a: \(\dfrac{9}{70}=\dfrac{9\cdot3}{70\cdot3}=\dfrac{27}{210};\dfrac{5}{42}=\dfrac{5\cdot5}{42\cdot5}=\dfrac{25}{210}\)

mà 27>25

nên \(\dfrac{9}{70}>\dfrac{5}{42}\)

b: \(\dfrac{-4}{27}=\dfrac{-4\cdot7}{27\cdot7}=\dfrac{-28}{189};\dfrac{15}{-63}=\dfrac{-15}{63}=\dfrac{-15\cdot3}{63\cdot3}=\dfrac{-45}{189}\)

mà -28>-45

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

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

b: \(\dfrac{-9}{21}=\dfrac{-9:3}{21:3}=\dfrac{-3}{7};\dfrac{27}{-63}=\dfrac{27:\left(-9\right)}{-63:\left(-9\right)}=\dfrac{-3}{7}\)

Do đó: \(\dfrac{-9}{21}=\dfrac{27}{-63}\)

a: \(\dfrac{7}{8}=\dfrac{7\cdot3}{8\cdot3}=\dfrac{21}{24};\dfrac{11}{12}=\dfrac{11\cdot2}{12\cdot2}=\dfrac{22}{24}\)

mà 21<22

nên \(\dfrac{7}{8}< \dfrac{11}{12}\)

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>-40

nên \(-\dfrac{5}{8}>\dfrac{7}{-10}\)

a: \(\dfrac{998}{555}=\dfrac{555+443}{555}=1+\dfrac{443}{555}\)

\(\dfrac{999}{556}=\dfrac{556+443}{556}=1+\dfrac{443}{556}\)

mà \(\dfrac{443}{555}>\dfrac{443}{556}\)

nên \(\dfrac{998}{555}>\dfrac{999}{556}\)

b: \(\dfrac{315}{380}=1-\dfrac{65}{380};\dfrac{316}{381}=1-\dfrac{65}{381}\)

Ta có: 380<381

=>\(\dfrac{65}{380}>\dfrac{65}{381}\)

=>\(-\dfrac{65}{380}< -\dfrac{65}{381}\)

=>\(-\dfrac{65}{380}+1< -\dfrac{65}{381}+1\)

=>\(\dfrac{315}{380}< \dfrac{316}{381}\)

=>\(-\dfrac{315}{380}>-\dfrac{316}{381}\)

a: \(\dfrac{15}{21}=\dfrac{15:3}{21:3}=\dfrac{5}{7};\dfrac{21}{49}=\dfrac{21:7}{49:7}=\dfrac{3}{7}\)

mà 5>3

nên \(\dfrac{15}{21}>\dfrac{21}{49}\)

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

\(\dfrac{-23}{47}=\dfrac{-23\cdot49}{49\cdot47}=\dfrac{-1127}{2303}\)

mà -893>-1127

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

a: \(\dfrac{-11}{6}< -1\)

\(-1=\dfrac{-9}{9}< \dfrac{8}{-9}\)

Do đó: \(\dfrac{-11}{6}< \dfrac{8}{-9}\)

b: \(-\dfrac{25}{20}< 0\)

\(0< \dfrac{20}{25}\)

Do đó: \(-\dfrac{25}{20}< \dfrac{20}{25}\)

\(a)3^{2x-1}+2\cdot9^{x-1}=405\\ =>3^{2x-1}+2\cdot\left(3^2\right)^{x-1}=405\\ =>3^{2x-1}+2\cdot3^{2x-2}=405\\ =>3^{2x-2}\cdot\left(3+2\right)=405\\ =>3^{2x-2}\cdot5=405\\ =>3^{2x-2}=\dfrac{405}{5}=81\\ =>3^{2x-2}=3^4\\ =>2x-2=4\\ =>2x=4+2=6\\ =>x=\dfrac{6}{2}\\ =>x=3\)

\(b)\left(\dfrac{1}{3}\right)^{x-1}+5\left(\dfrac{1}{3}\right)^{x+1}=\dfrac{14}{9^3}\\ =>\left(\dfrac{1}{3}\right)^{x-1}\left(1+5\cdot\dfrac{1}{3^2}\right)=\dfrac{14}{729}\\ =>\left(\dfrac{1}{3}\right)^{x-1}\cdot\dfrac{14}{9}=\dfrac{14}{729}\\ =>\left(\dfrac{1}{3}\right)^{x-1}=\dfrac{14}{729}:\dfrac{14}{9}\\ =>\left(\dfrac{1}{3}\right)^{x-1}=\dfrac{9}{729}=\dfrac{1}{81}\\ =>\left(\dfrac{1}{3}\right)^{x-1}=\left(\dfrac{1}{3}\right)^4\\ =>x-1=4\\ =>x=1+4\\ =>x=5\)

\(c)\dfrac{3}{5}\left(3x^3-\dfrac{8}{9}\right)-\dfrac{1}{2}\left(\dfrac{3}{2}-1\right)=-\dfrac{1}{4}\\ =>\dfrac{3}{5}\left(3x^3-\dfrac{8}{9}\right)-\dfrac{1}{2}\cdot\dfrac{1}{2}=-\dfrac{1}{4}\\ =>\dfrac{3}{5}\left(3x^3-\dfrac{8}{9}\right)-\dfrac{1}{4}=-\dfrac{1}{4}\\ =>\dfrac{3}{5}\left(3x^3-\dfrac{8}{9}\right)=0\\ =>3x^3-\dfrac{8}{9}=0\\ =>3x^3=\dfrac{8}{9}\\ =>x^3=\dfrac{8}{9}:3=\dfrac{8}{27}\\ =>x^3=\left(\dfrac{2}{3}\right)^3\\ =>x=\dfrac{2}{3}\)

a: \(3^{2x-1}+2\cdot9^{x-1}=405\)

=>\(\dfrac{3^{2x}}{3}+2\cdot3^{2x-2}=405\)

=>\(\dfrac{1}{3}\cdot3^{2x}+2\cdot3^{2x}\cdot\dfrac{1}{9}=405\)

=>\(3^{2x}\cdot\left(\dfrac{1}{3}+\dfrac{2}{9}\right)=405\)

=>\(3^{2x}\cdot\dfrac{5}{9}=405\)

=>\(3^{2x}=405:\dfrac{5}{9}=405\cdot\dfrac{9}{5}=81\cdot9=3^6\)

=>2x=6

=>x=3

b: \(\left(\dfrac{1}{3}\right)^{x-1}+5\cdot\left(\dfrac{1}{3}\right)^{x+1}=\dfrac{14}{9^3}\)

=>\(\left(\dfrac{1}{3}\right)^x\cdot3+5\cdot\left(\dfrac{1}{3}\right)^x\cdot\dfrac{1}{3}=\dfrac{14}{9^3}\)

=>\(\left(\dfrac{1}{3}\right)^x\cdot\left(3+\dfrac{5}{3}\right)=\dfrac{14}{9^3}\)

=>\(\left(\dfrac{1}{3}\right)^x=\dfrac{14}{3^6}:\dfrac{14}{3}=\dfrac{3}{3^6}=\dfrac{1}{3^5}\)

=>x=5

c: \(\dfrac{3}{5}\left(3x^3-\dfrac{8}{9}\right)-\dfrac{1}{2}\left(\dfrac{3}{2}-1\right)=-\dfrac{1}{4}\)

=>\(\dfrac{9}{5}x^3-\dfrac{24}{45}-\dfrac{1}{2}\cdot\dfrac{1}{2}+\dfrac{1}{4}=0\)

=>\(\dfrac{9}{5}x^3=\dfrac{24}{45}=\dfrac{8}{15}\)

=>\(x^3=\dfrac{8}{15}:\dfrac{9}{5}=\dfrac{8}{15}\cdot\dfrac{5}{9}=\dfrac{40}{135}=\dfrac{8}{27}=\left(\dfrac{2}{3}\right)^3\)

=>\(x=\dfrac{2}{3}\)

d: \(\dfrac{7}{x}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{29}{45}\)

=>\(\dfrac{7}{x}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{29}{45}\)

=>\(\dfrac{7}{x}+\dfrac{9}{45}-\dfrac{1}{45}=\dfrac{29}{45}\)

=>\(\dfrac{7}{x}=\dfrac{29}{45}-\dfrac{8}{45}=\dfrac{21}{45}=\dfrac{7}{15}\)

=>x=15

e: \(\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{5}{31}\)

=>\(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{10}{31}\)

=>\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2x+1}-\dfrac{1}{2x+3}=\dfrac{10}{31}\)

=>\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{10}{31}\)

=>\(\dfrac{1}{2x+3}=\dfrac{1}{3}-\dfrac{10}{31}=\dfrac{1}{93}\)

=>2x+3=93

=>2x=90

=>x=45

b: \(\dfrac{2}{5}-\left(\dfrac{4}{3}+\dfrac{4}{5}\right)-\left(-\dfrac{1}{9}-0,4\right)+\dfrac{11}{9}\)

\(=\dfrac{2}{5}-\dfrac{4}{3}-\dfrac{4}{5}+\dfrac{1}{9}+\dfrac{2}{5}+\dfrac{11}{9}\)

\(=\left(\dfrac{2}{5}-\dfrac{4}{5}+\dfrac{2}{5}\right)+\left(-\dfrac{4}{3}+\dfrac{1}{9}+\dfrac{11}{9}\right)\)

\(=-\dfrac{4}{3}+\dfrac{12}{9}=0\)

c: \(\dfrac{11}{8}\cdot\left[\left(-\dfrac{5}{11}:\dfrac{13}{8}-\dfrac{5}{11}:\dfrac{13}{5}\right)+\dfrac{-6}{33}\right]+\dfrac{3}{4}\)

\(=\dfrac{11}{8}\cdot\left[-\dfrac{5}{11}\cdot\dfrac{8}{13}-\dfrac{5}{11}\cdot\dfrac{5}{13}+\dfrac{-2}{11}\right]+\dfrac{3}{4}\)

\(=\dfrac{11}{8}\cdot\left[-\dfrac{5}{11}\left(\dfrac{8}{13}+\dfrac{5}{13}\right)-\dfrac{2}{11}\right]+\dfrac{3}{4}\)

\(=\dfrac{11}{8}\cdot\dfrac{-7}{11}+\dfrac{3}{4}=-\dfrac{7}{8}+\dfrac{3}{4}=-\dfrac{1}{8}\)

a:\(\widehat{BAC}+\widehat{xAC}=180^0\)(hai góc kề bù)

=> \(\widehat{BAC}+70^0=180^0\)

=>\(\widehat{BAC}=110^0\)

Ta có: \(\widehat{BAC}+\widehat{ABD}=180^0\)

mà hai góc này là hai góc ở vị trí trong cùng phía

nên AC//BD

b: Vì AC//BD

nên \(\widehat{yCx}=\widehat{CDB}\)(hai góc đồng vị)

=>\(\widehat{yCx}=60^0\)

Ta có: \(\widehat{yCx}+\widehat{ACD}=180^0\)(hai góc kề bù)

=>\(\widehat{ACD}+60^0=180^0\)

=>\(\widehat{ACD}=120^0\)

Ta có: \(\widehat{BAC}+\widehat{ABD}=180^0\)(AC//BD)

=>\(\widehat{BAC}+70^0=180^0\)

=>\(\widehat{BAC}=110^0\)

e: \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)

=>\(\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)=\left(x+1\right)\left(\dfrac{1}{13}+\dfrac{1}{14}\right)\)

=>\(\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)

=>x+1=0

=>x=-1