Question

Tìm x :

a) (x-2)³=(x-2)²

b) \(\dfrac{3x}{2.5}+\dfrac{3x}{5.8}+...+\dfrac{3x}{11.14}=\dfrac{1}{21}\)

Answer 1

a, dễ, tự làm

b, \(\dfrac{3x}{2.5}+\dfrac{3x}{5.8}+.........+\dfrac{3x}{11.14}=\dfrac{1}{21}\)

\(\Leftrightarrow x\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+.........+\dfrac{3}{11.14}\right)=\dfrac{1}{21}\)

\(\Leftrightarrow x\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+.....+\dfrac{1}{11}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)

\(\Leftrightarrow x\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)

\(\Leftrightarrow x.\dfrac{3}{7}=\dfrac{1}{21}\)

\(\Leftrightarrow x=\dfrac{1}{9}\)

Vậy ...

Answer 2

a) (x-2)³ = (x-2)²

<=> (x-2)³-(x-2)² = 0

<=> (x-2)²(x-2-1) = 0

<=> \(\left\{{}\begin{matrix}\left(x-2\right)^2=0\\x-3=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

b) \(\dfrac{3x}{2.5}+\dfrac{3x}{5.8}+...+\dfrac{3x}{11.14}=\dfrac{1}{21}\)

<=> \(x\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{11.14}\right)=\dfrac{1}{21}\)

<=> \(x\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)

<=> \(x=\dfrac{1}{21}:\dfrac{3}{7}\)

<=> \(x=\dfrac{1}{9}\)

Answer 3

a) \(\left(x-2\right)^3=\left(x-2\right)^2\)

=> x - 2 = 0 hoặc x - 2 = 1

Với x - 2 = 0

x = 0+2

x = 2

Với x - 2 = 1

x = 1 + 2

x = 3

Vậy x \(\in\){2;3}

b) \(\dfrac{3x}{2.5}+\dfrac{3x}{5.8}+...+\dfrac{3x}{11.14}=\dfrac{1}{21}\)

x .\(\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+....+\dfrac{3}{11.14}\right)=\dfrac{1}{21}\)

x . \(\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+....+\dfrac{1}{11}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)

x . \(\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)

x . \(\dfrac{3}{7}=\dfrac{1}{21}\)

x \(=\dfrac{1}{21}:\dfrac{3}{7}=\dfrac{1}{9}\)

Answer 4

a) (x-2)³=(x-2)² \(\Leftrightarrow\) (x-2)³-(x-2)²=0 \(\Leftrightarrow\) (x-2)²(x-2-1)=0 \(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\) b) \(\dfrac{3x}{2.5}+\dfrac{3x}{5.8}+.....+\dfrac{3x}{11.14}=\dfrac{1}{21}\) \(\Leftrightarrow x.\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{11.14}\right)=\dfrac{1}{21}\) \(\Leftrightarrow x.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{11}-\dfrac{1}{14}\right)=\dfrac{1}{21}\) \(\Leftrightarrow x.\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\) \(\Leftrightarrow x.\dfrac{3}{7}=\dfrac{1}{21}\) \(\Leftrightarrow x=\dfrac{1}{9}\)