80/103 = 0,77

8/10 = 0,8

9/10 = 0,9

90/99 = 0,909

0,95 ; 90/99 ; 9/10 ; 8/10 ; 80/103

rút gọn : lớp 9 nha

\(A=\frac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)\left(1+\sqrt{2}\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\frac{\sqrt{2017}-\sqrt{2016}}{\left(\sqrt{2017}-\sqrt{2016}\right)\left(\sqrt{2017}+\sqrt{2016}\right)}.\)

\(A=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2017}-\sqrt{2016}=\sqrt{2017}-1\)

\(\frac{x+2}{x+3}< \frac{x+4}{x+5}\)

<=> \(\left(x+2\right)\left(x+5\right)< \left(x+3\right)\left(x+4\right)\)

 <=> \(x^2+7x+10< x^2+7x+12\)

<=> \(x^2-x^2+7x-7x+10-12< 0\)

???

x = -3 ; x = -4/8/9

\(2x^2-3x=0\)

\(\Leftrightarrow x\left(2x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}}\)

\(2x^2-3x=0\Leftrightarrow x\left(2x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\2x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}}\)

Vậy \(S=\left\{0;\frac{3}{2}\right\}\)

4x - 4 - x +1 = 0

3x - 3 =0

3x=3

x=1

\(4\left(x-1\right)-x+1=0\)

\(\Leftrightarrow4\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow3\left(x-1\right)=0\)

\(\Rightarrow x-1=0\Leftrightarrow x=0\)