so sánh 2 số hữu tỉ
a)-22/35 và -103/177 b)-13/38 và 15/-88
c)22/-67 và -51/152 c)-17/35 và -43/85
mik cần gấp ạ!
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\(3x-\dfrac{3}{5}=\dfrac{-7}{10}\\ 3x=\dfrac{-7}{10}+\dfrac{3}{5}\\ 3x=\dfrac{-7}{10}+\dfrac{6}{10}\\ 3x=\dfrac{-1}{10}\\ x=\dfrac{-1}{30}\)
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\(\dfrac{2}{3}+\dfrac{1}{3}:x=\dfrac{3}{5}\\ \dfrac{1}{3}:x=\dfrac{3}{5}-\dfrac{2}{3}\\ \dfrac{1}{3}:x=\dfrac{9}{15}-\dfrac{10}{15}\\ \dfrac{1}{3}:x=\dfrac{-1}{15}\\ x=\dfrac{1}{3}:\dfrac{-1}{15}\\ x=-5\)
\(3x-\dfrac{3}{5}=-\dfrac{7}{10}\)
\(3x\) \(=-\dfrac{7}{10}+\dfrac{3}{5}\)
\(3x\) \(=-\dfrac{1}{10}\)
\(x\) \(=-\dfrac{1}{10}:3\)
\(x\) \(=-\dfrac{1}{30}\)
Vậy \(x=-\dfrac{1}{30}\)
\(\dfrac{2}{3}+\dfrac{1}{3}:x=\dfrac{3}{5}\)
\(\dfrac{1}{3}:x=\dfrac{3}{5}-\dfrac{2}{3}\)
\(\dfrac{1}{3}:x=-\dfrac{1}{15}\)
\(x=\dfrac{1}{3}:-\dfrac{1}{15}\)
\(x=-5\)
Vậy \(x=-5\)
\(\left(\dfrac{2}{7}-\dfrac{9}{4}\right)-\left(-\dfrac{3}{7}+\dfrac{5}{4}\right)-\left(\dfrac{2}{4}-\dfrac{9}{7}\right)\)
\(=\dfrac{2}{7}-\dfrac{9}{4}+\dfrac{3}{7}-\dfrac{5}{4}-\dfrac{2}{4}+\dfrac{9}{7}\)
\(=\left(\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{9}{7}\right)-\left(\dfrac{9}{4}+\dfrac{5}{4}+\dfrac{2}{4}\right)\)
\(=2-4\)
\(=-2\)
\(\dfrac{4}{7}+\dfrac{3}{7}x=\dfrac{1}{2}\\ \dfrac{3}{7}x=\dfrac{1}{2}-\dfrac{4}{7}\\ \dfrac{3}{7}x=-\dfrac{1}{14}\\ x=-\dfrac{1}{14}:\dfrac{3}{7}\\ x=-\dfrac{1}{6}\)
Vậy....
\(\dfrac{17}{6}-\left(x+\dfrac{7}{6}\right)=\dfrac{7}{4}\\x+\dfrac{7}{6}=\dfrac{17}{6}-\dfrac{7}{4}\\ x+\dfrac{7}{6}=\dfrac{13}{12}\\ x=\dfrac{13}{12}-\dfrac{7}{6}\\ x=-\dfrac{1}{12} \)
Vậy....
\(\dfrac{4}{7}+\dfrac{3}{7}x=\dfrac{1}{2}\\ \Rightarrow\dfrac{3}{7}x=\dfrac{1}{2}-\dfrac{4}{7}\\ \Rightarrow\dfrac{3}{7}x=-\dfrac{1}{14}\\ \Rightarrow x=-\dfrac{1}{14}:\dfrac{3}{7}\\ \Rightarrow x=-\dfrac{1}{6}\)
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\(\dfrac{17}{6}-\left(x+\dfrac{7}{6}\right)=\dfrac{7}{4}\\ \Rightarrow\dfrac{17}{6}-x-\dfrac{7}{6}=\dfrac{7}{4}\\ \Rightarrow\dfrac{5}{3}-x=\dfrac{7}{4}\\ \Rightarrow x=\dfrac{5}{3}-\dfrac{7}{4}\\ \Rightarrow x=-\dfrac{1}{12}\)
\(\left(\dfrac{1}{2}-\dfrac{1}{3}\right)-\left(\dfrac{5}{3}-\dfrac{3}{2}\right)+\left(\dfrac{7}{3}-\dfrac{5}{2}\right)\\ =\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{5}{3}+\dfrac{3}{2}+\dfrac{7}{3}-\dfrac{5}{2}\\ =\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)+\left(-\dfrac{1}{3}-\dfrac{5}{3}+\dfrac{7}{3}\right)\\ =-\dfrac{1}{2}+\dfrac{1}{3}\\ =-\dfrac{1}{6}\)
a)
\(\dfrac{-7}{20}\) = \(\dfrac{4}{5}\) + \(\left(\dfrac{-23}{20}\right)\)
b)
\(\dfrac{-7}{20}\) = \(\dfrac{-1}{5}\) + \(\left(\dfrac{-3}{20}\right)\)
Do x và y là hai đại lượng tỉ lệ nghịch nên:
\(a=xy=2\cdot\left(-15\right)=-30\)
x;y tỉ lệ nghịch với 4;5 nên :
\(k=\dfrac{x}{4}=\dfrac{y}{5}\)
Theo TCDSTLBN ta có :
\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{x+y}{4+5}=\dfrac{18}{9}=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{4}=2\\\dfrac{y}{5}=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=8\\y=10\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(8;10\right)\)
\(\dfrac{x}{y}=\dfrac{5}{4}\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}\)
ADTC dãy tỉ số bằng nhau
\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{x+y}{5+4}=\dfrac{18}{9}=2\Rightarrow x=10;y=8\)