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baif2 : a ) 49+x=51
x=51-49
x=2
b) x-17=15-x
x+x=17+15
2x=32
=>x=32:2
=>x=16
c) 56-6.(13-x)=116
6(13-x)=56-116
6(13-x) = -60
=>13-x=-60:6
13-x = -10
=> x=13-(-10) = 13+10
=> x=23
Ta có:
\(\dfrac{1}{20}=\dfrac{1}{4}-\dfrac{1}{5}\)
\(\dfrac{1}{14}=\dfrac{1}{10}-\dfrac{1}{35}\)
\(\dfrac{1}{56}=\dfrac{1}{7}-\dfrac{1}{8}\)
Thay vào ta được:
\(2 + \dfrac{1}{4}+ \dfrac{3}{4} - \dfrac{3}{5}+\dfrac{1}{10} + \dfrac{1}{10} - \dfrac{1}{35}+ \dfrac{3}{7} - \dfrac{3}{8} + \dfrac{1}{24} \)
\(= 3 - \dfrac{2}{5} - \dfrac{1}{35} + \dfrac{3}{7} - \dfrac{3}{8}+ \dfrac{1}{24} \)
\(= 3 + \dfrac{1}{35} - \dfrac{1}{35} + \dfrac{1}{24} - \dfrac{3}{8} \)
\(= 3 + \dfrac{1}{24} - \dfrac{3}{8}\)
\(= 2 + \dfrac{5}{8}+ \dfrac{1}{24} \)
\(= 2 + \dfrac{1}{8} (5 + \dfrac{1}{3}) \)
\(= 2 + \dfrac{16}{24} \)
\(=2+\dfrac{2}{3}\)
\(=\dfrac{8}{3}\)
\(1,0,75-\dfrac{2}{3}-0,5=\dfrac{3}{4}-\dfrac{2}{3}-\dfrac{1}{2}=\dfrac{9}{12}-\dfrac{8}{12}-\dfrac{1}{2}=\dfrac{1}{12}-\dfrac{1}{2}\)
\(=\dfrac{2}{24}-\dfrac{12}{24}=\dfrac{-10}{24}=\dfrac{-5}{12}\)
\(2,\dfrac{1}{5}-0,125-\dfrac{5}{4}=\dfrac{1}{5}-\dfrac{1}{8}-\dfrac{5}{4}=\dfrac{8}{40}-\dfrac{5}{40}-\dfrac{5}{4}=\dfrac{3}{40}-\dfrac{5}{4}\)
\(=\dfrac{3}{40}-\dfrac{50}{40}=\dfrac{-47}{40}\)
\(3,1,25-\dfrac{3}{4}+\dfrac{4}{3}=\dfrac{5}{4}-\dfrac{3}{4}+\dfrac{4}{3}=\dfrac{2}{4}+\dfrac{4}{3}=\dfrac{6}{12}+\dfrac{16}{12}=\dfrac{22}{12}=\dfrac{11}{6}\)
\(4,0,15-\dfrac{1}{4}+\dfrac{2}{5}=\dfrac{3}{20}-\dfrac{1}{4}+\dfrac{2}{5}=\dfrac{3}{20}-\dfrac{5}{20}+\dfrac{2}{5}=\dfrac{-2}{20}+\dfrac{2}{5}\)
\(=\dfrac{-2}{20}+\dfrac{8}{20}=\dfrac{6}{20}=\dfrac{3}{10}\)
\(5,5-3,4+\dfrac{1}{5}=\dfrac{5}{1}-\dfrac{17}{5}+\dfrac{1}{5}=\dfrac{25}{5}-\dfrac{17}{5}+\dfrac{1}{5}=\dfrac{25-17+1}{5}=\dfrac{9}{5}\)
\(6,\dfrac{1}{4}-0,3+\dfrac{4}{3}=\dfrac{1}{4}-\dfrac{3}{10}+\dfrac{4}{3}=\dfrac{10}{40}-\dfrac{12}{40}+\dfrac{4}{3}=\dfrac{-2}{40}+\dfrac{4}{3}\)
\(=\dfrac{-1}{20}+\dfrac{4}{3}=\dfrac{-3}{60}+\dfrac{80}{60}=\dfrac{77}{60}\)
\(7,0,2-3,25+4,7=\dfrac{1}{5}-\dfrac{13}{4}+\dfrac{47}{10}=\dfrac{4}{20}-\dfrac{65}{20}+\dfrac{47}{10}=\dfrac{-61}{20}+\dfrac{47}{10}\)
\(=\dfrac{-61}{20}+\dfrac{94}{20}=\dfrac{33}{20}=1,65\)
\(8,5,4+\dfrac{-7}{3}-\dfrac{-5}{7}=\dfrac{27}{5}+\dfrac{-7}{3}-\dfrac{-5}{7}=\dfrac{81}{15}+\dfrac{-35}{15}-\dfrac{-5}{7}\)
\(=\dfrac{46}{15}-\dfrac{-5}{7}=\dfrac{322}{105}-\dfrac{-75}{105}=\dfrac{397}{105}\)
\(9,\dfrac{-4}{2}+\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{-12}{6}+\dfrac{2}{6}-\dfrac{1}{4}=\dfrac{-10}{6}-\dfrac{1}{4}=\dfrac{-5}{3}-\dfrac{1}{4}\)
\(=\dfrac{-20}{12}-\dfrac{3}{12}=\text{ }\dfrac{-23}{12}\)
\(10,5,4-1,5-\left(7,2-1\right)=3,9-6,2=-2,3\)
\(11,4,9-\left(1,5-7,7+3\right)=4,9-\left(-3,2\right)=8,1\)
\(12,7,8-4,7+\left(5,3-1,4\right)=3,1+3,9=7\)
\(14,\dfrac{1}{2}-0,4+\dfrac{1}{5}\text{=}0,5-0,4+0,2=0,3\)
\(15,4,2-\dfrac{4}{5}+\dfrac{1}{2}=4,2-0,8+0,5=3,9\)
a)<=>A>hoặc=|x-2016+2015-x| <=>A>hoặc= 1 với mọi x
=>Amin=1 khi (x-2016).(2015-x)>hoặc =0
=>2015<hoặc=x<hoặc=2016
= (1 + 789) + ( 4 + 56) + (2+3)
= 790 + 60 + 5
= 850 + 5
= 855
1 + 2 + 3 + 4 + 56 + 789
= ( 1+789 ) + ( 4 + 56 ) + ( 2 + 3 )
= 790 + 60 + 5
= 850 + 5
= 855