F= 21x⁸ - 24x⁶ + 9x⁵ + 3x³ + 6x² + 2006 

  = 3x²( 7x⁶ - 8x⁴ + 3x³ + x +2) +2006 

  = 0 + 2006 

  = 0

sorry cái kquả ban nãy mình viết nhầm

Kquả là 2006

a)    x³-x²-21x+45=0

<=> x³+5x²-6x²-30x+9x+45=0

<=> (x+5)(x²-6x+9)=0

<=> (x+5)(x²-3x-3x+9)=0

<=> (x+5)(x-3)²=0

 Vậy S={-5;3}

b)    X³+3X²+4X+2=0

<=>  X³+X²+2X²+2X+2X+2=0

<=> (X+1)(X²+2X+2)=0

VÌ  X²+2X+2 >=0

NÊN S={-1}

C)    X⁴+7X-8=0

<=> X⁴-X³+X³-X²+X²-X+8X-8=0

<=> (X-1)(X³+X²+X+8)=0

VÌ X³+X²+X+8>=0

NÊN S={1}

D)     6X⁴-X³-7X²+X+1=0

<=>  6X⁴-6X³+5X³-5X²-2X²+2X-X+1=0

<=>  (X-1)(6X³+5X²-2X-1)=0

<=> (X-1)(6X³-3X²+8X²-4X+2X-1)=0

<=> (X-1)(2X-1)(3X^2_4X+1)=0

<=>  (X-1)(2X-1)(3X²-3x-x+1)=0

<=> (X-1)²(2X-1)(3x-1)=0

vậy S={1/3;1/2;1}

Ta có: (3x-3)(5-21x)+(7x+4)(6x-5)=45

\(\Leftrightarrow15x-63x^2-15+63x+42x^2-35x+24x-20-45=0\)

\(\Leftrightarrow-21x^2+67x-80=0\)

\(\Leftrightarrow-21\left(x^2-\frac{67}{21}x-\frac{80}{21}\right)=0\)

mà -21<0

nên \(x^2-\frac{67}{21}x-\frac{80}{21}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\frac{67}{42}+\frac{4489}{1764}-\frac{11209}{1764}=0\)

\(\Leftrightarrow\left(x-\frac{67}{42}\right)^2=\frac{11209}{1764}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{67}{42}=\frac{\sqrt{11209}}{42}\\x-\frac{67}{42}=-\frac{\sqrt{11209}}{42}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{67+\sqrt{11209}}{42}\\x=\frac{67-\sqrt{11209}}{42}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{67+\sqrt{11209}}{42};\frac{67-\sqrt{11209}}{42}\right\}\)

a)

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

=> x + 4 = 0 hoặc 3x-5 = 0

     x = -4                 x = 5/3

b)

  2x² + 7x + 3 = 0

  2x² + 6x + x + 3= 0

  (2x+1)(x+3) = 0

=> 2x+1 = 0 hoặc x + 3 = 0

    x = -1/2              x = -3

Lời giải:

$45-5(x+4)=10$

$5(x+4)=45-10=35$

$x+4=35:5=7$

$x=7-4=3$

------------------------

$(7x-15):3-6=17$

$(7x-15):3=17+6=23$
$7x-15=23\times 3=69$

$7x=69+15=84$

$x=84:7=12$

------------------

$3^{x+2}-5.3^x=108$

$3^x(3^2-5)=108$

$3^x.4=108$

$3^x=108:4=27=3^3$

$\Rightarrow x=3$

\(45-5\left(x+4\right)=10\)

\(\Rightarrow5\left(x+4\right)=45-10\)

\(\Rightarrow5\left(x+4\right)=35\)

\(\Rightarrow x+4=35:5\)

\(\Rightarrow x+4=7\)

\(\Rightarrow x=7-4\)

\(\Rightarrow x=3\)

______________

\(\left(7x-15\right):3-6=17\)

\(\Rightarrow\left(7x-15\right):3=17+6\)

\(\Rightarrow\left(7x-15\right):3=23\)

\(\Rightarrow7x-15=23\cdot3\)

\(\Rightarrow7x-15=69\)

\(\Rightarrow7x=69+15\)

\(\Rightarrow7x=84\)

\(\Rightarrow x=12\)

______________

\(3^{x+2}-5\cdot3^x=108\)

\(\Rightarrow3^x\cdot\left(3^2-5\right)=108\)

\(\Rightarrow3^x\cdot\left(9-5\right)=108\)

\(\Rightarrow3^x\cdot4=108\)

\(\Rightarrow3^x=108:4\)

\(\Rightarrow3^x=27\)

\(\Rightarrow3^x=3^3\)

\(\Rightarrow x=3\)

là m hả T.kim !!!!