CM : ( 3x - 2y )^2010 = 0 ,  / 5y - 6z /^2011 = 0 

=> 3x - 2y = 0    ,  5y - 6z = 0 

=> 3x = 2y     , 5y = 6z 

=> x/2 = y/3    , y/6 = z/5 

=> x/4 = y/6    , y/6 =z/5 

=> x/4 = y/6 = z/5 

=> 2x/ 8 , 5y/30 , 3z/15

Áp dụng tính chất DTSBN , ta có : 

2x/8 = 5y /30 = 3z / 15 = 2x - 5y + 3z / 8 - 30 + 15 = 54/-7 = -54 /7 

Rồi tính ra là xong 

\(a,4x=5y\:\Rightarrow\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{12}\)

\(4y=6z\Rightarrow\frac{y}{6}=\frac{z}{4}\Rightarrow\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{x}{15}=\frac{2y}{24}=\frac{3z}{24}\)

\(\Rightarrow\frac{x-2y+3z}{15-24+24}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{5}{15}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{1}{3}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\cdot15=5\\y=\frac{1}{3}\cdot12=4\\z=\frac{1}{3}\cdot8=\frac{8}{3}\end{cases}}\)

mọi người giúp mk câu b, c, d còn lại nha

nhầm đoạn cuối 54/-7 = -54/7

=> x= -216/7 ; y=-324/7 ; z= -270/7

\(\left(3x-2y\right)^{2014}\ge0\) ; \(\left|5y-6z\right|^{2015}\ge0\)

\(\Rightarrow\left(3x-2y\right)^{2014}+\left|5y-6z\right|^{2015}\ge0\)

mà \(\left(3x-2y\right)^{2014}+\left|5y-6z\right|^{2015}=0\)

\(\Rightarrow\left(3x-2y\right)^{2014}=\left|5y-6z\right|^{2015}=0\Rightarrow3x-2y=5y-6z=0\)

\(\Rightarrow3x=2y;5y=6z\)

\(3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{4}=\frac{y}{6}\left(1\right)\)

\(5y=6z\Rightarrow\frac{y}{6}=\frac{z}{5}\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\frac{x}{4}=\frac{y}{6}=\frac{z}{5}=\frac{2x}{8}=\frac{5y}{30}=\frac{3z}{15}=\frac{2x-5y+3z}{8-30+15}=\frac{54}{-7}=-\frac{7}{54}\) [áp dụng dãy tỉ số bằng nhau]

=> x= -14/27 ; y= -7/9 ; z= -35/51

khó quá 

k nhé tớ k lại cho 

hihihiihih ^_^ ~ hihihihihih 

 Vì \(\left(3x-2y\right)^{100}\ge0\forall x,y\inℤ\)

       \(|5y-6z|\ge0\forall y,z\inℤ\Rightarrow|5y-6z|^{153}\ge0\forall y,z\inℤ\)

Nên \(\Rightarrow\hept{\begin{cases}(3x-2y)^{100}=0\\|5y-6z|^{153}=0\end{cases}}\Rightarrow\hept{\begin{cases}3x-2y=0\\5y-6z=0\end{cases}}\Rightarrow\hept{\begin{cases}3x=2y\\5y=6z\end{cases}\Rightarrow\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{y}{6}=\frac{z}{5}\end{cases}}}\)

Từ \(\frac{x}{2}=\frac{y}{3};\frac{y}{6}=\frac{z}{5}\)suy ra\(\frac{x}{4}=\frac{y}{6}=\frac{z}{5}\)

 Ta có

 \(\frac{x}{4}=\frac{y}{6}=\frac{z}{5}=\frac{2x}{8}=\frac{5y}{30}=\frac{3z}{15}=\frac{2x-5y+3z}{8-30+15}=\frac{56}{-7}=-8\)

Do đó 

\(\frac{x}{4}=-8\Rightarrow x=-32\)

\(\frac{y}{6}=-8\Rightarrow y=-48\)

\(\frac{z}{5}=-8\Rightarrow z=-40\)

    Vậy \(x=-32;y=-48;z=-40\)

a) Ta có : \(\frac{x-1}{2}=\frac{y+3}{4}\Leftrightarrow\left(x-1\right).4=\left(y+3\right).2\Leftrightarrow4x-4=2y+6\Leftrightarrow4x-2y=10\Leftrightarrow x=\frac{10+2y}{4}\left(1\right)\)

 \(\frac{y+3}{4}=\frac{z-5}{6}\Leftrightarrow\left(y+3\right).6=\left(z-5\right).4\Leftrightarrow6y+18=4z-20\Leftrightarrow6y-4z=-38\Rightarrow z=\frac{6y+38}{4}\left(2\right)\)Thay (1) và (2) vào biểu thức \(5x-3y-4z=20\); ta được : 

\(\frac{5.\left(10+2y\right)}{4}-3y-\frac{4.\left(6y+38\right)}{4}=20\)

\(\Leftrightarrow50+10y-12y-24y-152=80\)

\(\Leftrightarrow-26y=182\Rightarrow y=-7\)

Với \(y=-7\Rightarrow x=\frac{10+2.-7}{4}=-1;z=\frac{6.-7+38}{4}=-1\)

Vậy .... 

mk ko bt 

bạn cute quá ; 

tặng bạn , tk mk nhé ; 

@Phạm thu phương sao lại cute ạ ><

Giải:

\(2x=5y=6z\Rightarrow\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{2}=\dfrac{2y}{10}=\dfrac{3z}{18}_{\left(1\right)}\)và \(3x-z+2y=24_{\left(2\right)}.\)

Từ \(_{\left(1\right)}\) và \(_{\left(2\right)}\), kết hợp tính chất dãy tỉ số bằng nhau có:

\(\dfrac{x}{2}=\dfrac{2y}{10}=\dfrac{3z}{18}=\dfrac{3z-x+2y}{18-2+10}=\dfrac{24}{26}=\dfrac{12}{13}.\)

Từ đó:

\(\dfrac{x}{2}=\dfrac{12}{13}\Rightarrow x=\dfrac{2.12}{13}=\dfrac{24}{13}.\)

\(\dfrac{2y}{10}=\dfrac{12}{13}\Rightarrow2y=\dfrac{10.12}{13}=\dfrac{120}{13}\Rightarrow y=\dfrac{60}{13}.\)

\(\dfrac{3z}{18}=\dfrac{12}{13}\Rightarrow3x=\dfrac{18.12}{13}=\dfrac{216}{13}\Rightarrow z=\dfrac{72}{13}.\)

Vậy: \(\left[{}\begin{matrix}x=\dfrac{24}{13}.\\y=\dfrac{60}{13}.\\z=\dfrac{72}{13}.\end{matrix}\right.\)