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

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 \(\dfrac{24}{35}>\dfrac{19}{30}\)

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

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

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

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

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

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

\(15+5x=50\)

\(5x=50-15\)

\(5x=35\)

\(x=7\)

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

\(x^3+6x^2+11x+6\)

\(=x^3+x^2+5x^2+5x+6x+6\)

\(=x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2+5x+6\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)

a: \(-\dfrac{15}{21}=\dfrac{-15:3}{21:3}=\dfrac{-5}{7}=\dfrac{-5\cdot11}{7\cdot11}=\dfrac{-55}{77}\)

\(\dfrac{-36}{44}=\dfrac{-36:4}{44:4}=\dfrac{-9}{11}=\dfrac{-9\cdot7}{11\cdot7}=\dfrac{-63}{77}\)

mà -55>-63

nên \(-\dfrac{15}{21}>-\dfrac{36}{44}\)

b: \(-\dfrac{16}{30}=\dfrac{-16:2}{30:2}=\dfrac{-8}{15}=\dfrac{-8\cdot4}{15\cdot4}=\dfrac{-32}{60}\)

\(\dfrac{-35}{84}=\dfrac{-35:7}{84:7}=\dfrac{-5}{12}=\dfrac{-5\cdot5}{12\cdot5}=\dfrac{-25}{60}\)

mà -32<-25

nên \(-\dfrac{16}{30}< -\dfrac{35}{84}\)

c: \(\dfrac{-5}{91}=\dfrac{-5\cdot101}{91\cdot101}=\dfrac{-505}{9191}\)

mà -505<-501

nên \(-\dfrac{5}{91}< -\dfrac{501}{9191}\)

a) Rút gọn:

\(-\dfrac{15}{21}=-\dfrac{15:3}{21:3}=-\dfrac{5}{7}\)

\(-\dfrac{36}{44}=-\dfrac{36:4}{44:4}=-\dfrac{9}{11}\)

Quy đồng(MSC:77)

\(-\dfrac{5}{7}=-\dfrac{5.11}{7.11}=-\dfrac{55}{77}\\ -\dfrac{9}{11}=-\dfrac{9.7}{11.7}=-\dfrac{63}{77}\)

Nhận thấy: \(-\dfrac{55}{77}>-\dfrac{63}{77}\Rightarrow-\dfrac{15}{21}>-\dfrac{36}{44}\)

b) Rút gọn:

\(-\dfrac{16}{30}=-\dfrac{16:2}{30:2}=-\dfrac{8}{15}\\ -\dfrac{35}{84}=-\dfrac{35:7}{84:7}=-\dfrac{5}{12}\)

Quy đồng (MSC:60)

\(-\dfrac{8}{15}=-\dfrac{8.4}{15.4}=-\dfrac{32}{60}\\ -\dfrac{5}{12}=-\dfrac{5.5}{12.5}=-\dfrac{25}{60}\)

Nhận thấy: \(-\dfrac{32}{60}< -\dfrac{25}{60}\Rightarrow-\dfrac{16}{30}< -\dfrac{35}{84}\)

c) \(-\dfrac{5}{91}=-\dfrac{5.101}{91.101}=-\dfrac{505}{9191}< -\dfrac{501}{9191}\)

Mình hướng dẫn bạn cách làm nhé:

Thời gian đi từ A đến B: Quãng đường AB : Vận tốc lúc đi (55)

Thời gian đi về từ B về A: Quãng đường AB : Vận tốc lúc về (60)

Ô tô về A lúc: 5 giờ 15p + Thời gian đi + 45p + Thời gian về

Đề thiếu dữ kiện rồi bạn nhé, quãng đường AB dài bao nhiêu km bạn nhỉ?