\(\dfrac{20}{x}=\dfrac{4}{15}\)
tìm x,y
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\(a,\dfrac{x}{4}=\dfrac{15}{20}\\ \Rightarrow\dfrac{x}{4}=\dfrac{3}{4}\\ \Rightarrow x=3\\ b,\dfrac{15}{x}=\dfrac{25}{35}\\ \Rightarrow\dfrac{15}{x}=\dfrac{5}{7}\\ \Rightarrow\dfrac{15}{x}=\dfrac{15}{21}\\ \Rightarrow x=21\\ c,\dfrac{x}{5}=\dfrac{26}{65}\\ \Rightarrow\dfrac{x}{5}=\dfrac{2}{5}\\ \Rightarrow x=2\\ d,\dfrac{3}{x}=\dfrac{51}{85}\\ \Rightarrow\dfrac{3}{x}=\dfrac{3}{5}\\ \Rightarrow x=5\)
a,x4=1520⇒x4=34⇒x=3b,15x=2535⇒15x=57⇒15x=1521⇒x=21c,x5=2665⇒x5=25⇒x=2d,3x=5185⇒3x=35⇒x=5
Đặt \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{40}=k\Leftrightarrow x=15k;y=20k;z=40k\)
\(xy=1200\\ \Leftrightarrow300k^2=1200\\ \Leftrightarrow k^2=4\Leftrightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=30;y=40;z=80\\x=-30;y=-40;z=-80\end{matrix}\right.\)
a, \(x\cdot\dfrac{-5}{8}=\dfrac{15}{32}\)
\(x=\dfrac{15}{32}:\dfrac{-5}{8}\)
\(x=\dfrac{-3}{4}\)
b, \(\dfrac{3}{10}:x=\dfrac{-9}{20}\)
\(x=\dfrac{3}{10}:\dfrac{-9}{20}\)
\(x=-\dfrac{2}{3}\)
c, \(\dfrac{-1}{4}x+\dfrac{4}{5}=\dfrac{3}{4}\)
\(\dfrac{-1}{4}x=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\dfrac{-1}{4}x=-\dfrac{1}{20}\)
\(x=-\dfrac{1}{20}:\dfrac{-1}{4}\)
\(x=\dfrac{1}{5}\)
d, \(\dfrac{-7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}\cdot\dfrac{-5}{12}\)
\(\dfrac{-7}{8}+\dfrac{2}{3}:x=-\dfrac{1}{4}\)
\(\dfrac{2}{3}:x=-\dfrac{1}{4}+\dfrac{-7}{8}\)
\(\dfrac{2}{3}:x=\dfrac{5}{8}\)
\(x=\dfrac{2}{3}:\dfrac{5}{8}\)
\(x=\dfrac{16}{15}\)
#YVA6
\(a,x.\dfrac{-5}{8}=\dfrac{15}{32}\)
\(\Leftrightarrow x=\dfrac{15}{32}:\dfrac{-5}{8}\)
\(\Leftrightarrow x=\dfrac{15}{32}.\dfrac{-8}{5}\)
\(\Leftrightarrow x=-\dfrac{3}{4}\)
\(b,\dfrac{3}{10}:x=-\dfrac{9}{20}\)
\(\Leftrightarrow x=\dfrac{3}{10}:\dfrac{-9}{20}\)
\(\Leftrightarrow x=\dfrac{3}{10}.\dfrac{-20}{9}\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)
\(c,-\dfrac{1}{4}x+\dfrac{4}{5}=\dfrac{3}{4}\)
\(\Leftrightarrow-\dfrac{1}{4}x=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\Leftrightarrow-\dfrac{1}{4}x=-\dfrac{1}{20}\)
\(\Leftrightarrow x=-\dfrac{1}{20}\times\left(-4\right)\)
\(\Leftrightarrow x=\dfrac{1}{5}\)
\(d,-\dfrac{7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}.\dfrac{-5}{12}\)
\(\Leftrightarrow-\dfrac{7}{8}+\dfrac{2}{3}:x=-\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2}{3}:x=-\dfrac{1}{4}+\dfrac{7}{8}\)
\(\Leftrightarrow\dfrac{2}{3}:x=\dfrac{5}{8}\)
\(\Leftrightarrow x=\dfrac{2}{3}:\dfrac{5}{8}\)
\(\Leftrightarrow x=\dfrac{16}{15}\)
\(\Leftrightarrow-\dfrac{16}{279}< \dfrac{x}{9}< =\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{x}{9}=0\)
hay x=0
7) 5x=4y ⇒\(\dfrac{x}{4}=\dfrac{y}{5}\)
Nhân cả hai vế với \(\dfrac{x}{4}\), ta có: \(\left(\dfrac{x}{4}\right)^2=\dfrac{x}{4}.\dfrac{y}{5}=\dfrac{xy}{20}=\dfrac{20}{20}=1\)
\(\left(\dfrac{x}{4}\right)^2=1\Rightarrow\left[{}\begin{matrix}\dfrac{x}{4}=1\\\dfrac{x}{4}=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}y=5\\y=-5\end{matrix}\right.\)
4) áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{0,5}=\dfrac{y}{0,3}=\dfrac{z}{0,2}=\dfrac{z-y+x}{0,2-0,3+0,5}=\dfrac{1}{\dfrac{2}{5}}=\dfrac{5}{2}\)
\(\dfrac{x}{0,5}=\dfrac{5}{2}\Rightarrow x=\dfrac{5}{4}\)
\(\dfrac{y}{0,3}=\dfrac{5}{2}\Rightarrow y=\dfrac{3}{4}\)
\(\dfrac{z}{0,2}=\dfrac{5}{2}\Rightarrow z=\dfrac{1}{2}\)
6) áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x+11}{13}=\dfrac{y+12}{14}=\dfrac{z+13}{15}=\dfrac{x+11+y+12+z+13}{13+14+15}=\dfrac{42}{42}=1\)
\(\dfrac{x+11}{13}=1\Rightarrow x=2\)
\(\dfrac{y+12}{13}=1\Rightarrow y=1\)
\(\dfrac{z+13}{15}=1\Rightarrow z=2\)
7) \(5x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{5}=k\)
\(\Rightarrow x=4k,y=5k\)
\(x.y=20\\ \Rightarrow4k.5k=20\\ \Rightarrow20k^2=20\\ \Rightarrow k^2=1\\ \Rightarrow\left[{}\begin{matrix}k=-1\\k=1\end{matrix}\right.\)
\(x=4k\Rightarrow\left[{}\begin{matrix}x=-4\\x=4\end{matrix}\right.\)
\(y=5k\Rightarrow\left[{}\begin{matrix}y=-5\\y=5\end{matrix}\right.\)
Vậy \(\left(x,y\right)=\left\{\left(-4;-5\right);\left(4;5\right)\right\}\)
\(\begin{array}{l}a)\dfrac{x}{6} = \dfrac{{ - 3}}{4}\\x = \dfrac{{( - 3).6}}{4}\\x = \dfrac{{ - 9}}{2}\end{array}\)
Vậy \(x = \dfrac{{ - 9}}{2}\)
\(\begin{array}{l}b)\dfrac{5}{x} = \dfrac{{15}}{{ - 20}}\\x = \dfrac{{5.( - 20)}}{{15}}\\x = \dfrac{{ - 20}}{3}\end{array}\)
Vậy \(x = \dfrac{{ - 20}}{3}\)
\(\Rightarrow x=\dfrac{20\cdot15}{4}=75\)
y đâu mà tìm :/