a) So với học sinh của cả lớp, 2 tổ chiếm:

\(\frac24\times100\%=50\%\)

b) Số học sinh của 2 tổ là:

\(36\times50\%=18\) (học sinh)

9hs

a;  \(\dfrac{4}{27}\) = \(\dfrac{4\times7}{27\times7}\) = \(\dfrac{28}{189}\)

     \(\dfrac{15}{63}\) = \(\dfrac{15\times3}{63\times3}\) = \(\dfrac{45}{189}\)

     \(\dfrac{28}{189}\) < \(\dfrac{45}{189}\)

     - \(\dfrac{28}{189}\) > - \(\dfrac{45}{189}\)

Vậy - \(\dfrac{4}{27}\) > \(\dfrac{15}{-63}\) 

b;   \(\dfrac{13}{15}\) = \(\dfrac{13\times9}{15\times9}\) =  \(\dfrac{117}{135}\)

     \(\dfrac{9}{11}\) = \(\dfrac{9\times13}{11\times13}\) = \(\dfrac{117}{143}\)

     \(\dfrac{117}{135}\) > \(\dfrac{117}{143}\) 

Vậy \(\dfrac{13}{15}\) > \(\dfrac{9}{11}\) 

Tổng 3 phân số là:

\(\dfrac{31}{90}\times3=\dfrac{31}{30}\)

Ba lần phân số thứ hai bằng:

\(\dfrac{31}{30}+\dfrac{2}{15}-\dfrac{1}{6}=1\)

Phân số thứ nhất là:

\(1+\dfrac{1}{6}=\dfrac{7}{6}\)

Phân số thứ ba là:

\(1-\dfrac{2}{15}=\dfrac{13}{15}\)

Đáp số:...

\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\\\Leftrightarrow \frac{x+4}{2000}+\frac{x+3}{2001}-\frac{x+2}{2002}-\frac{x+1}{2003}=0\\\Leftrightarrow \left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)-\left(\frac{x+2}{2002}+1\right)-\left(\frac{x+1}{2003}+1\right)=0\\\Leftrightarrow \frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\\\Leftrightarrow (x+2024)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\\\Leftrightarrow x+2024=0(\text{vì }\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003} \ne0)\\\Leftrightarrow x=-2024\)

Vậy phương trình có 1 nghiệm duy nhất là $x=-2024$.

a: \(-\dfrac{19}{49}=\dfrac{-19\cdot47}{49\cdot47}=\dfrac{-893}{2303}\)

\(\dfrac{-23}{47}=\dfrac{-23\cdot49}{47\cdot49}=\dfrac{-1127}{2303}\)

mà -893>-1127

nên \(-\dfrac{19}{49}>-\dfrac{23}{47}\)

b: \(\dfrac{-5}{8}=\dfrac{-5\cdot5}{8\cdot5}=\dfrac{-25}{40}\)

\(\dfrac{7}{-10}=\dfrac{-7}{10}=\dfrac{-7\cdot4}{10\cdot4}=\dfrac{-28}{40}\)

mà -25>-28

nên \(-\dfrac{5}{8}>\dfrac{7}{-10}\)

c: \(\dfrac{24}{35}=\dfrac{24\cdot6}{35\cdot6}=\dfrac{144}{210};\dfrac{19}{30}=\dfrac{19\cdot7}{30\cdot7}=\dfrac{133}{210}\)

mà 144>133

nên \(\dfrac{24}{35}>\dfrac{19}{30}\)

\(\dfrac{1}{4}\times X=\dfrac{8}{5}-\dfrac{6}{2}\)

=>\(X\times\dfrac{1}{4}=\dfrac{8}{5}-3=\dfrac{8}{5}-\dfrac{15}{5}=-\dfrac{7}{5}\)

=>\(X=-\dfrac{7}{5}\times4=-\dfrac{28}{5}\)

\(6\times(15+5x)=300\\15+5x=300:6\\15+5x=50\\5x=50-15\\5x=35\\x=35:5\\x=7\)

\(6\times\left(15+5x\right)=300\)

\(15+5x=300:6\)

\(15+5x=50\)

\(5x=50-15\)

\(5x=35\)

\(x=7\)

\(-1,25=-\dfrac{125}{100}=-\dfrac{5}{4}\)