a: \(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\)

=>\(\frac{2x}{3}\cdot60=\frac{3y}{4}\cdot60=\frac{4z}{5}\cdot60\)

=>40x=45y=48z

=>\(\frac{40x}{720}=\frac{45y}{720}=\frac{48z}{720}\)

=>\(\frac{x}{18}=\frac{y}{16}=\frac{z}{15}\)

mà x+y-z=38

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

\(\frac{x}{18}=\frac{y}{16}=\frac{z}{15}=\frac{x+y-z}{18+16-15}=\frac{38}{34-15}=\frac{38}{19}=2\)

=>\(\begin{cases}x=2\cdot18=36\\ y=2\cdot16=32\\ z=2\cdot15=30\end{cases}\)

b: 7x=10y=12z

=>\(\frac{7x}{420}=\frac{10y}{420}=\frac{12z}{420}\)

=>\(\frac{x}{60}=\frac{y}{42}=\frac{z}{35}\)

mà x+y+z=685

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

\(\frac{x}{60}=\frac{y}{42}=\frac{z}{35}=\frac{x+y+z}{60+42+35}=\frac{685}{102+35}=\frac{685}{137}=5\)

=>\(\begin{cases}x=5\cdot60=300\\ y=5\cdot42=210\\ z=5\cdot35=175\end{cases}\)

\(\left(5x^2-7x^2y^3+3y^4\right)-K=3x^2-7x^2y^3-3y^4\)

\(\Rightarrow K=\left(5x^2-7x^2y^3+3y^4\right)-\left(3x^2-7x^2y^3-3y^4\right)\)

\(\Rightarrow K=5x^2-7x^2y^3+3y^4-3x^2+7x^2y^3+3y^4\)

\(\Rightarrow K=2x^2+6y^4\)

________________

\(3x^2-8x+5-K=-2K+4x-6+x^2\)

\(\Rightarrow-K+2K=\left(4x-6+x^2\right)-\left(3x^2-8x+5\right)\)

\(\Rightarrow K=4x-6+x^2-3x^2+8x-5\)

\(\Rightarrow K=-2x^2+12x-11\)

cảm ơn bạn

\(K=5x^2-7x^2y^3+3y^4-3x^2+7x^2y^3+3y^4=2x^2+6y^4\)

\(\Rightarrow K=5x^2-7x^2y^3+3y^4-3x^2+7x^2y^3+3y^4\)

\(\Rightarrow K=2x^2+6y^4\)

a thay y =1 =>A=-14

với x=0 , y=1 

=>B=-3

2x^3-1=15

=>2x^3=16

=>x=2

(x+16)/9=(y-5)/16=(z+9)/25

=>(y-5)/16=(z+9)/25=2

=>y-5=32 và z+9=50

=>y=37 và z=41

B=x+y+z=2+37+41=80

\(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{4}\)\(\Leftrightarrow\frac{x}{10}=\frac{y}{15};\frac{y}{15}=\frac{z}{12}\Leftrightarrow\)\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)

\(\Rightarrow\frac{x-y+z}{10-15+12}=-\frac{49}{7}=-7\)

\(\Rightarrow x=-7.10=-70\)

\(y=-7.15=-105\)

\(z=-7.12=-84\)

\(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15};\dfrac{y}{5}=\dfrac{z}{4}\Rightarrow\dfrac{y}{15}=\dfrac{z}{12}\\ \Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-49}{7}=-7\\ \Rightarrow\left\{{}\begin{matrix}x=-70\\y=-105\\z=-84\end{matrix}\right.\)

\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{x}{20}=\dfrac{y}{15};\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{y}{15}=\dfrac{z}{18}\)=> \(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{18}\)

 Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{10}=\dfrac{x+y-z}{20+15-18}=\dfrac{49}{17}\)

=> x=  \(\dfrac{49}{17}.20\)= \(\dfrac{980}{17}\)

=> y= \(\dfrac{49}{17}.15\)= \(\dfrac{735}{17}\)

=> z= \(\dfrac{49}{17}.18=\dfrac{882}{17}\)

 Đây là kết quả mình tính đc, bạn tham khảo nhé !

\(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}\)

\(\dfrac{y}{5}=\dfrac{z}{4} \Rightarrow\dfrac{y}{15}=\dfrac{z}{12}\)

\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)

Áp dụng TCDTSBN ta có: 

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-49}{7}=-7\)

\(\dfrac{x}{10}=-7\Rightarrow x=-70\\ \dfrac{y}{15}=-7\Rightarrow y=-105\\ \dfrac{z}{12}=-7\Rightarrow z=-84\)