Đổi: \(50cm=0,5m\)

Gọi chiều rộng là \(x\left(x>0\right)\left(m\right)\)

Suy ra chiều dài là \(2x\left(m\right)\)

Diện tích sân phơi ban đầu: \(2x^2\left(m^2\right)\)

Diện tích sân phơi lúc sau: \(2x^2-0,5\left(m^2\right)\)

Theo đề bài ta có: \(\left(x-2\right)\left(2x-2\right)=2x^2-0,5\)

\(\Leftrightarrow x=0,75\)

Vậy chiều rộng ban đầu là \(0,75m\), chiều dài là \(1,5m\)

\(a,2x\left(3x^2+1\right)=6x^3+2x\)

\(b,\left(2x^3-5x^2+6x\right):2=x^3-\dfrac{5}{2}x^2+3x\)

\(c,\left(x-3\right)\left(x+5\right)-x\left(x+2\right)=x^2+5x-3x-15-x^2-2x=-15\)

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

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

\(S=\left(3x+1\right)\left(6-x\right)=18-3x^2+6-x=-3x^2+17x+6\)

Đặt \(A=-3x^2+17x+6\)

\(S_{max}\Leftrightarrow A_{max}=-3\left(x^2-\dfrac{17}{3}x-2\right)=-3\left[x^2-2.x.\dfrac{17}{6}+\left(\dfrac{17}{6}\right)^2-\left(-\dfrac{17}{6}\right)^2-2\right]\)

\(=-3\left[\left(x-\dfrac{17}{6}\right)^2-\dfrac{361}{36}\right]\)

\(=-3\left(x-\dfrac{17}{6}\right)^2+\dfrac{361}{12}\)

Vì \(-3\left(x-\dfrac{17}{6}\right)^2\le0\forall x\in R\)

\(\Rightarrow-3\left(x-\dfrac{17}{6}\right)^2+\dfrac{361}{12}\le\dfrac{361}{12}\forall x\in R\)

\(\Rightarrow A_{max}=\dfrac{361}{12}\Leftrightarrow\left(x-\dfrac{17}{6}\right)^2=0\Leftrightarrow x=\dfrac{17}{6}\)

\(\Rightarrow S_{max}\Leftrightarrow x=\dfrac{17}{6}\)

\(A=\dfrac{1}{\sqrt{3}+\sqrt{2}}-\dfrac{1}{\sqrt{3}-\sqrt{2}}\)

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

\(=\sqrt{3}-\sqrt{2}-\sqrt{3}-\sqrt{2}\)

\(=-2\sqrt{2}\)

\(\left(P\right):y=ax^2+bx+3\left(a\ne0\right)\)

\(\left(P\right)\) đi qua điểm \(A\left(-1;9\right)\Leftrightarrow a.\left(-1\right)^2+b.\left(-1\right)+3=9\Leftrightarrow a-b=6\left(1\right)\)

\(\left(P\right)\) có trục đối xứng \(x=-2\Leftrightarrow\dfrac{-b}{2a}=-2\Leftrightarrow-4a+b=0\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}a-b=6\\-4a+b=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-2\\b=-8\end{matrix}\right.\)

\(\Rightarrow\left(P\right):y=-2x^2-8x+3\)

\(\left\{{}\begin{matrix}3x-y=2m-1\\x+2y=3m+2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x-2y=4m-2\\x+2y=3m+2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7x=7m\\x+2y=3m+2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\m+2y=3m+2\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\y=m+1\end{matrix}\right.\)

\(x^2+y^2=5\Rightarrow m^2+\left(m+1\right)^2=5\)

\(\Leftrightarrow m^2+m^2+2m+1=5\)

\(\Leftrightarrow2m^2+2m-4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=1\\m=-2\end{matrix}\right.\)

\(P=\dfrac{3x^2+6x+3}{x+1}\)

\(a,\) Điều kiện xác định: \(x+1\ne0\Leftrightarrow x\ne-1\)

\(b,P=\dfrac{3x^2+6x+3}{x+1}=\dfrac{3\left(x^2+2x+1\right)}{x+1}=\dfrac{3\left(x+1\right)^2}{x+1}=3\left(x+1\right)=3x+3\)

\(c,x=1\Rightarrow P=3.1+3=6\)

\(a//b\Rightarrow\widehat{M_4}=\widehat{N}=70^o\) (2 góc so le trong)

\(\widehat{M_4}+\widehat{M_3}=180^o\) (2 góc kề bù)

\(\Rightarrow\widehat{M_3}=180^o-70^o=110^o\)