Bài 1:

\(54\left(\dfrac{km}{h}\right)=15\left(\dfrac{m}{s}\right);9\left(\dfrac{m}{s}\right)=32,4\left(\dfrac{km}{h}\right)\)

Baì 2:

\(t'=s':v'=5:\left(5.3,6\right)=\dfrac{5}{18}h\)

\(\Rightarrow v_{tb}=\dfrac{s'+s''}{t'+t''}=\dfrac{5+3,8}{\dfrac{5}{18}+\left(\dfrac{15}{60}\right)}\simeq16,67\left(\dfrac{km}{h}\right)\)

Câu 2: 

\(\Leftrightarrow\left(x+2\right)\left(10x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-\dfrac{3}{10}\end{matrix}\right.\)

a) \(3\left(x-y\right)^2+9y\left(y-x\right)^2\)

\(=3\left(x-y\right)^2+9y\left(x-y\right)^2\)

\(=\left(x-y\right)^2\left(3-9y\right)\)

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

b) \(3\left(x-y\right)^2+9y\left(y-x\right)\)

\(=3\left(y-x\right)^2+9y\left(y-x\right)\)

\(=\left(y-x\right)\left[3\left(y-x\right)+9y\right]\)

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

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

a: =3(x-y)^2+9y(x-y)^2

=(x-y)^2(3+9y)

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

b: =3(x-y)^2-9y(x-y)

=3(x-y)(x-y-9y)

=3(x-y)(x-10y)

x² + xy + 5x + 5y = ( x² + xy ) + ( 5x + 5y ) = x( x + y ) + 5( x + y ) = ( x + y )( x + 5 )

x² - y² + 3x - 3y = ( x² - y² ) + ( 3x - 3y ) = ( x - y )( x + y ) + 3( x - y ) = ( x - y )( x + y + 3 )

x² + xy + 5x + 5y 

= (x²+ xy) + ( 5x+5y)

= x(x+y) + 5(x+y)

= (x+y)(x+5)

x² - y² + 3x - 3y

= (x² - y²) + ( 3x -3y)

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

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

chúc bạn học tốt ^^

a) \(x^2+xy+5x+5y=\left(x^2+xy\right)+\left(5x+5y\right)\)

\(=x\left(x+y\right)+5\left(x+y\right)=\left(x+y\right)\left(x+5\right)\)

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

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

\(=x^2\left(x-6\right)-24\left(x-6\right)=\left(x^2-24\right)\left(x-6\right)\)

\(x^3-6x^2-24x+144=\left(x^3-6x^2\right)-\left(24x-144\right)=x^2\left(x-6\right)-24\left(x-6\right)=\left(x-6\right)\left(x^2-24\right)\)

b: \(\left(x^2+4\right)^2-16x^2\)

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

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

c: \(x^5-x^4+x^3-x^2\)

\(=x^4\left(x-1\right)+x^2\left(x-1\right)\)

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

Lời giải:

a. Bạn xem lại đề

b. \((x^2+4)^2-16x^2=(x^2+4)^2-(4x)^2=(x^2+4-4x)(x^2+4+4x)\)

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

c.

\(x^5-x^4+x^3-x^2=x^4(x-1)+x^2(x-1)=(x^4+x^2)(x-1)\)

\(=x^2(x^2+1)(x-1)\)

a) 7x.(-y)+2(y-x)2

=>-7xy+4y-4x

b)(x^2+4)-16x^2

=>x^2+4-16x^2

=>-15x^2+4

c)x^5-x^4+x^3-x^2

=>x^4(x-1)+x^2(x-1)

=>(x^4+x^2)(x-1)