Đặt 2x/3+3y/4=4z/5=k   (k khác 0 )

<=>x= 3/2.k ;  y=4/3.k    ; z=5/4.k

mà x+2y+4z = 220

suy ra 3/2.k+2.4/3.k+4.5/4k = 220

<=>k = 24

thay k vao tim dc x=36; y=32;  z=30

2x/3= 3y/4 => y= 8x/9;      2x/3=4z/5=> z= 10x/12

x+2y+4z= 220=> x+ 2(8x/9) + 4(10x/12) = 220 => x= 36=> y=32; z=30 

a) Áp dụng tính chất dãy tỉ số bằng nhau ta có : \(\frac{x}{20}=\frac{y}{9}=\frac{z}{6}=\frac{x-2y+4z}{20-2.9+4.6}=\frac{13}{26}=\frac{1}{2}\)

* \(\frac{x}{20}=\frac{1}{2}\Rightarrow x=\frac{1}{2}.20=10\)

*\(\frac{y}{9}=\frac{1}{2}\Rightarrow y=\frac{1}{2}.9=\frac{9}{2}\)

*\(\frac{z}{6}=\frac{1}{2}\Rightarrow z=\frac{1}{2}.6=3\)

b)c) đề bn viết ko rõ

a)\(x^2+2y^2+2xy-2y+1=0\)

\(\Leftrightarrow x^2+2xy+y^2+y^2-2y+1=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(y-1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}y-1=0\\x+y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=1\\x=-y=-1\end{cases}}\)

Vậy x=-1 y=1

a)  \(x^2+2y^2+2xy-2y+1=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(y-1\right)^2=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x+y\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+y=0\\y-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-y\\y=1\end{cases}\Rightarrow}x=-1;y=1}\)

b) \(5x^2+3y^2+z^2-4x+6xy+4z+6=0\)

\(\Leftrightarrow\left(2x^2-4x+2\right)+\left(3x^2+6xy+3y^2\right)+\left(z^2+4z+4\right)=0\)

\(\Leftrightarrow2.\left(x-1\right)^2+3.\left(x+y\right)^2+\left(z+2\right)^2=0\)

\(\Rightarrow\)  \(\left(x-1\right)^2=0\Rightarrow x-1=0\Rightarrow x=1\)

           \(\left(x+y\right)^2=0\Rightarrow x+y=0\Rightarrow y=-x=-1\) 

            \(\left(z+2\right)^2=0\Rightarrow z+2=0\Rightarrow z=-2\)

Áp dung... ta có

3x-2y/5=3x-1/4=(3x-2y+3x-1)/4+5=2y-1/9

=>2y-1/9=2y+5/7

=>(2y-1).7=(2y+5).9

=>14y-7=18y+45

=>y=-13

Thay vào,tìm x

Ko chac dung ko