a/ \(C=\left(x^3+x^2y-2x^2\right)-\left(xy+y^2-2y\right)+\left(x+y-1\right)\)

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

Mà x+y-2=0 => x+y-1=1 => C=1

b/  Với x=2; y=2 Ta nhận thấy \(x^3-2y^2=2^3-2.2^2=2^3-2^3=0\) => D=0

   x^2 + 4y^2 - 2x + 10+ 4xy - 4y

= (x^2 + 4xy + 4y^2) - (2x + 4y) + 10

= (x + 2y)^2 - 2 (x + 2y) + 10

Thay x + 2y = 5 vào biểu thức trên ,ta  được :

   5^2 - 2 . 5 + 10

= 25 - 10 + 10

= 25

\(1,P=\left(x+y+x-y\right)\left(x+y-x+y\right)+2\left(x^2-y^2\right)-4y^2\\ P=4xy+2x^2-6y^2\)

Bài 1: 

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

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

\(=2x^2-2y^2-x^2+2xy-y^2+x^2+2xy+y^2-4y^2\)

\(=2x^2+4xy-7y^2\)

Ta có

\(C=\left(x^2+2.x.2y+\left(2y\right)^2\right)-\left(2x+4y\right)+10\)

\(\Rightarrow C=\left(x+2y\right)^2-2\left(x+2y\right)+10\)

\(\Rightarrow C=5^2-2.5+10\)

\(\Rightarrow C=25-10+10=25\)

\(C=x^2+4y^2-2x+10+4xy-4y\)

    \(=\left[x^2+2.x.2y+\left(2y\right)^2\right]-\left(2x+4y\right)+10\)

    \(=\left(x+2y\right)^2-2\left(x+2\right)+10\)

    \(=5^2-2.5+10\)

    \(=5^2-10+10\)

    \(=25-10+10\)

    \(=25\)

CHÚC BẠN HỌC TỐT !!!

\(x^2+4y^2-5x+10y-4xy+20\)

\(=x^2-4xy+4y^2-2.\frac{5}{2}\left(x-2y\right)+\frac{25}{4}-\frac{25}{4}+20\)

\(=\left(x-2y\right)^2-2.\frac{5}{2}\left(x-2y\right)+\frac{25}{4}+\frac{55}{4}\)

\(=\left(x-2y-\frac{5}{2}\right)^2+\frac{55}{4}\)Thay x - 2y = 5 ta được : 

\(=\left(5-\frac{5}{2}\right)^2+\frac{55}{4}=20\)

\(B=x^2-2xy-2x+2y+y^2\)

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

\(=\left(x-y\right)^2-2\left(x-1\right)\)Thay x = y + 1 => x - y = 1 ta được : 

\(=1-2=-1\)

a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)

\(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\dfrac{x^2-4+10-x^2}{x+2}\)

\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{6}=\dfrac{-1}{x-2}\)

b: Khi x=1/2 thì \(B=\dfrac{-1}{\dfrac{1}{2}-2}=\dfrac{2}{3}\)

Khi x=-1/2 thì B=2/5

c: Để B nguyên thì \(x-2\in\left\{1;-1\right\}\)

hay \(x\in\left\{3;1\right\}\)

a, đk : x khác -2 ; 2 

\(B=\left(\dfrac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{x^2-4+10-x^2}{x+2}\right)\)

\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}:\dfrac{6}{x+2}=\dfrac{1}{2-x}\)

b, Ta có \(\left|x\right|=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{2};x=-\dfrac{1}{2}\)

Với x = 1/2 ta được \(B=\dfrac{1}{2-\dfrac{1}{2}}=\dfrac{2}{3}\)

Với x = -1/2 ta được \(B=\dfrac{1}{2+\dfrac{1}{2}}=\dfrac{2}{5}\)

c, \(\dfrac{1}{2-x}\Rightarrow2-x\inƯ\left(1\right)=\left\{\pm1\right\}\)