Ta có: \(C=-3x\left(x-4\right)\left(x-2\right)+x\left(3x-18\right)-25\)

\(=-3x\left(x^2-6x+8\right)+3x^2-18x-25\)

\(=-3x^3+18x^2-24x+3x^2-18x-25\)

\(=-3x^3+21x^2-42x-25\)

láo toét

láo toét

\(\dfrac{x+3}{x^2-3x}+\dfrac{3}{x^2+3x}+\dfrac{2x-18}{x^2-9}\\ =\dfrac{x^2+6x+9+3x-9+2x^2-18x}{x\left(x-3\right)\left(x+3\right)}=\dfrac{3x^2-9x}{x\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)

\(=\dfrac{x+3}{x\left(x-3\right)}+\dfrac{3}{x\left(x+3\right)}+\dfrac{2x-18}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{x^2+6x+9+3x-9+4x-36}{x\left(x+3\right)\left(x-3\right)}=\dfrac{x^2+13x-36}{x\left(x+3\right)\left(x-3\right)}\)

x² + 8x + 7  = x² + x + 7x + 7 = x(x + 1) + 7(x + 1)= (x + 7)(x + 1)

x² - 5x + 6 = x² - 2x - 3x + 6 = x(x - 2) - 3(x - 2) = (x - 3)(x - 2)

x² + 3x  - 18 = x² + 6x - 3x - 18 = x(x + 6) - 3(x + 6) = ((x - 3)(x + 6)

\(a,x^2+8x+7=x^2+7x+x+7=x\left(x+7\right)+\left(x+7\right)=\left(x+7\right)\left(x+1\right).\)

\(b,x^2-5x+6=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)

\(c,x^2+3x-18=x^2+6x-3x-18=x\left(x+6\right)-3\left(x+6\right)=\left(x+6\right)\left(x-3\right)\)

\(d,3x^2-16x+18=3x^2-4x-12x+18\)

trả lời

x^2+8x+7=(x+1)(x+7)

x^2-5x+6=(x-2)(x-3)

x^2+3x-18=(x-3)(x+6)

1) \(\left(x+1\right)\left(x+2\right)-3x\left(x-4\right)=x^2+3x+2-3x^2+12x=-2x^2+15x+2\)

2) \(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)\)

\(\Leftrightarrow3x^2-10x+8=3x^2-27x\)

\(\Leftrightarrow17x=-8\Leftrightarrow x=-\dfrac{8}{17}\)

3) \(-3\left(x-4\right)\left(x-2\right)-x^2\left(-3x+18\right)+24x-25\)

\(=-3x^3+6x^2+12x^2-24x+3x^3-18x^2+24x-25=-25\)

a)  3x – 15 = 25 – 5x 

=> 3x + 5x = 25 + 15

=> 8x = 40

=> x = 5

 b) 3x - 17 = 2x – 7     

=> 3x - 2x = -7 + 17

=> x = 10

 c) 2x – 17 =  – (3x – 18)

=> 2x - 17 = -3x + 18

=> 2x + 3x = 18 + 17

=> 5x = 35

=> x = 7

d) 3x – 14 = 2(x – 9) + 1

=> 3x - 14 = 2x - 18 + 1

=> 3x - 2x = -18 + 1 + 14

=> x = -3

f) (x – 5)² = 9          

\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

a) Ta có: \(3x-15=25-5x\)

\(\Leftrightarrow3x-15-25+5x=0\)

\(\Leftrightarrow8x-40=0\)

\(\Leftrightarrow8x=40\)

hay x=5

Vậy: x=5

b) Ta có: \(3x-17=2x-7\)

\(\Leftrightarrow3x-17-2x+7=0\)

\(\Leftrightarrow x-10=0\)

hay x=10

Vậy: x=10

c) Ta có: \(2x-17=-\left(3x-18\right)\)

\(\Leftrightarrow2x-17=-3x+18\)

\(\Leftrightarrow2x-17+3x-18=0\)

\(\Leftrightarrow5x-35=0\)

\(\Leftrightarrow5x=35\)

hay x=7

Vậy: x=7

d) Ta có: \(3x-14=2\left(x-9\right)+1\)

\(\Leftrightarrow3x-14=2x-18+1\)

\(\Leftrightarrow3x-14-2x+18-1=0\)

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)

Vậy: x=-3

f) Ta có: \(\left(x-5\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

Vậy: \(x\in\left\{2;8\right\}\)

Bài 2. Tìm x, biết :                                

a) \(3x-15=25-5x\)

\(\Leftrightarrow8x=40\)

\(\Leftrightarrow x=5\)

Vậy x = 5

b) \(3x-17=2x-7\)

\(\Leftrightarrow x=10\)

Vậy x = 10

c) \(2x-17=-\left(3x-18\right)\)

\(\Leftrightarrow2x-17=18-3x\)

\(\Leftrightarrow5x=35\)

\(\Leftrightarrow x=7\)

Vậy x = 7

d) \(3x-14=2\left(x-9\right)+1\)

\(\Leftrightarrow3x-14=2x-18+1\)

\(\Leftrightarrow3x-14=2x-17\)

\(\Leftrightarrow x=-3\)

Vậy x = -3

e) \(\left(x-5\right)^2=9\)

\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

Vậy x = {8; 2}

Ta có: \(6x\left(3x+5\right)-2x\left(3x-2\right)+\left(17-x\right)\left(x-1\right)+x\left(x-18\right)=0\)

\(\Leftrightarrow18x^2+30x-6x^2+4x+17x-17-x^2+x+x^2-18x=0\)

\(\Leftrightarrow12x^2-34x-17=0\)

\(\Leftrightarrow12\left(x^2-\frac{34}{12}x-\frac{17}{12}\right)=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\frac{17}{12}+\frac{289}{144}-\frac{493}{144}=0\)

\(\Leftrightarrow\left(x-\frac{17}{12}\right)^2=\frac{493}{144}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{17}{12}=\frac{\sqrt{493}}{12}\\x-\frac{17}{12}=-\frac{\sqrt{493}}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{17+\sqrt{493}}{12}\\x=\frac{17-\sqrt{493}}{12}\end{matrix}\right.\)

Vậy: \(S=\left\{\frac{17+\sqrt{493}}{12};\frac{17-\sqrt{493}}{12}\right\}\)

3x.(x^2 -2) - (3x^3 -18) =0 

suy ra x=0