Question

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Answer 1

Ответ:

$\displaystyle 10)\ \ \Big(x-1\frac{8}{23}\Big)+3\frac{19}{23}=5\frac{12}{23} \ \ \ \Rightarrow \ \ \ \Big(x-\frac{31}{23}\Big)+\frac{88}{23}=\frac{127}{23}\\\\\\x-\frac{31}{23}=\frac{127-88}{23}\ \ ,\ \ \ x-\frac{31}{23}=\frac{39}{23}\ \ \ ,\ \ \ x=\frac{39}{23}+\frac{31}{23}\ \ ,\ \ x=\frac{70}{23}\ ,\\\\\\x=3\frac{1}{23}$

$11)\ \ (8164-x)-2398=2557\ \ \ \to \ \ \ 8164-x=4955\ \ ,\ \ x=3209\\\\\\12)\ \ (20-a)-6\dfrac{7}{18}=3\dfrac{7}{18}\ \ \ \to \ \ \ 20-a=9\dfrac{14}{18}\ \ \ ,\ \ \ 20-a=\dfrac{88}{9}\ ,\\\\\\a=20-\dfrac{88}{9}\ \ ,\ \ \ a=\dfrac{180-88}{9}\ \ ,\ \ \ a=\dfrac{92}{9}\ \ ,\ \ \ a=10\dfrac{2}{9}\\\\\\13)\ \ 0,8-(x-0,326)=0,495\ \ \ \to \ \ \ x-0,326=0,8-0,495\ \ ,\\\\x-0,326=0,305\ \ \ ,\ \ x=0,631$

$14)\ \ 1,2-\Big (x-\dfrac{1}{4}\Big)=\dfrac{1}{2}\ \ \ \to \ \ \ x-\dfrac{1}{4}=1\dfrac{1}{5}-\dfrac{1}{2}\ \ ,\\\\\\x-\dfrac{1}{4}=\dfrac{6}{5}-\dfrac{1}{2}\ \ \ ,\ \ \ x-\dfrac{1}{4}=\dfrac{6\cdot 2-5}{10}\ \ ,\ \ \ x-\dfrac{1}{4}=\dfrac{7}{10}\ \ ,\ \ x=\dfrac{7}{10}+\dfrac{1}{4}\ \ ,\\\\\\x=\dfrac{7\cdot 2+5}{20}\ \ ,\ \ \ x=\dfrac{19}{20}\ \ ,\ \ \ x=0,95$