tìm x
\(\frac{15}{41}+\frac{-138}{41}\le x< \frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)
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\(\frac{15}{41}+\frac{-138}{41}\le x< \frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)
\(\Leftrightarrow-3\le x< 1\)
\(\Leftrightarrow x\in\left\{-3;-2;-1;0\right\}\)
\(\frac{15}{41}+\frac{-138}{41}\le x< \frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{15+(-138)}{41}\le x< \frac{1\cdot3}{6}+\frac{1\cdot2}{6}+\frac{1}{6}\)
\(\Rightarrow\frac{-123}{41}\le x< \frac{3}{6}+\frac{2}{6}+\frac{1}{6}\)
\(\Rightarrow-3\le x< 1\Leftrightarrow x\in\left\{-3;-2;-1;0\right\}\)
\(\frac{15}{41}+\frac{-138}{41}\le x< \frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow-3\le x< 1\)
\(\Rightarrow-3\le-3;-2;-1;0< 1\)
\(\Rightarrow x\in\left\{-3;-2;-1;0\right\}\)
~ Hok tốt ~
\(\frac{15}{41}+\frac{-138}{41}< x< \frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)
\(\Leftrightarrow\frac{-123}{41}< x< \frac{1.3+1.2+1}{6}\)
\(\Leftrightarrow-3< x< 1\)
\(\Rightarrow x\in\left\{-2;-1;0\right\}\)
\(\frac{x}{5}=\frac{15}{2}-\frac{51}{10}\)
\(\frac{x}{5}=\frac{15.5-51}{10}\)
\(\frac{x}{5}=\frac{24}{10}\)
\(\frac{x}{5}=\frac{12}{5}\)
\(x=12\)
\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
Bài làm
\(\frac{15}{41}+\frac{-138}{41}\le x< \frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)
\(\frac{123}{41}\le x< 1\)
\(\frac{123}{41}\le x< \frac{41}{41}\)
\(\Rightarrow123\le x< 41\)
\(\Rightarrow x\in\varnothing\)
=> -123 / 41 < hoặc = x < 1
=> -3 < hoặc = x <1
=>x = ( -3 ; -2 ; -1 ; 0 )