Ta có:

\(A\left(x\right)=x^2-5mx+10m-4\)

\(\Leftrightarrow\Delta=\left(5m-4\right)^2\) \(\Leftrightarrow\left\{{}\begin{matrix}x_1=5m-2\\x_2=2\end{matrix}\right.\)

Ta có 2 trường hợp:

Trường hợp 1: Nếu \(x_1=2x_2\)

\(\Leftrightarrow5m-2=4\Leftrightarrow5m=6\Leftrightarrow m=\dfrac{6}{5}\)

Trường hợp 2: Nếu \(x_2=2x_1\)

\(\Leftrightarrow2\left(5m-2\right)=2\Leftrightarrow5m-2=1\)

\(\Leftrightarrow5m=3\Leftrightarrow m=3\div5=\dfrac{3}{5}\)

Vậy \(m=\dfrac{3}{5}\) hoặc \(m=\dfrac{6}{5}\) là các giá trị cần tìm

I phoned the company, who _____ me that my goods had been dispatched.

A. guaranteed       

B. assured    

C. reassured 

D. confirmed

Dễ thấy:

\(\dfrac{1}{1^2}=\dfrac{1}{1.1}=\dfrac{1}{1}=1\)

\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)

\(.....................\)

\(\dfrac{1}{50^2}=\dfrac{1}{50.50}< \dfrac{1}{49.50}\)

Cộng các vế trên với nhau ta được:

\(A< 1+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)\)

\(\Rightarrow A< 1+\) \(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

\(\Rightarrow A< 1+\left(1-\dfrac{1}{50}\right)\)

\(\Rightarrow A< 1+1-\dfrac{1}{50}=2-\dfrac{1}{50}\)

Mà \(2-\dfrac{1}{50}< 2\Leftrightarrow A< 2\)

Vậy \(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}< 2\) (Đpcm)