Bài tập cuối chương 1 có đáp án

a) $\frac{2}{5}+\frac{3}{5}:(-\frac{3}{2})+\frac{1}{2}$

$=\frac{2}{5}+\frac{3}{5}.\frac{-2}{3}+\frac{1}{2}$

$=\frac{2}{5}+\frac{-2}{5}+\frac{1}{2}$

$=0+\frac{1}{2}=\frac{1}{2}$

b) $2\frac{1}{2}+{(-\frac{1}{3})}^{2}-\frac{3}{2}$

$=\frac{5}{2}+\frac{1}{9}-\frac{3}{2}$

$=\frac{42}{18}+\frac{2}{18}-\frac{27}{18}\ =\frac{17}{18}.$

c) $(\frac{7}{8}-0{,}25):{(\frac{5}{6}-0{,}75)}^{2}$

$=(\frac{7}{8}-\frac{1}{4}):{(\frac{5}{6}-\frac{3}{4})}^{2}$

$=(\frac{7}{8}-\frac{1}{4}):{(\frac{5}{6}-\frac{3}{4})}^{2}$

$=(\frac{7}{8}-\frac{2}{8}):{(\frac{10}{12}-\frac{9}{12})}^{2}$

$=\frac{5}{8}:{(\frac{1}{12})}^{2}$$=\frac{5}{8}:\frac{{1}^{2}}{{12}^{2}}\ =\frac{5}{8}{.12}^{2}=\frac{5.144}{8}=90.$

d) $(-0{,}75)-[(-2)+\frac{3}{2}]:1{,}5+(\frac{-5}{4})$

$=\frac{-3}{4}-[\frac{-4}{2}+\frac{3}{2}]:\frac{3}{2}+(\frac{-5}{4})$

$=\frac{-3}{4}-\frac{-1}{2}:\frac{3}{2}+(\frac{-5}{4})$

$=\frac{-3}{4}-\frac{-1}{2}:\frac{3}{2}+(\frac{-5}{4})$

$=\frac{-3}{4}+\frac{1}{3}+(\frac{-5}{4})$

$=(\frac{-3}{4}+\frac{-5}{4})+\frac{1}{3}$

$=\frac{-8}{4}+\frac{1}{3}$

$=\frac{-24}{12}+\frac{4}{12}$

$=\frac{-20}{12}=\frac{-5}{3}$

a) $\frac{5}{23}+\frac{7}{17}+0{,}25-\frac{5}{23}+\frac{10}{17}$

$=(\frac{5}{23}-\frac{5}{23})+(\frac{7}{17}+\frac{10}{17})+\frac{1}{4}$

$=0+1+\frac{1}{4}=\frac{4}{4}+\frac{1}{4}=\frac{5}{4}.$

b) $\frac{3}{7}.2\frac{2}{3}-\frac{3}{7}.1\frac{1}{2}$

$=\frac{3}{7}.\frac{8}{3}-\frac{3}{7}.\frac{3}{2}$$=\frac{3}{7}.(\frac{8}{3}-\frac{3}{2})$

$=\frac{3}{7}.(\frac{16}{6}-\frac{9}{6})$$=\frac{3}{7}.\frac{7}{6}=\frac{1}{2}$

c) $13\frac{1}{4}:(-\frac{4}{7})-17\frac{1}{4}:(-\frac{4}{7})$

$=(13+\frac{1}{4}).(\frac{-7}{4})-(17+\frac{1}{4})(\frac{-7}{4})$

$=(\frac{-7}{4}).(13+\frac{1}{4}-17-\frac{1}{4})\ =(\frac{-7}{4}).[(13-17)+(\frac{1}{4}-\frac{1}{4})]\ =(\frac{-7}{4}).(-4)=7.$

d) $\frac{100}{123}:(\frac{3}{4}+\frac{7}{12})+\frac{23}{123}:(\frac{9}{5}-\frac{7}{15})$

$=\frac{100}{123}:(\frac{9}{12}+\frac{7}{12})+\frac{23}{123}:(\frac{27}{15}-\frac{7}{15})$

$=\frac{100}{123}:\frac{16}{12}+\frac{23}{123}:\frac{20}{15}$

$=\frac{100}{123}:\frac{4}{3}+\frac{23}{123}:\frac{4}{3}$

$=\frac{100}{123}.\frac{3}{4}+\frac{23}{123}.\frac{3}{4}$

$=\frac{3}{4}.(\frac{100}{123}+\frac{23}{123})$

$=\frac{3}{4}.\frac{123}{123}=\frac{3}{4}$

a) $\frac{{5}^{16}{.27}^{7}}{{125}^{5}{.9}^{11}}=\frac{{5}^{16}.{({3}^{3})}^{7}}{{({5}^{3})}^{5}.{({3}^{2})}^{11}}$

$=\frac{{5}^{16}{.3}^{3.7}}{{5}^{3.5}{.3}^{2.11}}=\frac{{5}^{16}{.3}^{21}}{{5}^{15}{.3}^{22}}=\frac{5}{3}$

b) ${(-0{,}2)}^{2}.5-\frac{{2}^{3}{.27}^{3}}{{4}^{6}{.9}^{5}}$

$={(\frac{-1}{5})}^{2}.5-\frac{{2}^{3}.{({3}^{3})}^{3}}{{({2}^{2})}^{6}.{({3}^{2})}^{5}}$

$=\frac{1}{{5}^{2}}.5-\frac{{2}^{13}{.3}^{9}}{{2}^{12}{.3}^{10}}\ =\frac{1}{5.5}.5-\frac{{2}^{12}{.2.3}^{9}}{{2}^{12}{.3}^{9}.3}$

$=\frac{1}{5}-\frac{2}{3}$

$=\frac{3}{15}-\frac{10}{15}$

$=\frac{-7}{15}$

c) $\frac{{5}^{6}+{2}^{2}{.25}^{3}+{2}^{3}{.125}^{2}}{{26.5}^{6}}$

$=\frac{{5}^{6}+{2}^{2}.{({5}^{2})}^{3}+{2}^{3}.{({5}^{3})}^{2}}{{26.5}^{6}}$

$=\frac{{5}^{6}+{2}^{2}{.5}^{2.3}+{2}^{3}{.5}^{3.2}}{{26.5}^{6}}$

$=\frac{{5}^{6}+{2}^{2}{.5}^{6}+{2}^{3}{.5}^{6}}{{26.5}^{6}}$

$=\frac{{5}^{6}.(1+{2}^{2}+{2}^{3})}{{26.5}^{6}}$$=\frac{{5}^{6}.13}{{26.5}^{6}}=\frac{13}{26}=\frac{1}{2}$

a) $A=[(-0{,}5)-\frac{3}{5}]:(-3)+\frac{1}{3}-(-\frac{1}{6}):(-2)$

$A=[\frac{-1}{2}-\frac{3}{5}]:(-3)+\frac{1}{3}-\frac{-1}{6}.\frac{1}{-2}$

$A=[\frac{-5}{10}-\frac{6}{10}]:(-3)+\frac{1}{3}-\frac{1}{12}$

$A=\frac{-11}{10}.\frac{1}{-3}+\frac{1}{3}-\frac{1}{12}$

$A=\frac{11}{30}+\frac{1}{3}-\frac{1}{12}$

$A=\frac{37}{60}$

Vậy $A=\frac{37}{60}$

b) $B=(\frac{2}{25}-0{,}036):\frac{11}{50}-[(3\frac{1}{4}-2\frac{4}{9})].\frac{9}{29}$

$B=(\frac{2}{25}-\frac{9}{250}):\frac{11}{50}-[(\frac{13}{4}-\frac{22}{9})].\frac{9}{29}$

$B=(\frac{20}{250}-\frac{9}{250}):\frac{11}{50}-[(\frac{117}{36}-\frac{88}{36})].\frac{9}{29}$

$B=\frac{11}{250}:\frac{11}{50}-\frac{29}{36}.\frac{9}{29}$

$B=\frac{11}{250}.\frac{50}{11}-\frac{9}{36}$

$B=\frac{1}{5}-\frac{1}{4}$

$B=\frac{4}{20}-\frac{5}{20}=\frac{-1}{20}$

a) $-\frac{3}{5}x=\frac{12}{25}$

$x=\frac{12}{25}:(-\frac{3}{5})$

$x=\frac{12}{25}.\frac{-5}{3}$

$x=\frac{-4}{5}$

Vậy $x=\frac{-4}{5}$

b) $\frac{3}{5}x-\frac{3}{4}=-1\frac{1}{2}$

$\frac{3}{5}x-\frac{3}{4}=\frac{-3}{2}$

$\frac{3}{5}x=\frac{-3}{2}+\frac{3}{4}$

$\frac{3}{5}x=\frac{-6}{4}+\frac{3}{4}$

$\frac{3}{5}x=\frac{-3}{4}$

$x=\frac{-3}{4}:\frac{3}{5}$

$x=\frac{-3}{4}.\frac{5}{3}$

$x=\frac{-5}{4}$

Vậy $x=\frac{-5}{4}$

c) $\frac{2}{5}+\frac{3}{5}:x=0{,}5$

$\frac{2}{5}+\frac{3}{5}:x=\frac{1}{2}$

$\frac{3}{5}:x=\frac{1}{2}-\frac{2}{5}$

$\frac{3}{5}:x=\frac{5}{10}-\frac{4}{10}$

$\frac{3}{5}:x=\frac{1}{10}$

$x=\frac{3}{5}:\frac{1}{10}$

$x=\frac{3}{5}.\frac{10}{1}$

x = 6

Vậy x = 6

d) $\frac{3}{4}-(x-\frac{1}{2})=1\frac{2}{3}$

$\frac{3}{4}-(x-\frac{1}{2})=\frac{5}{3}$

$x-\frac{1}{2}=\frac{3}{4}-\frac{5}{3}\ x-\frac{1}{2}=\frac{9}{12}-\frac{20}{12}\ x-\frac{1}{2}=\frac{-11}{12}\ x=\frac{-11}{12}+\frac{1}{2}\ x=\frac{-11}{12}+\frac{6}{12}\ x=\frac{-5}{12}$

Vậy $x=\frac{-5}{12}$

e) $2\frac{2}{15}:(\frac{1}{3}-5x)=-2\frac{2}{5}$

$\frac{32}{15}:(\frac{1}{3}-5x)=\frac{-12}{5}\ \frac{1}{3}-5x=\frac{32}{15}:\frac{-12}{5}\ \frac{1}{3}-5x=\frac{32}{15}.\frac{5}{-12}\ \frac{1}{3}-5x=\frac{-8}{9}\ 5x=\frac{1}{3}-\frac{-8}{9}\ 5x=\frac{3}{9}+\frac{8}{9}\ 5x=\frac{11}{9}\ x=\frac{11}{9}:5\ x=\frac{11}{45}$

Vậy $x=\frac{11}{45}$

g) ${x}^{2}+\frac{1}{9}=\frac{5}{3}:3$

${x}^{2}+\frac{1}{9}=\frac{5}{9}\ {x}^{2}=\frac{5}{9}-\frac{1}{9}\ {x}^{2}=\frac{4}{9}\ {x}^{2}=\frac{4}{9}\ {x}^{2}={(\frac{2}{3})}^{2}={(\frac{-2}{3})}^{2}\ +)x=\frac{-2}{3}\ +)x=\frac{2}{3}\ Vậyx=\frac{2}{3};x=\frac{-2}{3}$