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1)(1+x)+(2+x)+(3+x)+.....+(22+x)=77
=>1+x+2+x+.....+22+x=77
=>22.x+(1+2+3+.....+21+22)=77
=>22.x+253=77
=>22x=77-253
=>22x=-176
=>x=-8
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B=[2+(-7)]+[12+(-17)]+...+[102+(-107)]+112
B=-5+-5+...+-5+112
B=-5.503+112=-2403
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\(-\dfrac{5}{6}-\dfrac{3}{10}=-\dfrac{17}{15}\)
\(-\dfrac{1}{3}+-\dfrac{3}{4}-\dfrac{2}{5}=-\dfrac{13}{12}-\dfrac{2}{5}=-\dfrac{89}{60}\)
\(-\dfrac{2}{9}-\dfrac{11}{6}-\dfrac{7}{12}=-\dfrac{37}{18}-\dfrac{7}{12}=-\dfrac{95}{36}\)
\(\dfrac{2}{5}--\dfrac{7}{10}-\dfrac{4}{15}=\dfrac{11}{10}-\dfrac{4}{15}=\dfrac{5}{6}\)
\(#YH\)
\(#NBaoNgoc\)
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a: \(=\dfrac{1}{7}\cdot\dfrac{17}{5}-\dfrac{3}{2}\cdot\dfrac{2-7}{12}\)
\(=\dfrac{17}{35}-\dfrac{-15}{24}\)
\(=\dfrac{17}{35}+\dfrac{15}{24}=\dfrac{311}{280}\)
b: \(=\dfrac{25}{77}-\dfrac{3-7}{15}\cdot\dfrac{10}{11}:\dfrac{3}{4}\)
\(=\dfrac{25}{77}+\dfrac{4}{15}\cdot\dfrac{10}{11}\cdot\dfrac{4}{3}\)
\(=\dfrac{25}{77}+\dfrac{32}{99}=\dfrac{449}{693}\)
Đặt A =\(\frac{154154154-2002002.77}{2+7+12+...+102}=\frac{154.1001001-2.1001001.77}{2+7+12+...+102}\)
\(=\frac{154.1001001-154.1001001}{2+7+12+...+102}=\frac{0}{2+7+12+...+102}=0\)
Vậy A = 0