Diện tích mỗi viên gạch men là:
20 x 20= 400 (cm)
Diện tích phòng học đó là:
12 x 5= 60 (m)
400cm = 4m
Số viên gạch cần dùng để lát kín nền phòng học đó là:
60 : 4= 15 ( viên gạch)
                  Đáp số: 15 viên gạch
 

l-xl+l3x+5l+5x-8/+3

Nếu (−x−7)⩾0; (3x+5)⩾0 thì−x−7+3x+5+5x=8⇔7x−2=8⇔7x=10⇔x=107  Nếu (−x−7)<0; (3x+5)<0 thìx+7−3x−5+5x=8⇔3x+2=8⇔3x=10⇔x=103Vậy x=103;x=107

c) ( x² + x + 1 )( x² + x + 2 ) - 12

Đặt t = x² + x + 1

<=> t( t + 1 ) - 12

= t² + t - 12

= t² - 3t + 4t - 12

= t( t - 3 ) + 4( t - 3 )

= ( t - 3 )( t + 4 )

= ( x² + x + 1 - 3 )( x² + x + 1 + 4 )

= ( x² + x - 2 )( x² + x + 5 )

= ( x² - x + 2x - 2 )( x² + x + 5 )

= [ x( x - 1 ) + 2( x - 1 ) ]( x² + x + 5 )

= ( x - 1 )( x + 2 )( x² + x + 5 )

d) ( x + 2 )( x + 3 )( x + 4 )( x + 5 ) - 24

= [ ( x + 2 )( x + 5 ) ][ ( x + 3 )( x + 4 ) ] - 24

= ( x² + 7x + 10 )( x² + 7x + 12 ) - 24

Đặt t = x² + 7x + 10

<=> t( t + 2 ) - 24

= t² + 2t - 24

= t² - 4t + 6t - 24

= t( t - 4 ) + 6( t - 4 )

= ( t - 4 )( t + 6 )

= ( x² + 7x + 10 - 4 )( x² + 7x + 10 + 6 )

= ( x² + 7x + 6 )( x² + 7x + 16 )

= ( x² + 6x + x + 6 )( x² + 7x + 16 )

= [ x( x + 6 ) + ( x + 6 ) ]( x² + 7x + 16 )

= ( x + 6 )( x + 1 )( x² + 7x + 16 )

a, Sửa đề:\(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)

Đặt \(t=x^2+x\)

\(\Rightarrow t^2-2t-15\)

\(=t^2-5t+3t-15\)

\(=t\left(t-5\right)+3\left(t-5\right)\)

\(=\left(t+3\right)\left(t-5\right)\)

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

b,\(x^2+2xy+y^2-x-y-12\)

\(=\left(x+y\right)^2-\left(x+y\right)-12\)

Đặt \(t=x+y\)

\(\Rightarrow t^2-t-12\)

\(=t^2-4t+3t-12\)

\(=t\left(t-4\right)+3\left(t-4\right)\)

\(=\left(t+3\right)\left(t-4\right)\)

\(=\left(x+y+3\right)\left(x+y-4\right)\)

c,\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

Đặt \(t=x^2+x+1\)

\(\Rightarrow t\left(t+1\right)-12\)

\(=t^2+t-12\)

\(=t^2-3t+4t-12\)

\(=t\left(t-3\right)+4\left(t-3\right)\)

\(=\left(t+4\right)\left(t-3\right)\)

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

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

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

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

d,\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt \(t=x^2+7x+10\)

\(\Rightarrow t\left(t+2\right)-24\)

\(=t^2+2t-24\)

\(=t^2-4t+6t-24\)

\(=t\left(t-4\right)+6\left(t-4\right)\)

\(=\left(t+6\right)\left(t-4\right)\)

\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)\)

\(=\left(x^2+7x+16\right)\left(x^2+x+6x+6\right)\)

\(=\left(x^2+7x+16\right)\left[x\left(x+1\right)+6\left(x+1\right)\right]\)

\(=\left(x^2+7x+16\right)\left(x+6\right)\left(x+1\right)\)