c: Ta có: x=16

nên x+1=17

Ta có: \(C=x^4-17x^3+17x^2-17x+20\)

\(=x^4-x^3\left(x+1\right)+x^2\left(x+1\right)-x\left(x+1\right)+20\)

\(=x^4-x^4-x^3+x^3+x^2-x^2-x+20\)

=20-x

=4

Ta có:

x+y=6          

=>  (x+y)²  = 36     

=> x² +2xy+ y² = 36      

=>20+2xy         =36    

=> 2xy             = 16

=> xy               =8

Ta lại có:

x³+y³= (x+y). ( x² + xy +y²)

         = 6 . (20 + 8)

         = 120 + 48

         = 168

Vậy x³+y³=168  

Ta có:

x+y=6          

=>  (x+y)2  = 36     

=> x2 +2xy+ y2 = 36      

=>20+2xy         =36    

=> 2xy             = 16

=> xy               =8

Ta lại có:

x3+y3= (x+y). ( x2 + xy +y2)

         = 6 . (20 + 8)

         = 120 + 48

         = 168

Vậy x3+y3=168  

\(\dfrac{4}{x}=\dfrac{y}{21}=\dfrac{28}{49}=\dfrac{28:7}{49:7}=\dfrac{4}{9}\\ Vậy:x=\dfrac{4.9}{4}=9\\ y=\dfrac{4.21}{9}=\dfrac{28}{3}\)

\(\dfrac{x}{2}=\dfrac{3}{y}\\ \Leftrightarrow x.y=2.3=6\\ Vậy:\left[{}\begin{matrix}\left(x;y\right)=\left(1;6\right)=\left(6;1\right)\\\left(x;y\right)=\left(2;3\right)=\left(3;2\right)\end{matrix}\right.\)

\(3,\\ \dfrac{42}{54}=\dfrac{42:6}{54:6}=\dfrac{7}{9}\\ \dfrac{42}{54}=\dfrac{7}{x}=\dfrac{7}{9}\\ Vậy:x=\dfrac{7.9}{7}=9\)

a/\(\dfrac{19}{20}-y=\dfrac{8}{5}-\dfrac{3}{4}\)
\(\dfrac{19}{20}-y=\dfrac{17}{20}\)
\(y=\dfrac{19}{20}-\dfrac{17}{20}\)
\(y=\dfrac{2}{20}=\dfrac{1}{10}\)
b/\(y:\dfrac{2}{3}=\dfrac{4}{9}\times3\)
\(y:\dfrac{2}{3}=\dfrac{4}{3}\)
\(y=\dfrac{4}{3}\times\dfrac{2}{3}\)
\(y=\dfrac{8}{9}\)

#データネ

a) 19/20 - y = 8/5 - 3/4

    19/20 - y =   17/20

                y = 17/20 + 19/20

                y =      36/20

b) y : 2/3 = 4/9 x 3

     y : 2/3 = 12/9

          y    =  12/9 x 2/3 

          y    =   24/27

tích cho mik nhaaaaaaaaaaa

\(\dfrac{4}{3}\) = \(\dfrac{20}{4}\)\(x\) - 1

5\(x\) - 1 = \(\dfrac{4}{3}\)

5\(x\)      = \(\dfrac{4}{3}\) + 1

5\(x\)     = \(\dfrac{7}{3}\)

 \(x\)      = \(\dfrac{7}{3}\) : 5

 \(x\)     = \(\dfrac{7}{15}\)

b,

2\(\times\)\(x\) = 3\(\times\) y 

     \(x\) = \(\dfrac{3}{2}\) \(\times\) y

y - \(\dfrac{3}{2}\) \(\times\) y = 5

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

\(y\) \(\times\) - \(\dfrac{1}{2}\) = 5

\(y\)            = 5 : (-\(\dfrac{1}{2}\))

\(y\)            = - 10

\(x\)  = y - 5 = -10 - 5 =-15

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+2y-3z}{2+2\cdot3-3\cdot4}=\dfrac{-20}{-4}=5\)

Do đó: x=10; y=15; z=20

\(\dfrac{x+y+z}{2+2.3-3.4}\)=\(\dfrac{-20}{-4}\)=5

⇒x=5.2=10

  y=5.3=15

   z=5.4=20

\(x\) - \(\dfrac{2}{3}\) không thể bằng \(x\) + \(\dfrac{3}{4}\)