\(xy^2-2xy+x+y^2=6\Leftrightarrow x\left(y^2-2y+1\right)+y^2-1=5\)

\(\Leftrightarrow x\left(y-1\right)^2+\left(y-1\right)\left(y+1\right)=5\)

\(\Leftrightarrow\left(y-1\right)\left(xy-x+y+1\right)=5\)

\(Ư\left(5\right)=\left(-5;-1;1;5\right)\)

Vì \(x;y\in Z\Rightarrow\left[{}\begin{matrix}\left(x;y\right)=\left(6;0\right)\\\left(x;y\right)=\left(2;2\right)\end{matrix}\right. \)

\(S_{ACDE}=\dfrac{30\left(20+40\right)}{2}=900\left(m^2\right)\)

\(S_{ABC}=\dfrac{30.40}{2}=600\left(m^2\right)\)

\(S=900+600=1500\left(m^2\right)\)

c. \(\left\{{}\begin{matrix}3x+5y=1\\2x-y=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+10y=2\\6x-3y=-24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13y=26\\2x-y=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=-3\end{matrix}\right.\)

d. \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=1\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\left(x\ne0\right)\\\dfrac{1}{y}=b\left(y\ne0\right)\end{matrix}\right.\)

hpt \(\Leftrightarrow\left\{{}\begin{matrix}a-b=1\\3a+4b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a-3b=3\\3a+4b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-7b=-2\\a-b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{2}{7}\\a=\dfrac{9}{7}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{9}{7}\\\dfrac{1}{y}=\dfrac{2}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{9}\\y=\dfrac{7}{2}\end{matrix}\right.\)

a. \(x\ne1,x\ne3\)

\(P=\left(1+\dfrac{1}{x-1}\right)\left(\dfrac{x^2-7}{x^2-4x+3}+\dfrac{1}{x-1}+\dfrac{1}{3-x}\right)\)

\(P=\left(\dfrac{x}{x-1}\right)\left(\dfrac{x^2-7}{\left(x-1\right)\left(x-3\right)}+\dfrac{1}{x-1}-\dfrac{1}{x-3}\right)\)

\(P=\dfrac{x}{x-1}\left(\dfrac{x^2-7+x-3-x+1}{\left(x-1\right)\left(x-3\right)}\right)=\dfrac{x}{x-1}.\dfrac{x^2-9}{\left(x-1\right)\left(x-3\right)}\)

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

b. \(\left|x+2\right|=5\Leftrightarrow\left[{}\begin{matrix}x=3\left(loại\right)\\x=-7\end{matrix}\right.\)

\(P=\dfrac{49-21}{64}=\dfrac{7}{16}\)

c. \(P>1\Leftrightarrow\dfrac{x^2+3x}{\left(x-1\right)^2}>1\left(x\ne1\right)\)

\(\Leftrightarrow\dfrac{x^2+3x-x^2+2x-1}{\left(x-1\right)^2}>0\Leftrightarrow\dfrac{5x-1}{\left(x-1\right)^2}>0\)

\(\Leftrightarrow5x-1>0\Leftrightarrow x>\dfrac{1}{5}\)

Bài 1:

a. 80cm

b. 10cm

Bài 2:

Chu vi HCN là (12 + 8).2 = 40 cm

Cạnh của hình thoi là 40 : 4 = 10 cm