\(S=\left(x+2012\right)+\left(2y-2013\right)+\left(3z+2014\right)=a+b+c\)

\(P=a^5+b^5+c^5\)

\(P-S=\left(a^5-a\right)+\left(b^5-b\right)+\left(c^5-c\right)\)

 \(=a\left(a-1\right)\left(a+1\right)\left(a^2+1\right)+b\left(b-1\right)\left(b+1\right)\left(b^2+1\right)+c\left(c-1\right)\left(c+1\right)\left(c^2+1\right)\)

 Ta chứng minh \(a\left(a-1\right)\left(a+1\right)\left(a^2+1\right)\) chia hết cho 30 .

tương tự => \(b\left(b-1\right)\left(b+1\right)\left(b^2+1\right);c\left(c-1\right)\left(c+1\right)\left(c^2+1\right)\)chia hết cho 30.

=> P -S chia hết cho 30 => (dpcm)

a)

\(\frac{x}{10}=\frac{y}{15}=>\frac{x}{2}=\frac{y}{3}\)

\(x=\frac{z}{2}=>\frac{x}{2}=\frac{z}{4}\)

=> x/2=y/3=z/4=\(\frac{x+2y-3z}{2+4-12}=\frac{-24}{-6}=4\)

x=4x2=8

y=4x3=12

z=4x4=16

\(\frac{x-2013}{2}=\frac{y-2014}{6}=\frac{z-2015}{8}\)

\(\Rightarrow\frac{x-2013}{2}=\frac{2y-4028}{12}=\frac{3z-6045}{24}\)

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

\(\frac{x-2013}{2}=\frac{2y-4028}{12}=\frac{3z-6045}{24}=\frac{\left(x-2013\right)+\left(2y-4028\right)-\left(3z-6045\right)}{2+12-24}=\frac{5}{-10}=\frac{-1}{2}\)

Từ đó suy ra :

\(\frac{x-2013}{2}=\frac{-1}{2}\Rightarrow x-2013=-1\Rightarrow x=2012\)

\(\frac{2y-4028}{12}=\frac{-1}{2}\Rightarrow2y-4028=-6\Rightarrow2y=4022\Rightarrow y=2011\)

\(\frac{3z-6045}{24}=\frac{-1}{2}\Rightarrow3z-6045=-12\Rightarrow3z=6033\Rightarrow z=2011\)

Ta có: \(\left\{{}\begin{matrix}x\left(x+2y+3z\right)=-5\\y\left(x+2y+3z\right)=27\\z\left(x+2y+3z\right)=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-5}=x+2y+3z\\\dfrac{y}{27}=x+2y+3z\\\dfrac{z}{5}=x+2y+3z\end{matrix}\right.\)

\(\Rightarrow\dfrac{x}{-5}=\dfrac{y}{27}=\dfrac{z}{5}\Rightarrow\left\{{}\begin{matrix}y=\dfrac{-27}{5}x\\z=-x\end{matrix}\right.\)

Ta có: \(x\left(x+2y+3z\right)=-5\Rightarrow x\left(x+2.\dfrac{-27}{5}x-3x\right)=-5\)

\(\Rightarrow\dfrac{-64}{5}x^2=-5\Rightarrow x^2=\dfrac{25}{64}\Rightarrow x=\dfrac{5}{8}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{8}\\y=-\dfrac{27}{5}x=-\dfrac{27}{8}\\z=-x=-\dfrac{5}{8}\end{matrix}\right.\)

Bài `10`

`a,` Ta có : `x/2=y/3=>(4x)/8 =(3y)/9`

ADTC dãy tỉ số bằng nhau ta có :

`(4x)/8 =(3y)/9=(4x-3y)/(8-9)=(-2)/(-1)=2`

`=> x/2=2=>x=2.2=4`

`=>y/3=2=>y=2.3=6`

`b,` Ta có : `2x=5y=>x/5=y/2`

ADTC dãy tỉ số bằng nhau ta có :

`x/5=y/2=(x+y)/(5+2)=-42/7=-6`

`=>x/5=-6=>x=-6.5=-30`

`=>y/2=-6=>y=-6.2=-12`

Bài `11`

`a,` Ta có : `x/3=y/4=z/6=>x/3=(2y)/8 =(3z)/18`

ADTC dãy tỉ số bằng nhau ta có :

`x/3=(2y)/8=(3z)/18=(x+2y-3z)/(3+8-18)=(-14)/(-7)=2`

`=>x/3=2=>x=2.3=6`

`=>y/4=2=>y=2.4=8`

`=>z/6=2=>z=2.6=12`

Bạn đăng lại `2` câu sau nhe , mình ko hiểu `x=y-z` với `15x-5y=3x=45`

`d,` Ta có :

`x/2=y/3=>x/4=y/6`

`y/2=z/3=>y/6=z/9`

`-> x/4=y/6=z/9=>x/4=(2y)/12 =(3z)/27`

ADTC dãy tỉ số bằng nhau ta có :

`x/4=(2y)/12=(3z)/27=(x-2y+3z)/(4-12+27)=19/19=1`

`=>x/4=1=>x=1.4=4`

`=>y/6=1=>y=1.6=6`

`=>z/9=1=>z=1.9=9`