y x 2/5 =8/21

y= 8/21:2/5

y= 20/21

12/7: y =7/5-2/5

12/7:y = 1

y = 12/7 :1

y= 12/7

\(y\times\dfrac{2}{5}=\dfrac{8}{21}\\ y=\dfrac{8}{21}:\dfrac{2}{5}\\ y=\dfrac{20}{21}\\ \dfrac{12}{7}:y+\dfrac{2}{5}=\dfrac{7}{5}\\ \dfrac{12}{7}:y=\dfrac{7}{5}-\dfrac{2}{5}\\ \dfrac{12}{7}:y=1\\ y=\dfrac{12}{7}:1=\dfrac{12}{7}\)

\(\dfrac{1}{5}+y=7\)

       \(y=7-\dfrac{1}{5}\)

      \(y=\dfrac{35}{5}-\dfrac{1}{5}\)

      \(y=\dfrac{34}{5}\)

_________________________

\(y:\dfrac{3}{4}=\dfrac{5}{7}\)

\(y=\dfrac{5}{7}\times\dfrac{3}{4}\)

\(y=\dfrac{15}{28}\)

Chúc bạn học tốt

\(\dfrac{1}{5}+Y=7\\ Y=7-\dfrac{1}{5}\\ Y=\dfrac{35}{5}-\dfrac{1}{5}\\ Y=\dfrac{34}{5}\)

\(Y\div\dfrac{3}{4}=\dfrac{5}{7}\\ Y=\dfrac{5}{7}\times\dfrac{3}{4}\\ Y=\dfrac{15}{28}\)

a: ta có: \(\dfrac{2x-5}{7x-1}=\dfrac{4x+3}{14x-9}\)

\(\Leftrightarrow\left(2x-5\right)\left(14x-9\right)=\left(7x-1\right)\left(4x+3\right)\)

\(\Leftrightarrow28x^2-18x-70x+45=28x^2+21x-4x-3\)

=>-88x+45=17x-3

=>-105x=-48

hay x=16/35

b: Sửa đề: \(\dfrac{x}{4}=\dfrac{y}{9}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{4}=\dfrac{y}{9}=\dfrac{x-y}{4-9}=\dfrac{105}{-5}=-21\)

Do đó: x=-84; y=-189

c: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{2x-5y}{2\cdot3-5\cdot4}=\dfrac{56}{-14}=-4\)

Do đó:x=-12; y=-16

e: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x^2}{2}=\dfrac{y^2}{3}=\dfrac{x^2+y^2}{2+3}=\dfrac{125}{5}=25\)

Do đó: \(x^2=50;y^2=75\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{5\sqrt{2};-5\sqrt{2}\right\}\\y\in\left\{5\sqrt{3};-5\sqrt{3}\right\}\end{matrix}\right.\)

a, y \(\times\) \(\dfrac{4}{3}\) = \(\dfrac{16}{9}\)

    y         =    \(\dfrac{16}{9}\) : \(\dfrac{4}{3}\)

    y         = \(\dfrac{4}{3}\)

b, ( y - \(\dfrac{1}{2}\)) + 0,5 = \(\dfrac{3}{4}\)

    y - 0,5 + 0,5 = \(\dfrac{3}{4}\)

   y                   = \(\dfrac{3}{4}\)

c, \(\dfrac{4}{5}-\dfrac{2}{5}y\) = 0,2

   0,8 - 0,4y = 0,2

           0,4y = 0,8 - 0,2

           0,4y  = 0,6

               y = 1,5

d, (y + \(\dfrac{3}{4}\)) \(\times\) \(\dfrac{5}{7}\) = \(\dfrac{10}{9}\)

    y + \(\dfrac{3}{4}\)           = \(\dfrac{10}{9}\) : \(\dfrac{5}{7}\)

   y + \(\dfrac{3}{4}\)            = \(\dfrac{14}{9}\)

y                    = \(\dfrac{14}{9}\) - \(\dfrac{3}{4}\)

 y                   =   \(\dfrac{29}{36}\)

e, y : \(\dfrac{5}{4}\)         = \(\dfrac{9}{5}\)  + \(\dfrac{1}{2}\)

   y : \(\dfrac{5}{4}\)         =   \(\dfrac{23}{10}\)

  y                =      \(\dfrac{23}{10}\)

  y               =   \(\dfrac{23}{8}\)

f, y \(\times\) \(\dfrac{1}{2}\) + \(\dfrac{3}{2}\) \(\times\) y   = \(\dfrac{4}{5}\)

   y \(\times\) ( \(\dfrac{1}{2}+\dfrac{3}{2}\))      =  \(\dfrac{4}{5}\)

   2y                       = \(\dfrac{4}{5}\)

    y                        = \(\dfrac{2}{5}\)

Bài 5 :

a) \(\dfrac{y}{4}=\dfrac{9}{y}\)

\(\Rightarrow y^2=36\left(y\ne0\right)\)

\(\Rightarrow y=\pm6\)

b) \(\dfrac{y+7}{20}=\dfrac{5}{y+7}\left(y\ne-7\right)\)

\(\Rightarrow\left(y+7\right)^2=100=10^2\)

\(\Rightarrow\left[{}\begin{matrix}y+7=10\\y+7=-10\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}y=3\\y=-17\end{matrix}\right.\)

c) \(\dfrac{4-5y}{3}=\dfrac{y+2}{5}\)

\(\Rightarrow5\left(4-5y\right)=3\left(y+2\right)\)

\(\Rightarrow20-25y=3y+6\)

\(\Rightarrow28y=14\)

\(\Rightarrow y=\dfrac{14}{28}=\dfrac{1}{2}\)

Bài 4 :

\(\dfrac{a}{5}=\dfrac{b}{7}=\dfrac{c}{10}\)

\(\Rightarrow\dfrac{2a}{10}=\dfrac{3b}{21}=\dfrac{4c}{40}=\dfrac{2a+3b-4c}{10+21-40}=\dfrac{81}{-9}=-9\)

\(\Rightarrow\left\{{}\begin{matrix}a=-9.5=-45\\b=-9.7=-63\\c=-9.10=-90\end{matrix}\right.\)

a) Ta có: \(\dfrac{x}{y}=\dfrac{10}{9}\Rightarrow\dfrac{x}{10}=\dfrac{y}{9}\)

               \(\dfrac{y}{z}=\dfrac{3}{4}\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{9}=\dfrac{z}{12}\)

\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{9}=\dfrac{z}{12}=\dfrac{x-y+z}{10-9+12}=\dfrac{78}{13}=6\)

\(\Rightarrow\left\{{}\begin{matrix}x=6.10=60\\y=6.9=54\\z=6.12=72\end{matrix}\right.\)

b)Ta có:  \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\)

               \(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\)

\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=-\dfrac{15}{5}=-3\)

\(\Rightarrow\left\{{}\begin{matrix}x=-3.9=-27\\y=-3.7=-21\\z=-3.3=-9\end{matrix}\right.\)

c) \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{3}\)

\(\Rightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{z^2}{9}=\dfrac{x^2+y^2+z^2}{9+16+9}=\dfrac{200}{34}=\dfrac{100}{17}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{900}{17}\\y^2=\dfrac{1600}{17}\\z^2=\dfrac{900}{17}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\pm\dfrac{30\sqrt{17}}{17}\\y=\pm\dfrac{40\sqrt{17}}{17}\\z=\pm\dfrac{30\sqrt{17}}{17}\end{matrix}\right.\)

Vậy\(\left(x;y;z\right)\in\left\{\left(\dfrac{30\sqrt{17}}{17};\dfrac{40\sqrt{17}}{17};\dfrac{30\sqrt{17}}{17}\right),\left(-\dfrac{30\sqrt{17}}{17};-\dfrac{40\sqrt{17}}{17};-\dfrac{30\sqrt{17}}{17}\right)\right\}\)

\(B(y) - A(y) = 2{y^3} - 9{y^2} + 4y\)

\(\begin{array}{l}A(y) =  - 5{y^4} - 4{y^2} + 2y + 7\\ \Rightarrow B(y) = 2{y^3} - 9{y^2} + 4y - 5{y^4} - 4{y^2} + 2y + 7\\ =  - 5{y^4} + 2{y^3} - 13{y^2} + 6y + 7\end{array}\)

a) \(\frac{7}{10}-y\cdot\frac{3}{4}=\frac{1}{5}\)

\(y\cdot\frac{3}{4}=\frac{7}{10}-\frac{1}{5}\)

\(y\cdot\frac{3}{4}=\frac{1}{2}\)

\(y=\frac{1}{2}:\frac{3}{4}\)

\(y=\frac{1}{2}\cdot\frac{4}{3}\)

\(y=\frac{2}{3}\)

b) \(\frac{5}{6}:\left(y+\frac{7}{9}\right)=\frac{3}{4}\)

\(y+\frac{7}{9}=\frac{5}{6}:\frac{3}{4}\)

\(y+\frac{7}{9}=\frac{5}{6}\cdot\frac{4}{3}\)

\(y+\frac{7}{9}=\frac{10}{9}\)

\(y=\frac{10}{9}-\frac{7}{9}\)

\(y=\frac{3}{9}=\frac{1}{3}\)

**** !

a, \(\frac{7}{10}-y\times\frac{3}{4}=\frac{1}{5}\)

\(\Rightarrow y\times\frac{3}{4}=\frac{7}{10}-\frac{1}{5}\)

\(\Rightarrow y\times\frac{3}{4}=\frac{1}{2}\)

\(\Rightarrow y=\frac{1}{2}\div\frac{3}{4}\)

\(\Rightarrow y=\frac{1}{3}\)

b, \(\frac{5}{6}\div\left(y+\frac{7}{9}\right)=\frac{3}{4}\)

\(\Rightarrow y+\frac{7}{9}=\frac{5}{6}\div\frac{3}{4}\)

\(\Rightarrow y+\frac{7}{9}=\frac{10}{9}\)

\(\Rightarrow y=\frac{10}{9}-\frac{7}{9}\)

\(\Rightarrow y=\frac{1}{3}\)