\(\frac{9}{11};\frac{7}{9};\frac{5}{7};\frac{3}{5};\frac{1}{3}\)

hok tốt ~

ta thấy : 1 - 1/3 = 2/3 , 1 - 3/5 = 2/5 , 1 - 5/7 = 2/7 , 1-7/9 = 2/9 , 1-9/11 = 2/11  

mà : 2/3 > 2/5 > 2/7 > 2/9 > 2/11 nên 1/3 < 1/5 < 1/7 < 1/9 < 1/11

`6/7 . 8/13 +6/7 . 9/13+3/13 . 6/7`

`=6/7 . (8/13+9/13+3/13)`

`=6/7 . 20/13`

`=120/91`

\(\dfrac{6}{7}.\dfrac{8}{13}+\dfrac{6}{7}.\dfrac{9}{13}+\dfrac{3}{13}.\dfrac{6}{7}\)

\(=\dfrac{6}{7}.\left(\dfrac{8}{13}+\dfrac{9}{13}+\dfrac{3}{13}\right)\)

\(=\dfrac{6}{7}.\left(\dfrac{8+9+3}{13}\right)\)

\(=\dfrac{6}{7}.\dfrac{20}{13}\)

\(=\dfrac{6.20}{7.13}\)

\(=\dfrac{120}{91}\)

1/-2000

2/4749

3/-1000

4/800

chúc bn hk tốt

có cần cóp nguyên bài thế k

=1 nha ban

Bài 22:

1: \(\sqrt{3-\sqrt5}=\frac{\sqrt{6-2\sqrt5}}{\sqrt2}\)

\(=\frac{\sqrt{\left(\sqrt5-1\right)^2}}{\sqrt2}=\frac{\sqrt5-1}{\sqrt2}=\frac{\sqrt{10}-\sqrt2}{2}\)

2: \(\sqrt{7+3\sqrt5}\)

\(=\frac{\sqrt{14+6\sqrt5}}{\sqrt2}\)

\(=\frac{\sqrt{\left(3+\sqrt5\right)^2}}{\sqrt2}=\frac{3+\sqrt5}{\sqrt2}=\frac{3\sqrt2+\sqrt{10}}{2}\)

3: \(\sqrt{9+\sqrt{17}}-\sqrt{9-\sqrt{17}}-2\)

\(=\frac{1}{\sqrt2}\left(\sqrt{18+2\sqrt{17}}-\sqrt{18-2\sqrt{17}}\right)-2\)

\(=\frac{1}{\sqrt2}\left(\sqrt{\left(\sqrt{17}+1\right)^2}-\sqrt{\left(\sqrt{17}-1\right)^2}\right)-2\)

\(=\frac{1}{\sqrt2}\left(\sqrt{17}+1-\sqrt{17}+1\right)-2=\frac{2}{\sqrt2}-2=\sqrt2-2\)

Bài 26:

1: \(\left|3-2x\right|=2\sqrt5\)

=>\(\left[\begin{array}{l}2x-3=2\sqrt5\\ 2x-3=-2\sqrt5\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=3+2\sqrt5\\ 2x=3-2\sqrt5\end{array}\right.\Rightarrow x=\frac{3\pm2\sqrt5}{2}\)

2: \(\sqrt{x^2}=12\)

=>|x|=12

=>x=12 hoặc x=-12

3: \(\sqrt{x^2-2x+1}=7\)

=>\(\sqrt{\left(x-1\right)^2}=7\)

=>|x-1|=7

=>\(\left[\begin{array}{l}x-1=7\\ x-1=-7\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=-6\end{array}\right.\)

`14/9:8/3=14/9xx3/8=[2xx7xx3]/[3xx3xx4xx2]=7/12`

`5/4+9/5xx5/4=5/4+9/4=14/4=7/2`

14/9:8/3=7/12
5/4+9/5x5/4=5/4+9/4=14/4=7/2