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tìm n N để \(\frac{n}{n+1}\) + \(\frac{n}{n+2}\) là số tự nhiên
giúp mik với sắp thi r

a) \(\frac{-8}{3}+\frac{7}{5}+\frac{-71}{15}\)< \(x\) < \(\frac{-13}{7}+\frac{19}{14}+\frac{-7}{2}\)
Ta có: \(\frac{-8}{3}+\frac{7}{5}+\frac{-71}{15}\)
=\(\frac{-40}{15}+\frac{21}{15}+\frac{-71}{15}\)
=\(\frac{-90}{15}\)
=\(-6\)
Ta có: \(\frac{-13}{7}+\frac{19}{14}+\frac{-7}{2}\)
=\(\frac{-26}{14}+\frac{19}{14}+\frac{-49}{14}\)
=\(\frac{-56}{14}\)
=\(-4\)
=> \(-6\)< \(x\)<\(-4\)
=> \(x=-5\)
b)\(\frac{5}{17}+\frac{-4}{9}+\frac{-20}{31}+\frac{12}{17}+\frac{-11}{31}\)< \(\frac{x}{9}\)<\(\frac{-3}{7}+\frac{7}{15}+\frac{4}{-7}+\frac{8}{15}+\frac{2}{3}\)
Ta có: \(\frac{5}{17}+\frac{-4}{9}+\frac{-20}{31}+\frac{12}{17}+\frac{-11}{31}\)
=\(\left(\frac{5}{17}+\frac{12}{17}\right)+\left(\frac{-20}{31}+\frac{-11}{31}\right)+\frac{-4}{9}\)
=\(1+\left(-1\right)+\frac{-4}{9}\)
=\(0+\frac{-4}{9}\)
=\(\frac{-4}{9}\)
Ta có: \(\frac{-3}{7}+\frac{7}{15}+\frac{4}{-7}+\frac{8}{15}+\frac{2}{3}\)
=\(\frac{-3}{7}+\frac{7}{15}+\frac{-4}{7}+\frac{8}{15}+\frac{2}{3}\)
=\(\left(\frac{-3}{7}+\frac{-4}{7}\right)+\left(\frac{7}{15}+\frac{8}{15}\right)+\frac{2}{3}\)
=\(\left(-1\right)+1+\frac{2}{3}\)
=\(0+\frac{2}{3}\)
=\(\frac{2}{3}\)
=> \(\frac{-4}{9}\)< \(\frac{x}{9}\)<\(\frac{2}{3}\)
=
=> \(\frac{-4}{9}\)<\(\frac{x}{9}\)<\(\frac{6}{9}\)
=> \(-4\)< \(x\)<\(6\)
=>\(x\in\left\{-3;-2;-1;0;1;2;3;4;5\right\}\)

\(\Leftrightarrow\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\Rightarrow x\in\left(-2;-1;0;1;2\right)\)
\(\Leftrightarrow\frac{-1}{24}\le\frac{x}{24}\le\frac{5}{24}\Rightarrow x\in\left(-1;0;1;2;3;4;5\right)\)
2 câu sau tự làm nha
\(-\frac{5}{17}+\frac{3}{17}\le\frac{x}{17}\le\frac{13}{17}+-\frac{11}{17}\)
\(\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\)
=> \(x\in\left\{-2;-1;0;1;2\right\}\)

=>(5/17+12/17)+(-20/31-11/31)-4/9<=x/9<=(-3/7-4/7)+(7/15+8/15)+2/3
=>-4/9<=x/9<=6/9
=>-4<=x<=6
hay \(x\in\left\{-4;-3;-2;-1;0;...;6\right\}\)

bài 1b)
\(8\frac{1}{14}-6\frac37\)
C1:\(\frac{113}{14}-\frac{45}{7}\) =\(\frac{113}{14}-\frac{90}{14}=\frac{23}{14}\)
C2:\(8\frac{1}{14}-6\frac37=\left(8-6\right)+\left(\frac{1}{14}-\frac37\right)=2+\left(\frac{1}{14}-\frac{6}{14}\right)\)
\(=2+\frac{-5}{14}=\frac{28}{14}-\frac{5}{14}=\frac{23}{14}\)
bài 1 c)\(7-3\frac67\)
C1:\(\) \(7-3\frac67=7-\frac{27}{7}=\frac{49}{7}-\frac{27}{7}=\frac{22}{7}\)
C2:\(7-3\frac67=\left(7-3\right)-\frac67=4-\frac67=\frac{28}{7}-\frac67=\frac{22}{7}\)

\(A=\left(\frac{5}{17}+\frac{12}{17}\right)+\left(-\frac{20}{31}-\frac{11}{31}\right)+-\frac{4}{9}\)
\(A=1+\left(-1\right)+\frac{-4}{9}\)
\(A=0+\frac{-4}{9}=\frac{-4}{9}\)
\(B=\left(\frac{-3}{7}+\frac{-4}{7}\right)+\left(\frac{7}{15}+\frac{8}{15}\right)+\frac{-2}{3}\)
\(B=-1+1+\frac{-2}{3}\)
\(B=\frac{-2}{3}\)

B1a)\(11\frac34-\left(6\frac56-4\frac12\right)+1\frac23\)
=\(11\frac34-6\frac56+4\frac12+1\frac23\)
=\(\left(11-6+4+1\right)+\left(\frac34-\frac56+\frac12+\frac23\right)\)
=\(10+\left(\frac{9}{12}-\frac{10}{12}+\frac{6}{12}+\frac{8}{12}\right)\)
=\(10+\left(-\frac{1}{12}+\frac{6}{12}+\frac{8}{12}\right)\)
=10+\(\frac{13}{12}\)
=\(\frac{120}{12}+\frac{13}{12}\)
=\(\frac{133}{12}\)
b)\(2\frac{17}{20}-1\frac{11}{5}+6\frac{9}{20}:3\)
= \(\frac{57}{20}-\frac{16}{5}+\frac{129}{20}\times\frac13\)
=\(\frac{57}{20}-\frac{16}{5}+\frac{129}{60}\)
=\(\frac{171}{60}-\frac{192}{60}+\frac{129}{60}\)
=\(\frac{108}{60}\)
=\(\frac95\)
\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
bạn ơi bạn giải câu b được ko. mk ko biết làm câu b