⇔x⋅6385=415

hay x=68189

Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

\(\Leftrightarrow\dfrac{3x}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow\dfrac{3x}{4}\cdot\dfrac{84}{85}=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)

hay \(x=\dfrac{68}{189}\)

Vậy: \(x=\dfrac{68}{189}\)

Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

\(\Leftrightarrow\dfrac{3x}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{81\cdot85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)

hay \(x=\dfrac{68}{189}\)

Vậy: \(x=\dfrac{68}{189}\)

Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

\(\Leftrightarrow\dfrac{3x}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{81\cdot85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)

hay \(x=\dfrac{68}{189}\)

Vậy: \(x=\dfrac{68}{189}\)

Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

\(\Leftrightarrow\dfrac{3x}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow\dfrac{3x}{4}\cdot\dfrac{84}{85}=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)

hay \(x=\dfrac{68}{189}\)

Vậy: \(x=\dfrac{68}{189}\)

Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

\(\Leftrightarrow\dfrac{3x}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{81\cdot85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)

hay \(x=\dfrac{68}{189}\)

Vậy: \(x=\dfrac{68}{189}\)

Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

\(\Leftrightarrow\dfrac{3x}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow\dfrac{3x}{4}\cdot\dfrac{84}{85}=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)

hay \(x=\dfrac{68}{189}\)

Vậy: \(x=\dfrac{68}{189}\)

Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

\(\Leftrightarrow\dfrac{3x}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{81\cdot85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)

hay \(x=\dfrac{68}{189}\)

Vậy: \(x=\dfrac{68}{189}\)

\(\left(3x+4\right)^2-\left(3x-1\right)\left(3x+1\right)=9x^2+24x+16-9x^2+1=24x+17\)

Đặt \(24x+17=0\Leftrightarrow x=-\frac{17}{24}\)

b: \(3x^2-2x-1=0\)

=>\(3x^2-3x+x-1=0\)

=>\(\left(x-1\right)\left(3x+1\right)=0\)

=>\(\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

a: Bạn ghi lại đề đi bạn