\(6x^2-3x-9=0\)

<=> \(6x^2+6x-9x-9=0\)

<=> \(6x.\left(x+1\right)-9.\left(x+1\right)=0\)

<=> \(\left(x+1\right).\left(6x-9\right)=0\)

<=> \(\orbr{\begin{cases}x+1=0\\6x-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{3}{2}\end{cases}}}\)

\(6x^2-3x-9=0\)

\(6x^2-9x+6x-9=0\)

\(\left(6x^2-9x\right)+\left(6x-9\right)=0\)

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

\(\left(2x-3\right)\left(3x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-3=0\\3x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=3\\3x=3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=1\end{cases}}\)

\(P\left(x\right)=3x^2-6x=0\)

            \(=3xx-6x=0\)

            \(=x\left(3x-6\right)=0\)

\(\Rightarrow\hept{\begin{cases}x=0\\3x-6=0\Rightarrow x=2\end{cases}}\)

Vậy P(x) có 2 nghiệm là 0 và 2

giúp mình vs mn ơi

Ta có f(x) = x(2 - 3x) + 3x² - 5x + 9 

= 2x - 3x² + 3x² - 5x + 9

= 9 - 3x

b) f(x) có nghiệm <=> 9 - 3x = 0 

<=> 9 = 3x

<=> x = 3

Vậy x = 3 là nghiệm của f(x) 

a) \(\left(3x^2+5x-3\right)+\left(x-3x^2-3\right)=0\)

\(\Leftrightarrow6x-6=0\)

\(\Leftrightarrow6x=6\Leftrightarrow x=1\)

b) \(\left(3x^2-5x\right)-\left(3x^2+x-12\right)=0\)

\(\Leftrightarrow3x^2-5x-3x^2-x+12=0\)

\(\Leftrightarrow-6x=-12\Leftrightarrow x=2\)

b: Ta có: \(\left(3x^2-5x\right)-\left(3x^2+x-12\right)=0\)

\(\Leftrightarrow12-6x=0\)

hay x=2

Mình nghĩ ra câu C rồi bạn nào giúp mình nghĩ nốt câu A,B hộ mình nhé mình cảm ơn!

a:6x-5-9x^2

=-(9x^2-6x+5)

=-(9x^2-6x+1+4)

=-(3x-1)^2-4<=-4

=>A>=2/-4=-1/2

Dấu = xảy ra khi x=1/3

b: \(B=\dfrac{4x^2-6x+4-1}{2x^2-3x+2}=2-\dfrac{1}{2x^2-3x+2}\)

2x^2-3x+2=2(x^2-3/2x+1)

=2(x^2-2*x*3/4+9/16+7/16)

=2(x-3/4)^2+7/8>=7/8

=>-1/2x^2-3x+2<=-1:7/8=-8/7

=>B<=-8/7+2=6/7

Dâu = xảy ra khi x=3/4

a) = x³ + 9x² + 27x + 27 - 9x³ -6x² - x + 8x³ +1 -3x² =54

26x +28 = 54

26x = 54-28 = 26

x = 1

b) = x³ - 9x² + 27x -27 - x³ +27 +6x² + 12x + 6 +3x² = -33

39x +6 = -33

39x = -33-6 = -39

x = -1

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

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

\(\Leftrightarrow\left(x+3\right)\left(\left(x-3\right)\left(x+2\right)-1\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-x-6-1\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\x^2-x-7=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{1\pm\sqrt{29}}{2}\end{cases}}\)

b)\(x^4-6x^2+4x=0\)

\(\Leftrightarrow x\left(x^3-6x+4\right)=0\)

\(\Leftrightarrow x\left[x^3+2x^2-2x-2x^2-4x+4\right]=0\)

\(\Leftrightarrow x\left[x\left(x^2+2x-2\right)-2\left(x^2+2x-2\right)\right]=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x^2+2x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0;x=2\\x=\pm\sqrt{3}-1\end{cases}}\)

c)\(\sqrt{x^2-3x+3}+\sqrt{x^2-3x+6}=3\)

Đặt \(a=\sqrt{x^2-3x+3}>0\Rightarrow a^2+3=x^2-3x+6\)

\(pt\Leftrightarrow a+\sqrt{a^2+3}=3\)\(\Leftrightarrow\sqrt{a^2+3}=3-a\)

\(\Leftrightarrow a^2+3=a^2-6a+9\)

\(\Leftrightarrow6a-6=0\Leftrightarrow6\left(a-1\right)=0\Rightarrow a=1\) (thỏa)

\(\sqrt{x^2-3x+3}=1\)\(\Rightarrow x^2-3x+3=1\)

\(\Rightarrow x^2-3x+2=0\Rightarrow\left(x-2\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\) (thỏa)