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1.
\(\left(\dfrac{-1}{8}+\dfrac{-5}{6}\right)\cdot\dfrac{6}{23}\\ =-\dfrac{23}{24}\cdot\dfrac{6}{23}\\ =-\dfrac{6}{24}=-\dfrac{1}{4}\)
2. Xem lại đề nha!
4.
\(x+0,75=-1\dfrac{1}{4}\\ x+\dfrac{3}{4}=-\dfrac{3}{4}\\ x=-\dfrac{3}{4}-\dfrac{3}{4}\\ x=-\dfrac{3}{4}+\left(-\dfrac{3}{4}\right)=-\dfrac{6}{4}=-\dfrac{3}{2}\)
5.
\(\dfrac{x}{28}=-\dfrac{4}{7}\\ \Leftrightarrow7x=-4.28\\ \Rightarrow7x=-112\\ \Rightarrow x=-112:7=-16\)
6.
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\\ \Leftrightarrow\left(3x-y\right).4=3\left(x+y\right)\\ \Rightarrow12x-4y=3x+3y\\ \Rightarrow12x-3x=4y+3y\\ \Rightarrow9x=7y\\ \Leftrightarrow\dfrac{x}{7}=\dfrac{y}{9}\Leftrightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
Vậy giá trị của tỉ số \(\dfrac{x}{y}=\dfrac{7}{9}\).
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\\ \Rightarrow4\cdot\left(3x-y\right)=3\cdot\left(x+y\right)\\ \Leftrightarrow12x-4y=3x+3y\\ \Leftrightarrow12x-3x=3y+4y\\ \Leftrightarrow9x=7y\\ \Rightarrow\dfrac{x}{7}=\dfrac{y}{9}\\ \Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
Vậy \(\dfrac{x}{y}=\dfrac{7}{9}\)
Ta có: \(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\)
\(\Rightarrow\left(3x-y\right)4=\left(x+y\right)3\)
\(\Rightarrow12x-4y=3x+3y\)
\(\Rightarrow12x-3x=4y+3y\)
\(\Rightarrow9x=7y\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
Vậy \(\dfrac{x}{y}=\dfrac{7}{9}.\)
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\)
\(\Leftrightarrow4\left(3x-y\right)=3\left(x+y\right)\)
\(\Leftrightarrow12x-4y=3x+3y\)
\(\Leftrightarrow12x=3x+7y\)
\(\Leftrightarrow9x=7y\)
\(\Leftrightarrow\dfrac{x}{7}=\dfrac{y}{9}\Leftrightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
1 a) \(\dfrac{\left(-2\right)}{5}\)= \(\dfrac{-6}{15}\); \(\dfrac{15}{-6}\)= \(\dfrac{5}{-2}\); \(\dfrac{-6}{-2}\)= \(\dfrac{15}{5}\); \(\dfrac{-2}{-6}\)= \(\dfrac{5}{15}\)
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\)
\(\Rightarrow4\left(3x-y\right)=3\left(x+y\right)\)
\(\Rightarrow12x-4y=3x+3y\)
\(\Rightarrow12x-4y-3y=3x\)
\(\Rightarrow12x-7y=3x\)
\(\Rightarrow12x-3x=7y\)
\(\Rightarrow9x=7y\)
\(\Rightarrow\dfrac{x}{7}=\dfrac{y}{9}\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
\(a,A=\dfrac{\dfrac{3}{4}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\\ A=\dfrac{\dfrac{405}{572}}{\dfrac{645}{1001}}+\dfrac{\dfrac{5}{12}}{\dfrac{25}{24}}\\ A=\dfrac{189}{172}+\dfrac{2}{5}\\ A=\dfrac{1289}{860}\)
Ta có:
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\\ \Rightarrow12x-4y=3x+3y\\ \Rightarrow9x=7y\\ \Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
Vậy.........
Có: \(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\)
\(\Leftrightarrow4\left(3x-y\right)=3\left(x+y\right)\)
\(\Leftrightarrow12x-4y=3x+3y\)
\(\Leftrightarrow12x+3x=3y+4y\)
\(\Leftrightarrow15x=7y\)
\(\Leftrightarrow\dfrac{x}{y}=\dfrac{7}{15}\)
Kết luận: ...