a) x²+2x-3xy-6y

= x(x+2) - 3y(x+2)

= (x+2)(x-3y)

b) x²-6x+9+4y²

= (x-3)²+4y²

Chiều rộng là:

\(\dfrac{5}{3}-\dfrac{1}{2}=\dfrac{7}{6}\left(dm\right)\)

Diện tích là:

\(\dfrac{5}{3}\times\dfrac{7}{6}=\dfrac{35}{18}\left(dm^2\right)\)

Vậy ...

\(A=x\left(x-1\right)-y\left(x-1\right)\)

\(=\left(x-y\right)\left(x-1\right)\)

Thay x=2001, y=1999, ta có:

\(A=\left(2001-1999\right)\left(2001-1\right)\)

\(=2.2000=4000\)

\(178-3x=5^2=25\)

\(3x=178-25=153\)

\(x=153:3=51\)

a) 3x(x+2)-(x+2)=0

(x+2)(3x-1)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)

b) \(\left(x-5\right)^2-9x^2=\left(x-5-3x\right)\left(x-5+3x\right)=\left(2x+5\right)\left(5-4x\right)\)

c) x-10x+25=0

-9x+25=0

\(x=\dfrac{25}{9}\)

\(A=\dfrac{5}{\sqrt{x}+1}\left(đk:x\ge0\right)\)

Vì x≥0 => A≤\(\dfrac{5}{2}\)

\(MaxA=\dfrac{5}{2}\Leftrightarrow x=0\)

\(B=\dfrac{15}{3\sqrt{x}+1}\left(đk:x\ge0\right)\)

Vì x≥0 => \(B\le\dfrac{15}{1}=15\)

\(MaxB=15\Leftrightarrow x=0\)

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

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

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

\(=5\left(2x-3\right)+x^2+2x+1\)

\(=x^2+12x-14\)

Đổi \(9\dfrac{2}{3}m=\dfrac{29}{3}m\)

Chiều rộng là:

\(\dfrac{29}{3}\times\dfrac{1}{2}=\dfrac{29}{6}m\)

Chu vi là:

\(\left(\dfrac{29}{3}+\dfrac{29}{6}\right)\times2=29\left(m\right)\)

Diện tích là:

\(\dfrac{29}{3}\times\dfrac{29}{6}=\dfrac{841}{18}\left(m^2\right)\)

Vậy ...

Số học sinh nam là:

\(253:\left(3+8\right)\times8=184\left(hs\right)\)

Số học sinh nữ là:

\(253-184=69\left(hs\right)\)

Vậy ...

Chiều dài là:

\(360:\left(7-4\right)\times7=840\left(m\right)\)

Chiều rộng là:

\(840-360=480\left(m\right)\)

Chu vi là:

\(\left(840+480\right)\times2=2640\left(m\right)\)

Diện tích là:

\(840\times480=403200\left(m^2\right)\)

Vậy ...