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\(a)\frac{7}{12}x+0,75=-2\frac{1}{6}\)
\(\Rightarrow\frac{7}{12}x+\frac{3}{4}=-\frac{13}{6}\)
\(\Rightarrow\frac{7}{12}x=-\frac{13}{6}-\frac{3}{4}\)
\(\Rightarrow\frac{7}{12}x=-\frac{26}{12}-\frac{9}{12}\)
\(\Rightarrow\frac{7}{12}x=-\frac{35}{12}\)
\(\Rightarrow x=-\frac{35}{12}:\frac{7}{12}\)
\(\Rightarrow x=-\frac{35}{12}.\frac{12}{7}\)
\(\Rightarrow x=-5\)
\(b)-1< \frac{x}{4}< \frac{1}{2}\)
\(\Rightarrow-\frac{4}{4}< \frac{x}{4}< \frac{2}{4}\)
\(\Rightarrow-4< x< 2\)
\(\Rightarrow x\in\left\{-3;-2;-1;0;1\right\}\)
Chúc bạn học tốt !!!
\(\frac{7}{12}x+0,75=-2\frac{1}{6}\)
\(\frac{7}{12}x=-\frac{13}{6}-\frac{3}{4}\)
\(\frac{7}{12}x=\frac{-35}{12}\)
\(x=\frac{-35}{12}:\frac{7}{12}\)
\(x=-5\)
b) \(-1< \frac{x}{4}< \frac{1}{2}\)
\(\frac{-4}{4}< \frac{x}{4}< \frac{2}{4}\)
\(\Rightarrow-4< x< 2\)
mà \(x\in Z\) nên \(x\in\left\{-3;-2;\pm1;0\right\}\)
\(1.\)\(M=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}\)
\(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}\)
\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{6}-\frac{1}{7}\)
\(M=1-\frac{1}{7}=\frac{6}{7}\)
Mình làm câu 1 thoi nha!
1.
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\)
=\(1-\frac{1}{7}\)
=\(\frac{6}{7}\)
\(|-1-\frac{-2}{3}|-\left(\frac{7}{-6}-x+\frac{1}{2}\right)=-|-\frac{7}{4}|\)
\(\frac{1}{3}+\frac{7}{6}+x-\frac{1}{2}=-\frac{7}{4}\)
\(\frac{3}{2}+x-\frac{1}{2}=-\frac{7}{4}\)
\(\frac{3}{2}+x=-\frac{5}{4}\)
\(x=-\frac{11}{4}\)
\(\frac{4}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)=\frac{4}{3}.\frac{-1}{3}=\frac{-4}{9}\)
k nha