Gọi vận tốc lượt đi là x(km/h)

(Điều kiện: x>0)

vận tốc lượt về là \(x\left(1+20\%\right)=1,2x\left(\dfrac{km}{h}\right)\)

Thời gian đi là \(\dfrac{120}{x}\left(giờ\right)\)

Thời gian về là \(\dfrac{120}{1,2x}=\dfrac{100}{x}\left(giờ\right)\)

Tổng thời gian cả đi lẫn về là 4h24p=4,4 giờ nên ta có:

\(\dfrac{120}{x}+\dfrac{100}{x}=4,4\)

=>\(\dfrac{220}{x}=4,4\)

=>\(x=\dfrac{220}{4,4}=50\left(nhận\right)\)

Vậy: vận tốc lượt đi là 50km/h

\(2x^3-5x-6\\ =2x^3-4x^2+4x^2-8x+3x-6\\ =2x^2\left(x-2\right)+4x\left(x-2\right)+3\left(x-2\right)\\=\left(2x^2+4x+3\right)\left(x-2\right)\)

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

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

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

Em ghi là đường  cao H là sai, phải ghi là BH mới đúng vì vậy Olm bảo em làm  sai em hiểu  chưa nhỉ?

\(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\\ =x^2y-x^2z+y^2z-xy^2+xz^2-yz^2\\ =\left(x^2y-x^2z\right)+\left(y^2z-yz^2\right)-\left(xy^2-xz^2\right)\\ =x^2\left(y-z\right)+yz\left(y-z\right)-x\left(y^2-z^2\right)\\ =x^2\left(y-z\right)+yz\left(y-z\right)-x\left(y+z\right)\left(y-z\right)\\ =\left(y-z\right)\left[x^2+yz-x\left(y+z\right)\right]\\ =\left(y-z\right)\left(x^2+yz-xy-xz\right)\\ =\left(y-z\right)\left[x\left(x-y\right)-z\left(x-y\right)\right]\\ =\left(y-z\right)\left(x-z\right)\left(x-y\right)\)

a: \(\left(x-2\right)\left(3x-1\right)\left(x^2-4x+1\right)\)

\(=\left(3x^2-x-6x+2\right)\left(x^2-4x+1\right)\)

\(=\left(3x^2-7x+2\right)\left(x^2-4x+1\right)\)

\(=3x^4-12x^3+3x^2-7x^3+28x^2-7x+2x^2-8x+2\)

\(=3x^4-19x^3+33x^2-15x+2\)

b: \(x\left(3-4x\right)\left(2x^2-3x\right)\)

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

\(=-8x^4+12x^3+6x^3-9x^2\)

\(=-8x^4+18x^3-9x^2\)

a) 

\(\left(x-2\right)\left(3x-1\right)\left(x^2-4x+1\right)\\ =\left(3x^2-6x-x+2\right)\left(x^2-4x+1\right)\\ =\left(3x^2-7x+2\right)\left(x^2-4x+1\right)\\ =3x^4-12x^3+3x^2-7x^3+28x^2-7x-8x+2\\ =3x^4-19x^3+31x^2-15x+2\) 

b) 

\(x\left(3-4x\right)\left(2x^2-3x\right)\\ =\left(3x-4x^2\right)\left(2x^2-3x\right)\\ =6x^3-9x^2-8x^4+12x^3\\ =-8x^4+18x^3-9x^2\)

\(x^7+x^2+1\)

\(=x^7+x^6+x^5-x^6-x^5-x^4+x^4+x^3+x^2-x^3-x^2-x+x^2+x+1\)

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

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

a) Thay x=2 vào ta có:

\(2^2-4m\cdot2+1=0\\ \Leftrightarrow4-8m+1=0\\ \Leftrightarrow5-8m=0\\ \Leftrightarrow8m=5\\ \Leftrightarrow m=\dfrac{5}{8}\)

b) Thay x=2 vào ta có:

\(3\cdot2^2-5m\cdot2+7\\ \Leftrightarrow12-10m+7=0\\ \Leftrightarrow19-10m=0\\ \Leftrightarrow10m=19\\\Leftrightarrow m=\dfrac{19}{10}\)

a:

Đặt \(x^2-4mx+1=0\left(1\right)\)

Thay x=2 vào (1), ta được:

\(2^2-4m\cdot2+1=0\)

=>\(4-8m+1=0\)

=>5-8m=0

=>8m=5

=>\(m=\dfrac{5}{8}\)

b: Đặt \(3x^2-5mx+7=0\left(2\right)\)

Thay x=2 vào (2), ta được:

\(3\cdot2^2-5m\cdot2+7=0\)

=>12-10m+7=0

=>19-10m=0

=>10m=19

=>\(m=\dfrac{19}{10}\)

a)5x+17-(2x+5)=0

=>5x+17-2x-5=0

=>3x+12=0

=>3x=-12

=>x=-12:3=-4

b)3(1-x)-(5-2x)=0

=>3-3x-5+2x=0

=>-2-x=0

=>x=-2

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

=>2x-2-3x+6=0

=>-x+4=0

=>x=4

d)(x-3)(2x-5)+(2x-4)(5-2x)=0

=>(x-3)(2x-5)-(2x-4)(2x-5)=0

=>(2x-5)(x-3-2x+4)=0

=>(2x-5)(1-x)=0

TH1: 2x - 5=0=>2x=5=>x=5/2

TH2: 1-x=0=>x=1

a: Đặt 5x+17-(2x+5)=0

=>\(5x+17-2x-5=0\)

=>\(3x+12=0\)

=>\(3x=-12\)

=>\(x=-\dfrac{12}{3}=-4\)

b: Đặt \(3\left(1-x\right)-\left(5-2x\right)=0\)

=>\(3-3x-5+2x=0\)

=>\(-x-2=0\)

=>x+2=0

=>x=-2

c: Đặt \(2\left(x-1\right)-3\left(x-2\right)=0\)

=>\(2x-2-3x+6=0\)

=>4-x=0

=>x=4

d: Sửa đề: (x-3)(2x-5)+(2x-4)*(5-x) 

Đặt \(\left(x-3\right)\left(2x-5\right)+\left(2x-4\right)\left(5-x\right)=0\)

=>\(2x^2-5x-6x+15+10x-2x^2-20+4x=0\)

=>3x-5=0

=>3x=5

=>\(x=\dfrac{5}{3}\)

Đặt 5x+17-(2x+5)=0

=>5x+17-2x-5=0

=>3x+12=0

=>3x=-12

=>\(x=-\dfrac{12}{3}=-4\)