a) \(A=x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right)\)

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

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

a) A = a) A = x² - xy + x - y= (x² - xy) + (x - y)=x(x-y)+(x-y)=(x+1)(x-y)
c) A = 3x - 3y + x² - y²=3(x-y)+(x-y)(x+y)=(3+x+y)(x-y)
d) A = x² - y² - 2x - 2y = (x-y)(x+y)-2(x+y)=(x+y)(x-y-2)

câu b bạn xem lại đúng đề ko
 

\(\)a, \(A=x^2-xy+x-y\)
        \(=x\left(x-y\right)+\left(x-y\right)\)
        \(=\left(x+1\right)\left(x-y\right)\)
 

a. Trừ vế theo vế \(\left(1\right)\) cho \(\left(2\right)\) ta được \(x^2-y^2=4x-4y\)

\(\Leftrightarrow\left(x-y\right)\left(x+y-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=4-y\end{matrix}\right.\)

TH1: \(x=y\)

Phương trình \(\left(1\right)\) tương đương:

\(x^2=2x\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y=0\\x=y=2\end{matrix}\right.\)

TH2: \(x=4-y\)

Phương trình \(\left(2\right)\) tương đương:

\(y^2=4y-4\)

\(\Leftrightarrow y^2-4y+4=0\)

\(\Leftrightarrow\left(y-2\right)^2=0\)

\(\Leftrightarrow y=2\)

\(\Rightarrow x=2\)

Vậy hệ đã cho có nghiệm \(\left(x;y\right)\in\left\{\left(0;0\right);\left(2;2\right)\right\}\)

b. \(\left\{{}\begin{matrix}x+y+xy=5\\x^2+y^2=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=5-\left(x+y\right)\\\left(x+y\right)^2-2xy=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=5-\left(x+y\right)\\\left(x+y\right)^2-10+2\left(x+y\right)=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=5-\left(x+y\right)\\\left(x+y\right)^2+2\left(x+y\right)-15=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=5-\left(x+y\right)\\\left(x+y+5\right)\left(x+y-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=5-\left(x+y\right)\\\left[{}\begin{matrix}x+y=-5\\x+y=3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=-5\\xy=10\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+y=-5\\xy=10\end{matrix}\right.\Leftrightarrow\) vô nghiệm

TH2: \(\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\end{matrix}\right.\)

Vậy ...

a) \(\dfrac{1}{x^3-8}=\dfrac{1}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{2}{2\left(x-2\right)\left(x^2+2x+4\right)}\)

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

b) \(\dfrac{x}{x^2-1}=\dfrac{x}{\left(x+1\right)\left(x-1\right)}=\dfrac{x\left(x+1\right)}{\left(x+1\right)^2\left(x-1\right)}\)

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

c) \(\dfrac{1}{x+2}=\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)^2}\)

\(\dfrac{1}{x^2-4x+4}=\dfrac{1}{\left(x-2\right)^2}=\dfrac{x+2}{\left(x+2\right)\left(x-2\right)^2}\)

\(\dfrac{5}{2-x}=\dfrac{-5}{x-2}=\dfrac{-5\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)^2}\)

d) \(\dfrac{1}{3x+3y}=\dfrac{1}{3\left(x+y\right)}=\dfrac{\left(x-y\right)^2}{3\left(x+y\right)\left(x-y\right)^2}\)

\(\dfrac{2x}{x^2-y^2}=\dfrac{2x}{\left(x+y\right)\left(x-y\right)}=\dfrac{6x\left(x-y\right)}{3\left(x+y\right)\left(x-y\right)^2}\)

\(\dfrac{x^2-xy+y^2}{x^2-2xy+y^2}=\dfrac{x^2-xy+y^2}{\left(x-y\right)^2}=\dfrac{3\left(x^2-xy+y^2\right)\left(x+y\right)}{3\left(x+y\right)\left(x-y\right)^2}=\dfrac{3\left(x^3+y^3\right)}{3\left(x+y\right)\left(x-y\right)^2}\)

a: Ta có: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)

\(=4x^2-4x+1-2\left(4x^2-12x+9\right)+4\)

\(=4x^2-4x+5-8x^2+24x-18\)

\(=-4x^2+20x-13\)

b: \(\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(1-2y\right)^2\)

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

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

a: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)

\(=4x^2-4x+1+4-2\left(4x^2-12x+9\right)\)

\(=4x^2-4x+5-8x^2+24x-18\)

\(=-4x^2+20x-13\)

e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)

x y + ( 1 + x 2 ) ( 1 + y 2 ) = 1 ⇔ ( 1 + x ) 2 ( 1 + y ) 2 = 1 − x y ⇒ ( 1 + x 2 ) ( 1 + y 2 ) = 1 - x y 2 ⇔ 1 + x 2 + y 2 + x 2 y 2 = 1 − 2 x y + x 2 y 2 ⇔ x 2 + y 2 + 2 x y = 0 ⇔ x + y 2 = 0 ⇔ y = − x ⇒ x 1 + y 2 + y 1 + x 2 = x 1 + x 2 − x 1 + x 2 = 0

CMR: \(\frac{1}{x}+\frac{1}{y}\le2\)  biết \(^{x^3+y^3+3\left(x^2+y^2\right)+4\left(x+y\right)+4=0}\) và xy>0

tôi quên mât CMR: 1/x+1/y<=-2