Đáp án:
Bài 1 :
$a. = \frac{7}{6}$
$b. = \frac{27}{125}$
Bài 2 :
$a.$ \(\left[ \begin{array}{l}x=\frac{-1}{15}\\x=\frac{1}{3}\end{array} \right.\)
$b.$ \(\left[ \begin{array}{l}x=\frac{7}{10}\\x=\frac{3}{10}\end{array} \right.\)
$c. a = - 60 , b = - 24 , c = - 40$
Giải thích các bước giải:
Bài 1 :
$a. 5.| - \frac{1}{10} + \frac{7}{15} | - \frac{2}{13}.4\frac{1}{3}$
$= 5.| - \frac{3}{30} + \frac{14}{30} | - \frac{2}{13}.\frac{13}{3}$
$= 5.| \frac{11}{30} | - \frac{2}{3}$
$= \frac{11.5}{6.5} - \frac{4}{6}$
$= \frac{11}{6} - \frac{4}{6}$
$= \frac{7}{6}$
$b. ( - \frac{3}{8} + \frac{5}{32} ) : ( - \frac{4}{5} ) - ( - \frac{3}{5} )^{3} + ( - \frac{5}{8} + \frac{27}{32} ).( - \frac{5}{4} )$
$= ( \frac{-12}{32} + \frac{5}{32} ).\frac{-5}{4} - (-\frac{3}{5})^{3} + ( - \frac{20}{32} + \frac{27}{32} ).\frac{-5}{4}$
$= \frac{-5}{4}.( \frac{-12}{32} + \frac{5}{32} - \frac{20}{32} + \frac{27}{32} ) - \frac{-27}{125}$
$= \frac{-5}{4}.0 + \frac{27}{125}$
$= \frac{27}{125}$
Bài 2 :
$a. ( \frac{2}{5} - 3x )^{2} - \frac{1}{5} = \frac{4}{25}$
⇔ $( \frac{2}{5} - 3x )^{2} = \frac{4}{25} + \frac{1}{5}$
⇔ $( \frac{2}{5} - 3x )^{2} = \frac{4}{25} + \frac{5}{25}$
⇔ $( \frac{2}{5} - 3x )^{2} = \frac{9}{25}$
⇔ \(\left[ \begin{array}{l}\frac{2}{5}-3x=\frac{3}{5}\\\frac{2}{5}-3x=-\frac{3}{5}\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}3x=\frac{-1}{5}\\3x=1\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{-1}{15}\\x=\frac{1}{3}\end{array} \right.\)
$b. \frac{4}{3} - \frac{1}{3} : | 2x - 1 | = \frac{1}{2}$
⇔ $\frac{1}{3} : | 2x - 1 | = \frac{4}{3} - \frac{1}{2}$
⇔ $\frac{1}{3|2x-1|} = \frac{8}{6} - \frac{3}{6}$
⇔ $\frac{1}{3|2x-1|} = \frac{5}{6}$
⇔ $3| 2x - 1 |.5 = 6.1$
⇔ $15| 2x - 1 | = 6$
⇔ $| 2x - 1 | = \frac{6}{15}$
⇔ $| 2x - 1 | = \frac{2}{5}$
⇔ \(\left[ \begin{array}{l}2x-1=\frac{2}{5}\\2x-1=-\frac{2}{5}\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}2x=\frac{7}{5}\\2x=\frac{3}{5}\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{7}{10}\\x=\frac{3}{10}\end{array} \right.\)
$c. 2a = 5b = 3c = k$
⇒ $a = \frac{k}{2} , b = \frac{k}{5} , c = \frac{k}{3}$
Ta co : $a + b - c = - 44$
⇔ $\frac{k}{2} + \frac{k}{5} - \frac{k}{3} = - 44$
⇔ $\frac{15k}{30} + \frac{6k}{30} - \frac{10k}{30} = - 44$
⇔ $\frac{11k}{30} = - 44$
⇔ $k = - 120$
⇔ $a = - 60 , b = - 24 , c = - 40$
