Đặt

\(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)

\(S=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(S=\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)\)

\(S=\dfrac{49}{99}\)

\(\Rightarrow\dfrac{1}{x}-\dfrac{1}{9999}=\dfrac{49}{99}\)

\(\Rightarrow\dfrac{1}{x}=\dfrac{50}{101}\Leftrightarrow50x=101\Leftrightarrow x=\dfrac{101}{50}\)

Đặt \(M=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)

\(M=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(M=\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)\)

\(M=\dfrac{1}{2}\cdot\dfrac{98}{99}\)

\(M=\dfrac{49}{99}\)

a) \(\dfrac{1+\dfrac{1}{4}}{1-\dfrac{1}{4}}:\dfrac{1+\dfrac{1}{8}}{1-\dfrac{1}{8}}\\ =\dfrac{\dfrac{5}{4}}{\dfrac{3}{4}}:\dfrac{\dfrac{9}{8}}{\dfrac{7}{8}}\\ =\dfrac{5}{3}:\dfrac{9}{7}\\ =\dfrac{5}{3}.\dfrac{9}{7}\\ =\dfrac{35}{27}=1\dfrac{8}{27}\)

b) \(0,25+37\%-2\dfrac{1}{4}\\ =\dfrac{1}{4}+\dfrac{37}{100}-\dfrac{9}{4}\\ =\dfrac{25+37-225}{100}\\ =-\dfrac{163}{100}=-1\dfrac{63}{100}\)

\(=\dfrac{-5}{12}+\dfrac{5}{12}+\dfrac{6}{11}+\dfrac{5}{11}+\dfrac{7}{18}=1+\dfrac{7}{18}=\dfrac{25}{18}\)

Ta có: \(\frac{2}{3}a=\frac{1}{4}b\)

\(\Leftrightarrow\frac{2a}{3}=\frac{b}{4}\)

\(\Leftrightarrow2a=\frac{3b}{4}\)

hay \(a=\frac{3b}{4}:2=\frac{3b}{8}\)

Ta có: \(\frac{1}{2}b=\frac{1}{3}c\)

\(\Leftrightarrow\frac{b}{2}=\frac{c}{3}\)

hay \(c=\frac{3b}{2}\)

Ta có: a+b+c=90

\(\Leftrightarrow\frac{3b}{8}+b+\frac{3b}{2}=90\)

\(\Leftrightarrow b\left(\frac{3}{8}+1+\frac{3}{2}\right)=90\)

\(\Leftrightarrow b\cdot\frac{23}{8}=90\)

hay \(b=90:\frac{23}{8}=\frac{720}{23}\)

Ta có: \(a=\frac{3b}{8}\)(cmt)

hay \(a=3\cdot\frac{720}{23}:8=\frac{270}{23}\)

Ta có: a+b+c=90

\(\Leftrightarrow c=90-a-b=90-\frac{270}{23}-\frac{720}{23}=\frac{1080}{23}\)

Vậy: \(\left(a,b,c\right)=\left(\frac{270}{23};\frac{720}{23};\frac{1080}{23}\right)\)

Giải:

Ta có:

\(\left(a+b+c+d\right)-\left(a+c+d\right)._{\left(1\right)}\)

\(=a+b+c+d-a-c-d.\)

\(=\left(a-a\right)+\left(c-c\right)+\left(d-d\right)+b.\)

\(=0+0+0+b=b.\)

Thay số vào \(_{\left(1\right)}\)\(\Rightarrow1-2=b\Rightarrow b=-1\in Z.\)

\(\left(a+b+c+d\right)-\left(a+b+d\right)._{\left(2\right)}\)

\(=a+b+c+d-a-b-d.\)

\(=\left(a-a\right)+\left(b-b\right)+\left(d+d\right)+c.\)

\(=0+0+0+c=c.\)

Thay số vào \(_{\left(2\right)}\)\(\Rightarrow1-3=c\Rightarrow c=-2\in Z.\)

\(\left(a+b+c+d\right)-\left(a+b+c\right)_{\left(3\right)}.\)

\(=a+b+c+d-a-b-c.\)

\(=\left(a-a\right)+\left(b-b\right)+\left(c-c\right)+d.\)

\(=0+0+0+d=d.\)

Thay số vào \(_{\left(3\right)}\)\(\Rightarrow1-4=d\Rightarrow d=-3\in Z.\)

\(\Rightarrow a+b+c+d=1.\)

\(a+\left(-1\right)+\left(-2\right)+\left(-3\right)=1.\)

\(\Rightarrow a=1-\left(-1\right)-\left(-2\right)-\left(-3\right).\)

\(\Rightarrow a=1+1+2+3=7\in Z.\)

Vậy \(\left\{a;b;c;d\right\}=\left\{7;-1;-2;-3\right\}.\)

Do a + b + c + d = 1 mà a + c + d = 2
=> b = 1 - 2 = -1
=> c = 1 - 3 = -2
=> d = 1 - 4 = -3
=> a = 1 - (-1 - 2 - 3) = 7
@Valentine