Tim y biet:7/4-yx5/6=1/2=1/3
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7/4-y*5/6=1/2+1/3
\(\frac{7}{4}-y\cdot\frac{5}{6}=\frac{5}{6}\)
=>\(\frac{7}{4}-y=1\)
=> \(y=\frac{7}{4}-1\)
=> \(y=\frac{3}{4}\)
\(\frac{9}{4}-y.\frac{5}{6}=\frac{1}{2}+\frac{1}{3}\)
\(\frac{9}{4}-y.\frac{5}{6}=\frac{5}{6}\)
\(y.\frac{5}{6}=\frac{9}{4}-\frac{5}{6}\)
\(y.\frac{5}{6}=\frac{17}{12}\)
\(y=\frac{17}{12}:\frac{5}{6}\)
\(y=\frac{17}{10}\)
\(\frac{9}{4}\)- y x \(\frac{5}{6}\)= \(\frac{1}{2}\)+ \(\frac{2}{3}\) = \(\frac{9}{4}\)- \(\frac{1}{2}\)- \(\frac{2}{3}\)= y x \(\frac{5}{6}\)
y x \(\frac{5}{6}\)= \(\frac{13}{12}\)
y = \(\frac{13}{12}\)x \(\frac{6}{5}\)= \(\frac{13}{10}\)
\(\left(y-6\right)^3-\left(y-6\right)^2=0\)
\(\left(y-6\right)^2.\left[\left(y-6\right)-1\right]=0\)
\(=>\orbr{\begin{cases}y^2-6=0\\y-7=0\end{cases}}\) \(=>\orbr{\begin{cases}y=6\\y=7\end{cases}}\)
Vậy ...
-x-2/3=-6/7
=>-x=-6/7+(-2/3)
=>-x=-32/21
=>x=32/21
4/7-x=1/3
=>x=4/7-1/3
=>x=5/21
-x-2/3=-6/7
=>-x=-6/7+(-2/3)
=>-x=-32/21
=>x=32/21
4/7-x=1/3
=>x=4/7-1/3
=>x=5/21
Vì x,y tỉ lệ thuận nên \(\dfrac{x_1}{x_2}=\dfrac{y_1}{y_2}\)
a: \(\dfrac{x_1}{x_2}=\dfrac{y_1}{y_2}\)
nên \(\dfrac{x_1}{3}=\dfrac{-2}{\dfrac{3}{8}}=-2\cdot\dfrac{8}{3}=-\dfrac{16}{3}\)
=>\(x_1=-16\)
b: \(\dfrac{x_1}{x_2}=\dfrac{y_1}{y_2}\)
\(\Leftrightarrow\dfrac{x_2}{x_1}=\dfrac{y_2}{y_1}\)
\(\Leftrightarrow\dfrac{x_2}{-6}=\dfrac{y_2}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x_2}{-6}=\dfrac{y_2}{4}=\dfrac{y_2-x_2}{4-\left(-6\right)}=\dfrac{-5}{10}=-\dfrac{1}{2}\)
Do đó: \(x_2=3;y_2=-2\)