\(\dfrac{9}{8}>\dfrac{3}{9};\dfrac{7}{9}=\dfrac{7}{9}\)

>;=

9/8 >3/9

7/9=7/9

Nếu tử số = nhau thì ta so sánh mẫu số. Mẫu số nào bé hơn thì lớn hơn

-9/7^4<-9/7^9

Cho mk

Bn có thể giải thích tại sao -9/\(7^4\)<-9/\(7^9\)đc ko

\(7\dfrac{4}{5}và9\dfrac{1}{2}\\ Tacó:7< 9\\ \Rightarrow7\dfrac{4}{5}< 9\dfrac{1}{2}\\ 7\dfrac{1}{6}và3\dfrac{4}{5}\\ Tacó:7>3\\ \Rightarrow7\dfrac{1}{6}>3\dfrac{4}{5}\)

Câu cuối không phải hỗn số

Lời giải:
a. $\frac{3}{-7}=\frac{-27}{63}$

$\frac{-5}{9}=\frac{-35}{63}$

Do $\frac{27}{63}< \frac{35}{63}$ nên $\frac{-27}{63}> \frac{-35}{63}$

$\Rightarrow \frac{3}{-7}> \frac{-5}{9}$

---------

b.

$-0,625=\frac{-625}{1000}=\frac{-5}{8}=\frac{-125}{200}$

$\frac{-19}{50}=\frac{-76}{200}> \frac{-125}{200}$

$\Rightarrow -0,625> \frac{-19}{50}$

c.

$-2\frac{5}{9}=-(2+\frac{5}{9})=\frac{-23}{9}=-(\frac{-23}{-9})$

\(\dfrac{1}{4444}< 1,\dfrac{3}{7}< 1,\dfrac{9}{5}>1,\dfrac{7}{3}>1,\dfrac{14}{15}< 1,\dfrac{16}{16}=1,\dfrac{14}{11}>1\)

1/4 < 1

3/7 < 1

9/5 > 1

7/3 > 1

14/15 < 1

16/16 = 1

14/11 >1

=-2/13

\(=\left(\dfrac{2}{9}+\dfrac{7}{9}\right)+\left(\dfrac{3}{7}+\dfrac{4}{7}\right)-\dfrac{2}{13}\)

\(=1+1-\dfrac{2}{13}=2-\dfrac{2}{13}=\dfrac{26}{13}-\dfrac{2}{13}=\dfrac{24}{13}\)

= ( 2/9 + 7/9 ) + ( 3/7 + 4/7 ) - 2/13

= 1 + 1 - 2/13

= 2-2/13

= 26/13 - 2/13 = 23/13

đấy nhá

b) Ta có: \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)

\(\Leftrightarrow\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}=\dfrac{20}{12}+\dfrac{9}{12}=\dfrac{29}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{12}\\x=-\dfrac{29}{12}\end{matrix}\right.\)

c) Ta có: \(\left|2x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)

\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{1}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=\dfrac{1}{6}\\2x-\dfrac{1}{3}=\dfrac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1}{2}\\2x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{1}{12}\end{matrix}\right.\)

3: Ta có: \(\sqrt{4x+1}=x+1\)

\(\Leftrightarrow x^2+2x+1=4x+1\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

4: Ta có: \(2\sqrt{x-1}+\dfrac{1}{3}\sqrt{9x-9}=15\)

\(\Leftrightarrow3\sqrt{x-1}=15\)

\(\Leftrightarrow x-1=25\)

hay x=26

5: Ta có: \(\sqrt{4x^2-12x+9}=7\)

\(\Leftrightarrow\left|2x-3\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)