\(\frac{1}{4}+\frac{8}{9}\le\frac{x}{36}< 1-\frac{3}{8}+\frac{5}{6}\)
tìm x ; x e Z
K
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Những câu hỏi liên quan
15 tháng 3 2019
\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
9 tháng 8 2019
\(a,\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)< x< \left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}\)
\(taco:\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)=\frac{35}{36}\cdot\frac{-36}{35}=-1\)
\(\left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}=\frac{13}{8}\cdot\frac{8}{13}=1\)
\(=>x=0\)
\(b,\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}< x< \frac{-1}{2}+2+\frac{5}{2}\)(dau <co dau gach ngang o duoi nha)
\(taco:\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}=\frac{-5}{6}+\frac{8}{3}+\frac{-29}{3}=\frac{-5}{6}+\frac{16}{6}+\frac{-58}{6}=\frac{-47}{6}=-7,8\)
\(\frac{-1}{2}+2+\frac{5}{2}=\frac{3}{2}+\frac{5}{2}=4\)
tu do \(=>x=-7,8;...;0;1;2;3;4\)
KT
11 tháng 8 2018
\(\frac{1}{4}+\frac{8}{9}\le\frac{x}{36}\le1-\left(\frac{3}{8}-\frac{5}{6}\right)\)
<=> \(\frac{41}{36}\le\frac{x}{36}\le\frac{35}{24}\)
<=> \(\frac{82}{72}\le\frac{2x}{72}\le\frac{105}{72}\)
<=> \(82\le2x\le105\)
<=> \(41\le x\le52,5\)
Do \(x\in N\)nên \(x=\left\{x\in N|41\le x\le52,5\right\}\)
C =22−3 x 4− x =12−3 x +104− x =3(4− x )4− x +104− x =3+104− x ... +199.100 ). x =201251 +201252 +. .... =12√3 (4+2√3√3+1+4−2√3√3−1 )=12√3 .4√3−4+6−2√3+4√3+4−6−2√3(√3−1)(√3+1).
\(\frac{1}{4}+\frac{8}{9}< \frac{x}{36}< 1-\frac{3}{8}+\frac{5}{6}\)
\(=\frac{41}{36}< \frac{x}{36}< \frac{35}{24}\)
=>\(x=35< x< 41\)