Giải thích các bước giải:
Bài 1:
a,5$\frac{1}{3}$-(4$\frac{1}{3}$+$\frac{5}{6}$)
=5$\frac{1}{3}$-4$\frac{1}{3}$+$\frac{5}{6}$
=1+$\frac{5}{6}$
=$\frac{11}{6}$
b,$\frac{-13}{14}$.$\frac{7}{12}$+$\frac{5}{12}$.$\frac{-13}{14}$+5$\frac{13}{14}$
=$\frac{-13}{14}$.($\frac{7}{12}$+$\frac{5}{12}$)+5$\frac{13}{14}$
=$\frac{-13}{14}$.$\frac{12}{12}$+5$\frac{13}{14}$
=$\frac{-13}{14}$+5+$\frac{13}{14}$
=0+5
=5
Bài 2:
Ta có:
a=$\frac{5.6}{9.25}$
a=$\frac{5.3.2}{3.3.5.5}$
a=$\frac{2}{3.5}$
a=$\frac{2}{15}$
b=$\frac{18.4-18}{8.9+7.9}$
b=$\frac{18.4-18.1}{9.(8+7)}$
b=$\frac{18.(4-1)}{9.(8+7)}$
b=$\frac{18.3}{9.15}$
b=$\frac{3.6.3}{3.3.3.5}$
b=$\frac{6}{3.5}$
b=$\frac{6}{15}$=$\frac{2}{5}$
Vì $\frac{2}{15}$<$\frac{2}{5}$
⇒a<b
Bài 3:
a,$3^{3}$.($x^{}$+$125^{}$)=$9^{2}$
$27^{}$.($x^{}$+$125^{}$)=$81^{}$
$x^{}$+$125^{}$=$81^{}$÷$27^{}$
$x^{}$+$125^{}$=$3^{}$
$x^{}$=$3^{}$-$125^{}$
$x^{}$=$-122^{}$
b, ($1,1^{}$.$x^{}$-$9^{}$).|$\frac{-1}{5}$|=$40%^{}$%
($\frac{11}{10}$.$x^{}$-$9^{}$).$\frac{1}{5}$=$\frac{40}{100}$
($\frac{11}{10}$.$x^{}$-$9^{}$).$\frac{1}{5}$=$\frac{2}{5}$
($\frac{11}{10}$.$x^{}$-$9^{}$)=$\frac{2}{5}$÷$\frac{1}{5}$
($\frac{11}{10}$.$x^{}$-$9^{}$)=$\frac{2}{5}$.$5^{}$
($\frac{11}{10}$.$x^{}$-$9^{}$)=$2^{}$
$\frac{11}{10}$.$x^{}$=$2^{}$+$9^{}$
$\frac{11}{10}$.$x^{}$=$11^{}$
$x^{}$=$11^{}$÷$\frac{11}{10}$
$x^{}$=$11^{}$.$\frac{10}{11}$
$x^{}$=$10^{}$
