(x-5) (x-7)=0

\(x\left(1-y\right)+2y-3=0\)

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

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

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

\(y-1;2-x\inƯ\left(1\right)\)

\(\left(x^2+cx+2\right)\left(ax+b\right)=x^3-x^2+2\)

\(\Leftrightarrow x^3a+x^2b+x^2ac+xbc+2xa+2b=x^3-x^2+2\)

\(\Leftrightarrow x^3a+\left(b+ac\right)x^2+\left(bc+2a\right)x+2b=x^3-x^2+2\)

\(\Leftrightarrow\hept{\begin{cases}x^3a=x^3\\\left(b+ac\right)x^2=-x^2\\\left(bc+2a\right)x=0\end{cases}}\)

            2b=2

\(\Leftrightarrow\hept{\begin{cases}a=1\\ac+b=-1\\2a+bc=0\end{cases}}\)

         b=1

\(\Leftrightarrow\hept{\begin{cases}a=1\\1.c+b=c+b=-1\\2.1+1.c=2+c=0\end{cases}}\)

         b=1

\(\Leftrightarrow\hept{\begin{cases}a=1\\c=-2\\c=1\end{cases}}\)

\(9x^2-9xy-4y^2\)

\(=9x\left(x-y\right)-4y^2\)

\(=\left(3\sqrt{x\left(x-y\right)}-2y\right)\left(3\sqrt{x\left(x-y\right)}+2y\right)\)

(x-3)(x²+3x+9)+x(5-x²)=6x

x(x²+3x+9)-3(x²+3x+9)+x(5-x²)=6x

x³+3x²+9x-3x²-9x-27+5x-x³-6x=0

(x³-x³)+(3x²-3x²)+(9x-9x+5x-6x)=27

-x=27

x=-27

Ta có

x^2+5y^2+2x-4xy-10y+14 
=[x^2+2x(1-2y)+(1-2y)^2]+y^2-6y+13 
=(x+1-2y)^2+(y^2-2y.3+9)+4 
=(x+1-2y)^2+(y-3)^2+4. 
mà

(x+1-2y)^2 > hoặc=0 với mọi x,y thuộc R 
và (y-3)^2 > hoặc=0 với mọi y thuộc R 
=> (x+1-2y)^2+(y-3)^2+4 > hoặc =4 với mọi x,y thuộc R 
=> (x+1-2y)^2+(y-3)^2+4 >0 với mọi x,y thuộc R.

Akai Shuichi:chép sai đề rồi con ơi :D

a) Ta có \(x^2+y^2+2x-4y+5=0\Leftrightarrow\left(x^2+2x+1\right)+\left(y^2-4y+4\right)=0\Leftrightarrow\left(x+1\right)^2+\left(y-2\right)^2=0\)

<=> x=-1;y=2

b)Ta có:\(x^2+4y^2-x+4y+\frac{5}{4}=0\Leftrightarrow\left(x^2-x+\frac{1}{4}\right)+\left(4y^2+4y+1\right)=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\left(2y+1\right)^2=0\)

<=> x=1/2 ;y=-1/2

a, \(x^2+y^2+2x-4y+5=0\Rightarrow\left(x^2+2x+1\right)+\left(y^2-4y+4\right)=0.\)

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

   \(\Rightarrow x+1=0\)và \(y-2=0\)

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

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

Vậy x=-1 ; y=2 

b, \(x^2+4y^2-x+4y+\frac{5}{4}=0\)

\(\Rightarrow\left(x^2-x+\frac{1}{4}\right)+\left(4y^2+4y+\frac{4}{4}\right)=0\)

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

\(\Rightarrow x-\frac{1}{2}=0\) và \(2y+1=0\)

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

\(\left(+\right)2y+1=0\Rightarrow2y=-1\Rightarrow y=-\frac{1}{2}\)

Vậy \(x=\frac{1}{2};y=-\frac{1}{2}\)

\(2x^{n+1}\left(x^{n-1}-y^{n-1}\right)+y^{n-1}\left(2x^{n+1}-y^{n+1}\right)\)

\(=2x^{n+1}.x^{n-1}-2x^{n+1}.y^{n-1}+y^{n-1}.2x^{n+1}-y^{n-1}.y^{n+1}\)

\(=2x^{2n}-2x^{n+1}.y^{n-1}+2x^{n+1}.y^{n-1}-y^{2n}\)

\(=2x^{2n}-\left(2x^{n+1}.y^{n-1}-2x^{n+1}.y^{n-1}\right)-y^{2n}\)

\(=2x^{2n}-y^{2n}\)

Có gì sai sót mong cậu bỏ qua.