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\(\frac{2}{3}\left(x-1\right)-x-\frac{3}{4}=1\)
<=> \(\frac{2}{3}x-\frac{2}{3}-x-\frac{3}{4}=1\)
<=> \(-\frac{1}{3}x-\frac{17}{12}=1\)
<=> \(-\frac{1}{3}x=\frac{29}{12}\)
<=> \(x=-\frac{29}{4}\)
\(\frac{5}{6}\left(x+2\right)-x-\frac{1}{2}=\frac{1}{3}\)
<=> \(\frac{5}{6}x+\frac{5}{3}-x-\frac{1}{2}=\frac{1}{3}\)
<=> \(-\frac{1}{6}x+\frac{7}{6}=\frac{1}{3}\)
<=> \(-\frac{1}{6}x=-\frac{5}{6}\)
<=> \(x=5\)
học tốt
Cách 1 :
\(M=\left(7-\frac{1}{3}+\frac{3}{4}\right)-\left(6+\frac{2}{3}-\frac{1}{4}\right)\)
\(M=\left(\frac{84}{12}-\frac{4}{12}+\frac{9}{12}\right)-\left(\frac{72}{12}+\frac{8}{12}-\frac{3}{12}\right)\)
\(M=\frac{89}{12}-\frac{77}{12}\)
\(M=\frac{12}{12}\)
\(M=1\)
Cách 2 :
\(M=\left(7-\frac{1}{3}+\frac{3}{4}\right)-\left(6+\frac{2}{3}-\frac{1}{4}\right)\)
\(M=7-\frac{1}{3}+\frac{3}{4}-6-\frac{2}{3}+\frac{1}{4}\)
\(M=\left(7-6\right)-\left(\frac{2}{3}+\frac{1}{3}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)\)
\(M=1-1+1\)
\(M=1\)
Cách 1:
\(M=\left(7-\frac{1}{3}+\frac{3}{4}\right)-\left(6+\frac{2}{3}-\frac{1}{4}\right)\)
\(M=\frac{89}{12}-\frac{77}{12}\)
\(M=1\)
Cách 2:
\(M=\left(7-\frac{1}{3}+\frac{3}{4}\right)-\left(6+\frac{2}{3}-\frac{1}{4}\right)\)
\(M=7-\frac{1}{3}+\frac{3}{4}-6-\frac{2}{3}+\frac{1}{4}\)
\(M=\left(7-6\right)-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)\)
\(M=1-1+1\)
\(M=1\)
\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)
\(A=1+\frac{1+2}{2}+\frac{1+2+3}{3}+\frac{1+2+3+4}{4}+...+\frac{1+2+3+...+16}{16}\)
\(A=1+\frac{2\left(2+1\right):2}{2}+\frac{3\cdot\left(3+1\right):2}{3}+\frac{4\left(4+1\right):2}{4}+...+\frac{16\left(16+1\right):2}{16}\)
\(A=1+\frac{2+1}{2}+\frac{3+1}{2}+\frac{4+1}{2}+...+\frac{16+1}{2}\)
\(A=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{17}{2}\)
\(A=\frac{2+3+4+5+...+17}{2}\)
\(A=\frac{152}{2}\)
\(A=76\)
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\(\left|\frac{13}{4}-2x\right|=\frac{2}{5}+\frac{5}{2}=\frac{29}{10}\left(1\right)\)
+ Nếu \(\frac{13}{4}-2x\ge0\Leftrightarrow x\le\frac{13}{8}\)
\(\Rightarrow\left(1\right)\Leftrightarrow\frac{13}{4}-2x=\frac{29}{10}\Rightarrow x=\frac{7}{40}\) so với điều kiện \(x\le\frac{13}{8}\) nên thoả mãn
+ Nếu \(\frac{13}{4}-2x< 0\Leftrightarrow x>\frac{13}{8}\)
\(\Rightarrow\left(1\right)\Leftrightarrow2x-\frac{13}{4}=\frac{29}{10}\Leftrightarrow x=\frac{123}{40}\) so với điều kiện \(x>\frac{13}{8}=\frac{65}{40}\) nên thoả mãn