Chiều dài sân là: 720 : 60 = 20 (m)

Chu vi sân là: (60 + 20)x2=160(m)

Nửa chu vi là: \(\dfrac{5}{3}:2=\dfrac{5}{6}\left(m\right)\)

Chiều dài là: \(\left(\dfrac{5}{6}+\dfrac{1}{2}\right):2=\dfrac{2}{3}\left(m\right)\)

Chiều rộng là: \(\dfrac{5}{6}-\dfrac{2}{3}=\dfrac{1}{6}\left(m\right)\)

Diện tích miếng bìa là: \(\dfrac{2}{3}\times\dfrac{1}{6}=\dfrac{1}{9}\left(m^2\right)\)

a) Ta có \(x^2\ge0;\left(y-\dfrac{1}{3}\right)^2\ge0\)

\(x^2+\left(y-\dfrac{1}{3}\right)^2=0\Leftrightarrow\left\{{}\begin{matrix}x=0\\y-\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{3}\end{matrix}\right.\)

b) Ta có \(\left(\dfrac{1}{2}x-3\right)^4\ge0;\left(y^2-\dfrac{1}{4}\right)^6\ge0\)

\(\left(\dfrac{1}{2}x-3\right)^4+\left(y^2-\dfrac{1}{4}\right)^6\le0\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x-3=0\\y^2-\dfrac{1}{4}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=6\\\left[{}\begin{matrix}y=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=6\\y=\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=6\\y=-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

a) Nửa chu vi mảnh vườn là: 320 : 2 = 160 (m)

Chiều dài mảnh vườn là: (160 + 20) : 2 = 90 (m)

Chiều rộng mảnh vườn là: 160 - 90 = 70 (m)

Diện tích mảnh vườn là: 90 . 70 = 6300 (m²)

b) Cạnh bồn hoa là: 20 : 4 = 5 (m)

Diện tích bồn hoa là: 5 .5 = 25 (m²)

Diện tích phần còn lại là: 6300 - 25 = 6275 (m²)

Câu 3.

a) \(x^2-7x=0\Rightarrow x\left(x-7\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)

b) \(-3x^2+5x=0\Rightarrow x\left(-3x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{3}\end{matrix}\right.\)

c) \(x^2-4x+4=0\Rightarrow\left(x-2\right)^2=0\Rightarrow x-2=0\Rightarrow x=2\)

d) \(2x^2-72=0\Rightarrow x^2-36=0\Rightarrow x^2=36\Rightarrow x=\pm6\)

Câu 2.

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

b) \(64x^3+27=\left(4x+3\right)\left(16x^2-12x+9\right)\)

c) \(a^4-b^2=\left(a^2+b\right)\left(a^2-b\right)\)

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

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

f) \(x^3+6x^2+12x+8=x^3+3.x^2.2+3.x.2^2+2^3=\left(x+2\right)^3\)

Câu 1.

a) \(49a^2-\left(2a-b\right)^2=\left(7a+2a-b\right)\left(7a-2a+b\right)=\left(9a-b\right)\left(5a+b\right)\)

b) \(\left(a-2b\right)^2-\left(3a+b\right)^2=\left(a-2b+3a+b\right)\left(a-2b-3a-b\right)=\left(4a-b\right)\left(-2a-3b\right)\)

c) \(36a^2-\left(3a-2b\right)^2=\left(6a+3a-2b\right)\left(6a-3a+2b\right)=\left(9a-2b\right)\left(3a+2b\right)\)

d) \(9\left(a+b\right)^2-4\left(a-2b\right)^2=\left(3a+3b+2a-4b\right)\left(3a+3b-2a+4b\right)=\left(5a-b\right)\left(a+7b\right)\)

e) \(16\left(x+2y\right)^2-9\left(x-3y\right)^2=\left(4x+8y+3x-9y\right)\left(4x+8y-3x+9y\right)=\left(7x-y\right)\left(x+17y\right)\)

Ta có \(\left\{{}\begin{matrix}\left(3x-1\right)^2\ge0\\\left(x-y+8\right)^4\ge0\end{matrix}\right.\Rightarrow B\ge5\)

Dấu = xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}3x-1=0\\x-y+8=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=x+8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=\dfrac{25}{3}\end{matrix}\right.\)

Vậy \(B_{min}=5\) đạt được khi \(x=\dfrac{1}{3};y=\dfrac{25}{3}\)

3cm trên bản đồ ứng với \(3\times1000000=3000000\left(cm\right)\) trên thực tế.