\(\frac{1.5.18+2.10.36+3.15.54}{1.3.9+2.6.18+3.9.27}=\frac{1.5.18.\left(1+2.2.2+3.3.3\right)}{1.3.9.\left(1+2.2.2+3.3.3\right)}\)

\(=\frac{1.5.18}{1.3.9}=\frac{10}{3}\)

dmm

a: \(\left(x+1\right)^3+\left(x-2\right)^3=2x^3+2\left(2x-1\right)^2-9\)

\(\Leftrightarrow x^3+3x^2+3x+1+x^3-6x^2+12x-8=2x^3+2\left(4x^2-4x+1\right)-9\)

\(\Leftrightarrow2x^3-3x^2+15x-7=2x^3+8x^2-8x-7\)

\(\Leftrightarrow-11x^2+23x=0\)

\(\Leftrightarrow x\left(-11x+23\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{23}{11}\end{matrix}\right.\)

a, \(\left(2x-1\right)\left(x+3\right)-2x^2+5x=7\)

\(\Leftrightarrow2x^2+6x-x-3-2x^2+5x=7\)

\(\Leftrightarrow2x^2+5x-3-2x^2+5x=7\)

\(\Leftrightarrow10x-10=0\Leftrightarrow x=1\)

b, \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-4\right)\left(x+4\right)=54\)

\(\Leftrightarrow\left(x^3+27\right)-x\left(x^2-16\right)=54\)

\(\Leftrightarrow x^3+27-x^3+16x=54\)

\(\Leftrightarrow-27+16x=0\Leftrightarrow x=\frac{27}{16}\)

(x+3)(x^2-3x+9)-(54+x^3)=0

=>x^3+27-54-x^3=0

=>-27=0(vô lý)

a: =>x/-3=3

hay x=-9

b: =>x/9=-1/9

hay x=-1

c: =>x+1/5=-1/3

hay x=-8/15

d: =>-7/x=-7/9

hay x=9

a, \(\dfrac{x}{-3}=3\Leftrightarrow x=-9\)

b, \(\dfrac{x}{9}=-\dfrac{1}{9}\Rightarrow x=-1\)

c, \(\dfrac{x+3}{15}=-\dfrac{6}{15}\Rightarrow x=-9\)

d, \(\dfrac{42}{-54}=-\dfrac{42}{6x}\Rightarrow6x=54\Leftrightarrow x=9\)

a) (x + 1) + (x + 2) + (x + 3) + ... + (x + 9)   = 54

(x + x + x + ... + x) + (1 + 2 + 3 + ... + 9) = 54

9x + 45 = 54

9x          = 54 - 45

9x          = 9

  x          = 1

b) (x + 1) + (x + 2) + (x + 3) + ... + (x + 10)   = 2010

     (x + x + x + ... + x) + (1 + 2 + 3 + ... + 10) = 2010

      10x + 55 = 2010

      10x          = 2010 - 55

      10x          = 1955

          x          = 391/2.

1)\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+....+\left(x+9\right)=54\)

\(\Rightarrow x+1+x+2+x+3+.....+x+9=54\)

\(\Rightarrow\left(x+x+x+....+x\right)+\left(1+2+3+....+9\right)=54\)

\(\Rightarrow9x+45=54\)\(\Rightarrow9x=9\Rightarrow x=1\)

2)\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+....+\left(x+10\right)=2010\)

\(\Rightarrow x+1+x+2+x+3+.....+x+10=2010\)

\(\Rightarrow\left(x+x+x+....+x\right)+\left(1+2+3+....+10\right)=2010\)

\(\Rightarrow10x+55=2010\Rightarrow10x=1995\Rightarrow x=195,5\)

Gọi (x^54 + x^45 +......+ x^9 + 1) =f(x) 

Đặt f(x) = (x^2 -1 )* Q(x) +R(x)

Do đa thức có bậc không quá 2 nên đa thức dư có bậc không quá 1 nên đặt R(x) = ax +b 

Thay vào ta có (x^54 + x^45 +x^36+......+x^9+1) = x^2 -1* Q(x) +ax+b

Lần lượt gán x=1 và x= -1 ta có 

F(1) = ( 1^54+1^45+.....,,+1^9+1)= 1^2-1 *Q(x) +a*1+ b 

=> 7 = a+b

Tương tự gán x =-1 ta dược 1= b-a

=> b= 7+1/2 =4

a= 7-4=3

Do đó dư là 3x +4