\(1\frac{1}{7}-\frac{5}{7}< \frac{x}{7}< 2\frac{1}{14}-1\frac{3}{14}\)
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\(1\frac{1}{7}-\frac{5}{7}< \frac{x}{7}< 2\frac{1}{14}-1\frac{3}{14}\)
=> \(\frac{8}{7}-\frac{5}{7}< \frac{x}{7}< \frac{29}{14}-\frac{17}{14}\)
=> \(\frac{3}{7}< \frac{x}{7}< \frac{12}{14}\)
=> \(\frac{3}{7}< \frac{x}{7}< \frac{6}{7}\)
=> 3 < x < 6
=> x thuộc { 3 ; 4 ; 5 ; 6 }
\(a)x+30\%x=-1,31\)
\(\Leftrightarrow x+\frac{3x}{10}=-1,31\)
\(\Leftrightarrow10x+3x=-13,1\)
\(\Leftrightarrow13x=-13,1\Leftrightarrow x=-\frac{131}{130}\)
\(b)\left(x-\frac{1}{2}\right):\frac{1}{3}+\frac{5}{7}=9\frac{5}{7}\)
\(\Leftrightarrow\frac{2x-1}{2}.3+\frac{5}{7}=\frac{68}{7}\)
\(\Leftrightarrow\frac{6x-3}{2}=\frac{63}{7}\)
\(\Leftrightarrow\frac{6x-3}{2}=9\)
\(\Leftrightarrow6x-3=18\)
\(\Leftrightarrow x=\frac{7}{2}\)
\(A=\left(\frac{-4}{5}+\frac{4}{3}\right)+\left(\frac{-5}{4}+\frac{14}{5}\right)-\frac{7}{3}\)
\(A=\frac{-4}{5}+\frac{4}{3}+\frac{-5}{4}+\frac{14}{5}-\frac{7}{3}\)
\(A=\left(\frac{-4}{5}+\frac{14}{5}\right)+\left(\frac{4}{3}-\frac{7}{3}\right)+\frac{-5}{4}\)
\(A=2+\left(-1\right)+\frac{-5}{4}\)
\(A=\frac{-1}{4}\)
A=\(\left(\frac{-4}{5}+\frac{4}{3}\right)+\left(\frac{-5}{4}+\frac{14}{5}\right)-\frac{7}{3}\)\(\frac{7}{3}\)
=\(\frac{-4}{5}+\frac{4}{3}+\frac{-5}{4}+\frac{14}{5}+\frac{-7}{3}\)=\(\left(\frac{-4}{5}+\frac{14}{5}\right)+\left(\frac{4}{3}+\frac{-7}{3}\right)+\frac{-5}{4}\)
=\(\frac{10}{5}+\frac{-3}{3}+\frac{-5}{4}\)=\(2-1+\frac{-5}{4}\)=\(1+\frac{-5}{4}\)=\(\frac{4}{4}+\frac{-5}{4}\)=\(\frac{4-5}{4}\)=\(\frac{-1}{4}\)
Mình làm như thế này nek
\(\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{0,75+\frac{9}{7}-2\frac{2}{5}}+\frac{\frac{3}{14}-\frac{2}{10}+\frac{5}{18}+\frac{7}{66}}{\frac{6}{7}-\frac{4}{5}+\frac{10}{9}+\frac{14}{33}}\)
\(=\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{\frac{2}{4}+\frac{9}{7}-\frac{12}{5}}+\frac{\frac{1}{2}\cdot\left(\frac{3}{7}-\frac{2}{5}+\frac{5}{9}+\frac{7}{33}\right)}{2\cdot\left(\frac{3}{7}-\frac{2}{5}+\frac{5}{9}+\frac{7}{33}\right)}\)
\(=\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{3\cdot\left(\frac{1}{4}+\frac{3}{7}-\frac{4}{5}\right)}+\frac{\frac{1}{2}}{2}\)
\(=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
\(x\in\left\{4;5\right\}\)
Thiếu một điều kiện \(x\) là số tự nhiên nữa nhé
Ta có :
\(1\frac{1}{7}-\frac{5}{7}< \frac{x}{7}< 2\frac{1}{14}-1\frac{3}{14}\)
\(\Leftrightarrow\)\(\frac{8}{7}-\frac{5}{7}< \frac{x}{7}< \frac{29}{14}-\frac{17}{14}\)
\(\Leftrightarrow\)\(\frac{3}{7}< \frac{x}{7}< \frac{12}{14}\)
\(\Leftrightarrow\)\(\frac{3}{7}< \frac{x}{7}< \frac{6}{7}\)
\(\Leftrightarrow\)\(3< x< 6\)
\(\Rightarrow\)\(x\in\left\{4;5\right\}\)
Vậy \(x\in\left\{4;5\right\}\)
Chúc bạn học tốt ~