Bài tập toán 6 : So sánh phân số

a) $\frac{-12}{17}<\frac{-11}{17}<\frac{-10}{17}<\frac{-9}{17}<\frac{-8}{17}$

b) $\frac{5}{11}<\frac{6}{11}<\frac{7}{11}<\frac{8}{11}<\frac{9}{11}$

a) $\frac{1}{3}<\frac{2}{3}$

b) $\frac{3}{4}<\frac{3}{2}$

a) $\frac{2}{-5}=\frac{-2}{5}<\frac{3}{5}$

b) $\frac{7}{3}>\frac{7}{4}\Rightarrow \frac{-7}{3}<\frac{-7}{4}$

a) $\frac{12}{36}<\frac{11}{36}<\frac{10}{36}<\frac{9}{36}\Rightarrow \frac{1}{3}<\frac{11}{36}<\frac{5}{18}<\frac{1}{4}$

b) $\frac{-12}{24}<\frac{-11}{24}<\frac{-10}{24}<\frac{-9}{24}$$\Rightarrow \frac{-1}{2}<\frac{-11}{24}<\frac{-5}{12}<\frac{-3}{8}$

a) $\frac{1}{3}=\frac{2}{6};\frac{2}{6}<\frac{5}{6}\Rightarrow \frac{1}{3}<\frac{5}{6}$

b) $\frac{4}{5}=\frac{28}{35};\frac{3}{7}=\frac{15}{35};\frac{28}{35}>\frac{15}{35}\Rightarrow \frac{4}{5}>\frac{3}{7}$

a) $\frac{-3}{11}=\frac{-39}{143};\frac{-4}{13}=\frac{-44}{143};$$\frac{-39}{143}>\frac{-44}{143}\Rightarrow \frac{-3}{11}>\frac{-4}{13}$

b) $\frac{-63}{70}=\frac{-9}{10}$

$\frac{-9}{10}=\frac{-27}{30};\frac{-5}{6}=\frac{-25}{30}\Rightarrow \frac{-5}{6}>\frac{-63}{70}$

a) $\frac{3}{4}=\frac{6}{8}<\frac{6}{7}$

b) $\frac{17}{-21}=\frac{51}{-63}>\frac{51}{-31}$

a) $\frac{-4}{9}=\frac{-12}{27}<\frac{-12}{52}=\frac{-3}{13}$

b) $\frac{-4}{-11}=\frac{12}{33}>\frac{12}{38}=\frac{-6}{-19}$

a) Ta có: $1-\frac{26}{27}=\frac{1}{27};1-\frac{96}{97}=\frac{1}{97}$. Vì $\frac{1}{27}>\frac{1}{97}$nên $\frac{26}{27}<\frac{96}{97}$

b) Ta có: $1-\frac{102}{103}=\frac{1}{103};1-\frac{103}{105}=\frac{2}{105}$. Vì $\frac{1}{103}=\frac{2}{206}<\frac{2}{105}$ nên $\frac{102}{103}>\frac{103}{105}$

a) Ta có: $\frac{2017}{2016}=1+\frac{1}{2016};\frac{2019}{2018}=1+\frac{1}{2018}$. Vì $\frac{1}{2016}>\frac{1}{2018}$ nên $\frac{2017}{2016}>\frac{2019}{2018}$

b) Ta có: $\frac{73}{64}=1+\frac{9}{64};\frac{51}{45}=1+\frac{6}{45}$. Vì $\frac{9}{64}=\frac{18}{128}>\frac{6}{45}=\frac{18}{135}$ nên $\frac{73}{64}>\frac{51}{45}$

a) $\frac{16}{-19}<0<\frac{15}{17}$

b) $\frac{419}{-723}<0<\frac{-697}{-313}$

a) $\frac{311}{256}>1>\frac{199}{203}$

b) $\frac{30}{235}=\frac{6}{47}>\frac{6}{48}=\frac{8}{64}>\frac{8}{63}=\frac{168}{1323}$

a) $\frac{19}{60}<\frac{20}{60}=\frac{30}{90}<\frac{31}{90}$

b) $\frac{15}{23}>\frac{14}{23}=\frac{70}{115}>\frac{70}{117}$

Thực hiện quy đồng: $\frac{a}{b}=\frac{a\left(b+m\right)}{b\left(b+m\right)}=\frac{ab+am}{b^2+bm}$

$\frac{a+m}{b+m}=\frac{b\left(a+m\right)}{b\left(b+m\right)}=\frac{ab+bm}{b^2+bm}$

Vì $\frac{a}{b}<1\Rightarrow a<b\Rightarrow ab+am<ab+bm$

Từ đó ta thu được $\frac{a}{b}<\frac{a+m}{b+m}$

$\frac{437}{564}<\frac{437+9}{564+9}=\frac{446}{573}$

Do $A=\frac{{98}^{99}+1}{{98}^{89}+1}>1$ nên $A=\frac{{98}^{99}+1}{{98}^{89}+1}>\frac{{98}^{99}+1+97}{{98}^{89}+1+97}=\frac{98\left({98}^{98}+1\right)}{98\left({98}^{88}+1\right)}=\frac{{98}^{98}+1}{{98}^{88}+1}=B$

Vậy A > B

${\left(\frac{1}{243}\right)}^9={\left(\frac{1}{3^4}\right)}^9=\frac{1}{3^{36}}$

${\left(\frac{1}{83}\right)}^{13}<{\left(\frac{1}{81}\right)}^{13}={\left(\frac{1}{3^4}\right)}^{13}=\frac{1}{3^{52}}$

Mà $\frac{1}{3^{36}}<\frac{1}{3^{42}}\Rightarrow {\left(\frac{1}{81}\right)}^{13}<{\left(\frac{1}{243}\right)}^9$

Vậy ${\left(\frac{1}{83}\right)}^{13}<{\left(\frac{1}{243}\right)}^9$

a) $x\in \left\{2;3\right\}$

b) $x\in \left\{-10;-9\right\}$

a) $\frac{9}{21}<\frac{x}{21}<\frac{14}{21}\Rightarrow x\in \left\{10;11;12;13\right\}$

b) $\frac{-536}{168}<\frac{x}{168}<\frac{-63}{168}\Rightarrow x\in \left\{-535;-564;\dots ;-65;-64\right\}$

a) $x\in \left\{4;3\right\}$

b) $\frac{-19}{23}<\frac{-19}{x}<\frac{-19}{29}\Rightarrow x\in \left\{24;25;26;27;28\right\}$

a) $\frac{88}{132}<\frac{88}{x}<\frac{88}{128}\Rightarrow x\in \left\{131;130;129\right\}$

b) $\frac{-60}{140}<\frac{-60}{x}<\frac{-60}{148}\Rightarrow x\in \left\{141;\dots ;147\right\}$

a) $\frac{3}{4}=\frac{9}{12}<\frac{10}{12}=\frac{5}{6}\Rightarrow \frac{3}{4}h<\frac{5}{6}h$

b) $1-\frac{8}{9}=\frac{1}{9};1-\frac{13}{14}=\frac{1}{14}$ mà $\frac{1}{9}>\frac{1}{14}\Rightarrow \frac{8}{9}<\frac{13}{14}\Rightarrow \frac{8}{9}m<\frac{13}{14}m$

a) $\frac{6}{7}<1<\frac{11}{10}\Rightarrow \frac{6}{7}kg<\frac{11}{10}kg$

b) $\frac{8}{15}=\frac{80}{150}<\frac{80}{136}=\frac{10}{17}\Rightarrow \frac{8}{15}d{m}^2<\frac{10}{17}d{m}^2$

a) $36ph=\frac{3}{5}h$ mà $\frac{2}{5}<\frac{3}{5}\Rightarrow \frac{2}{5}h<36ph$

b) $7dm=\frac{7}{10}m$ mà $\frac{7}{10}<\frac{8}{10}=\frac{4}{5}\Rightarrow 7dm<\frac{4}{5}m$

a) $400g=\frac{2}{5}kg$ mà $\frac{2}{5}>\frac{2}{7}\Rightarrow 400g>\frac{2}{7}kg$

b) $\frac{2500}{6}{m}^2>75d{m}^2$

a) $\frac{3}{8}+\frac{5}{-8}=\frac{3}{8}+\frac{-5}{8}=\frac{3+\left(-5\right)}{8}=\frac{-2}{8}=\frac{-1}{4}$

b) $\frac{-3}{7}+\frac{5}{-7}=\frac{-3}{7}+\frac{-5}{7}=\frac{\left(-3\right)+\left(-5\right)}{7}=\frac{-8}{7}$

a) $\frac{1}{-5}+\frac{9}{-5}=\frac{1+9}{-5}=\frac{10}{-5}=-2$

b) $\frac{5}{15}+\frac{-8}{24}=\frac{1}{3}+\frac{-1}{3}=\frac{1+\left(-1\right)}{3}=0$

a) $\frac{7}{21}+\frac{8}{-40}=\frac{1}{3}+\frac{-1}{5}=\frac{5}{15}+\frac{-3}{15}=\frac{2}{15}$

b) $\frac{-12}{36}+\frac{-24}{40}=\frac{-1}{3}+\frac{-3}{5}=\frac{-5}{15}+\frac{-9}{15}=\frac{-14}{15}$

a) $-3+\frac{10}{15}=-3+\frac{2}{3}=\frac{-9}{3}+\frac{2}{3}=\frac{-7}{3}$

b) $\frac{7}{12}+\frac{13}{32}=\frac{56}{96}+\frac{39}{96}=\frac{95}{96}$

a) $\frac{-4}{5}+\frac{-6}{7}=\frac{-28}{35}+\frac{-30}{35}=\frac{-58}{35}<\frac{-35}{35}=-1$

Vậy $\frac{-4}{5}+\frac{-6}{7}<-1$

b) $\frac{1}{6}+\frac{-10}{15}=\frac{1}{6}+\frac{-2}{3}=\frac{1}{6}+\frac{-4}{6}=\frac{-3}{6}=\frac{-1}{2}=\frac{-15}{30}$

$\frac{1}{5}+\frac{-12}{36}=\frac{1}{5}+\frac{-1}{3}=\frac{3}{15}+\frac{-5}{15}=\frac{-2}{15}=\frac{-4}{30}$

 Vậy $\frac{1}{6}+\frac{-10}{15}<\frac{1}{5}+\frac{-12}{36}$

a) $\frac{-5}{7}+\frac{-6}{8}=\frac{-5}{7}+\frac{-3}{4}=\frac{-20}{28}+\frac{-21}{28}=\frac{-41}{28}=\frac{-123}{84}$

$\frac{-1}{3}=\frac{-28}{84}$

Vậy $\frac{-5}{7}+\frac{-6}{8}<\frac{-28}{84}$

b) $\frac{14}{20}+\frac{4}{-5}=\frac{7}{10}+\frac{-4}{5}=\frac{7}{10}+\frac{-8}{10}=\frac{-1}{10}$

Vậy $\frac{14}{20}+\frac{4}{-5}=\frac{-1}{10}$

Trong 1 giờ, vòi thứ nhất chảy được: $\frac{1}{6}$ bể

Trong 1 giờ, vòi thứ hai chảy được: $\frac{1}{8}$ bể

Trong 1 giờ, cả hai vòi chảy được : $\frac{1}{6}+\frac{1}{8}=\frac{7}{24}$ bể

Ta có $\frac{-1}{7}=\frac{-3}{21};\frac{-1}{8}=\frac{-3}{24}$

Tổng các phân số lớn hơn $\frac{-1}{7}$, nhỏ hơn $\frac{-1}{8}$ và có tử là -3 là $\frac{-3}{22}+\frac{-3}{23}=\frac{-135}{506}$

$\frac{1}{6}=\frac{3}{18}=\frac{2+1}{18}=\frac{2}{18}+\frac{1}{18}=\frac{1}{9}+\frac{1}{18}$

$\frac{7}{16}=\frac{1+6}{16}=\frac{1}{16}+\frac{6}{16}=\frac{1}{16}+\frac{3}{8}$

$\frac{7}{16}=\frac{2+5}{16}=\frac{2}{16}+\frac{5}{16}=\frac{1}{8}+\frac{5}{16}$

$\frac{7}{16}=\frac{3+4}{16}=\frac{3}{16}+\frac{4}{16}=\frac{3}{16}+\frac{1}{4}$

$\frac{7}{16}=\frac{8+(-1)}{16}=\frac{8}{16}+\frac{-1}{16}=\frac{1}{2}+\frac{-1}{16}$

$A=\frac{2018}{2019}+\frac{2019}{2020}>\frac{2018}{2020}+\frac{2019}{2020}=\frac{2018+2019}{2020}>\frac{2018+2019}{2019+2020}=B$

Vậy A > B