x = 2012

Help me!

/ là phần har bn

Nếu x = 2012 nhé thì tổng trên = 4 

Ta có \(\left(3x-5\right)^{2010}+\left(y-1\right)^{2012}+\left(x-z\right)^{2014}=0\left(1\right)\)

Vì \(2010;2012;2014\) đều là số mủ chẵn (2)

Từ (1) và (2) 

\(\Rightarrow\left(3x-5\right)=0;\left(y-1\right)=0;\left(x-z\right)=0\)

\(\left(+\right)3x-5=0\Rightarrow3x=5\Rightarrow x=\frac{5}{3}\)

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

\(\left(+\right)x-z=0\Rightarrow z=x=\frac{5}{3}\)

Vậy \(x=z=\frac{5}{3};y=1\)

Ta có:

\(\frac{x}{x+1}=1-\frac{1}{x+1}\in Z\Rightarrow x+1\inƯ\left(1\right)\Rightarrow x+1\in\left\{-1;1\right\}\Rightarrow x\in\left\{-2;0\right\}\)

\(+,x=0;\Rightarrow\frac{x}{x+1}=0\left(tm\right);+,x=-2\Rightarrow\frac{x}{x+1}=\frac{-2}{-1}=2\left(tm\right)\)

Vậy: x E {0;2}

b,  \(\frac{a}{2010}=\frac{b}{2012}=\frac{c}{2014}\Rightarrow a=2010k;b=2012k;c=2014k\left(k\in Z\right)\)

\(\frac{\left(a-c\right)^2}{4}=\frac{\left(-4k\right)^2}{4}=\frac{16k^2}{4}=4k^2\)và: \(\left(a-b\right)\left(b-c\right)=\left(-2k\right)\left(-2k\right)=4k^2\)

\(\frac{\left(a-c\right)^2}{4}=\left(a-b\right)\left(b-c\right)\)\(\left(ĐPCM\right)\)

c, Ta có:

\(25-y^2=8.x^2\Rightarrow25-y^2⋮8\Rightarrow y^2:8\left(dư1\right)\left(y\le5\right)\Rightarrow y\in\left\{1;3;5\right\}\)

Ta lần lượt thử ta thấy:

\(25-y^2=8.x^2\left(tm\right)\Leftrightarrow y=5\Rightarrow x=0\)

Vậy: y=5;x=0

Ko thanks mk à

cảm ơn nha

#123 toán học #