1, \(x^2-\dfrac{2}{5}x< 0\)

\(\Rightarrow x\left(x-\dfrac{2}{5}\right)< 0\)

\(\Rightarrow\left\{{}\begin{matrix}x< 0\\x-\dfrac{2}{5}>0\end{matrix}\right.\) ( loại ) hoặc \(\left\{{}\begin{matrix}x>0\\x-\dfrac{2}{5}< 0\end{matrix}\right.\)

\(\Rightarrow0< x< \dfrac{2}{5}\)

Vậy...

b, \(\dfrac{x^2-1}{x^2}< 0\)

\(\Rightarrow\left\{{}\begin{matrix}x^2-1< 0\\x^2>0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x^2-1>0\\x^2< \end{matrix}\right.\) ( loại )

\(\Rightarrow0< x^2< 1\Rightarrow0< x< 1\)

Vậy...

a, \(a^2+b^2-2ab=\left(a-b\right)^2\ge0\)

b, Áp dụng bất đẳng thức AM-GM có:
\(\dfrac{a^2+b^2}{2}\ge\dfrac{2ab}{2}=ab\)

Dấu " = " khi a = b = 1

Ta có: \(\dfrac{a}{b}=\dfrac{b}{c}\Rightarrow\dfrac{a^2}{b^2}=\dfrac{b^2}{c^2}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{a^2}{b^2}=\dfrac{b^2}{c^2}=\dfrac{a^2+b^2}{b^2+c^2}\) (1)

\(\dfrac{a^2}{b^2}=\dfrac{a}{b}.\dfrac{b}{c}=\dfrac{a}{c}\) (2)

Từ (1), (2) \(\Rightarrow\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a}{c}\Rightarrowđpcm\)

Đề của bạn sai nhé!

\(M=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{2003.2005}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2003.2005}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2003}-\dfrac{1}{2005}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{2005}\right)\)

\(=\dfrac{1}{2}.\dfrac{2004}{2005}=\dfrac{1002}{2005}\)

\(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)

\(\Leftrightarrow-2\left(2x-5\right)=0\)

\(\Leftrightarrow x=\dfrac{5}{2}\)

Vậy...

Bài 1: Nhân chéo

Bài 2:

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)

\(\Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}\)

\(\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)

\(\Rightarrowđpcm\)

\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{n}\left(1+2+...+n\right)\)

\(=\dfrac{2}{2}+\dfrac{2.3}{2.2}+\dfrac{3.4}{3.2}+...+\dfrac{n\left(n+1\right)}{2n}\)

\(=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{n+1}{2}\)

\(=\dfrac{1}{2}\left(1+2+...+n\right)\)

\(=\dfrac{n\left(n+1\right)}{4}\)

P/s: \(1+...+n=\dfrac{n\left(n+1\right)}{2}\)

\(x^4-9x^3+x^2-9x=x^3\left(x-9\right)+x\left(x-9\right)\)

\(=\left(x^3+x\right)\left(x-9\right)=x\left(x^2+1\right)\left(x-9\right)\)