\(\frac{9}{4}-y\times\frac{5}{6}=\frac{1}{2}+\frac{2}{3}\)

\(\frac{9}{4}-y\times\frac{5}{6}=\frac{7}{6}\)

           \(y\times\frac{5}{6}=\frac{9}{4}-\frac{7}{6}\)

           \(y\times\frac{5}{6}=\frac{13}{12}\)

                      \(y=\frac{13}{12}:\frac{5}{6}\)

                      \(y=\frac{13}{10}\)

a) \(\left(3x-5\right)\left(5-3x\right)+9\left(x+1\right)^2=30\)

\(\Rightarrow15x-9x^2-25+15x+9\left(x^2+2x+1\right)-30=0\)

\(\Rightarrow30x-9x^2-25+9x^2+18x+9-30=0\)

\(\Rightarrow48x-46=0\)

\(\Rightarrow x=\frac{23}{24}\)

b) \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(\Rightarrow\left(x^2+8x+16\right)-\left(x^2-1\right)=16\)

\(\Rightarrow x^2+8x+16-x^2+1=16\)

\(\Rightarrow8x+17=16\)

\(\Rightarrow8x=-1\)

\(\Rightarrow x=\frac{-1}{8}\)

c) \(\left(y-2\right)^3-\left(y-3\right)\left(y^2+3y+9\right)+6\left(y+1\right)^2=49\)

\(\Rightarrow\left(y-2\right)^3-\left(y^3-3^3\right)+6\left(y^2+2y+1\right)=49\)

\(\Rightarrow y^3-6y^2+12y-8-y^3+27+6y^2+12y+6=49\)

\(\Rightarrow\left(y^3-y^3\right)+\left(-6y^2+6y^2\right)+\left(12y+12y\right)+\left(-8+27+6\right)=49\)

\(\Rightarrow24y+25=49\)

\(\Rightarrow24y=24\)

\(\Rightarrow y=1\)

d) \(\left(y+3\right)^3-\left(y+1\right)^3=56\)

\(\Rightarrow\left(y+3-y-1\right)[\left(y+3\right)^2+\left(y+3\right)\left(y+1\right)+\left(y+1\right)^2]=56\)

\(\Rightarrow2\left(y^2+6y+9+y^2+4y+3+y^2+2y+1\right)=56\)

\(\Rightarrow3y^2+12y+13=28\)

\(\Rightarrow\left(3y^2+15y\right)-\left(3y+15\right)=0\)

\(\Rightarrow3y\left(y+5\right)-3\left(y+5\right)=0\)

\(\Rightarrow3\left(y-1\right)\left(y+5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)

bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 =>  x-1/3=y-2/4=z-3/5 

áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1

do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự

Bài 1: 

a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )

cũng dễ thôi

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