a) \(xy+x-y=2\)

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

\(\Leftrightarrow\left(x-1\right)\left(y+1\right)=1=1.1=\left(-1\right).\left(-1\right)\)

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

b) \(x-2xy+y=0\)

\(\Leftrightarrow2x-4xy+2y=0\)

\(\Leftrightarrow2x\left(1-2y\right)-\left(1-2y\right)=-1\)

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

Tương tự nha

c) \(x\left(x-2\right)-\left(2-x\right)y-2\left(x-2\right)=3\)

\(\Leftrightarrow x\left(x-2\right)+\left(x-2\right)y-2\left(x-2\right)=3\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=0\end{cases}}\)

\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)=0\)

\(\Leftrightarrow a^2\left(b-c\right)+b^2[\left(c-b\right)-\left(a-b\right)]+c^2\left(a-b\right)\)

\(\Leftrightarrow a^2\left(b-c\right)+b^2\left(c-b\right)-b^2\left(a-b\right)+c^2\left(a-b\right)=0\)

\(\Leftrightarrow a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)=0\)

\(\Leftrightarrow\left(b-c\right)\left(a^2-b^2\right)-\left(a-b\right)\left(b^2-c^2\right)=0\)

\(\Leftrightarrow\left(b-c\right)\left(a-b\right)\left(a+b\right)-\left(a-b\right)\left(b-c\right)\left(b+c\right)=0\)

\(\Leftrightarrow\left(b-c\right)\left(a-b\right)\left(a+b-b-c\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(b-c\right)\left(a-c\right)=0\)

\(\Rightarrow a-b=0\)hoặc \(b-c=0\)hoặc \(a-c=0\)

\(\Rightarrow a=b\)hoặc \(b=c\) hoặc \(a=c\)

Vậy trong 3 số a,b,c tồn tại hai số bằng nhau. 

P/s : Không hiểu thì hiểu mình nha !

\(a^4\left(b-c\right)+b^4\left(c-a\right)+c^4\left(a-b\right)\)

\(=a^4\left(b-c\right)+b^4[\left(c-b\right)-\left(a-b\right)]+c^4\left(a-b\right)\)

\(=a^4\left(b-c\right)+b^4\left(c-b\right)-b^4\left(a-b\right)+c^4\left(a-b\right)\)

\(=a^4\left(b-c\right)-b^4\left(b-c\right)-b^4\left(a-b\right)+c^4\left(a-b\right)\)

\(=\left(b-c\right)\left(a^4-b^4\right)-\left(a-b\right)\left(c^4-b^4\right)\)

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

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

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

a) \(\left(x+2\right)\left(x-2\right)-\left(x-3\right)\left(x+1\right)\)

\(=x^2-4-\left(x^2+x-3x-3\right)\)

\(=x^2-4-x^2-x+3x+3\)

\(=2x-1\)

b) \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1\)

đặt ẩn phụ là ra

tích cho t đi

a) (2018x - 1) - 2019x (2018x - 1)=0

<=> (2018x - 1)(1 - 2019x)=0

<=> 2018x-1=0

1-2019x=0

<=> x=1/2018

x=1/2019