(x+9)(x+10)(x+11) -8x =0

<=>(x²+19x+90)(x+1)-8x=0

<=>x³+30x²+299x+990-8x=0

<=>(x+15)(x²+15x+66)=0

<=>x+15=0 hoặc x²+15x+66=0 (1)

<=>x=-15. Denta₍₁₎=15²-4(1.66)=-39<0

=>(1) vô nghiệm

Vậy nghiệm duy nhất thỏa mãn là x=-15

Ta có \(x^3+30x^2+291x+990=0\)
\(\Leftrightarrow\left(x^3+15x^2\right)+\left(15x^2+225x\right)+\left(66x+990\right)=0\)
\(\Leftrightarrow\left(x+15\right)\left(x^2+15x+66\right)=0\)
mà \(x^2+15x+66=x^2+2.x.\dfrac{15}{2}+\dfrac{225}{4}+\dfrac{39}{4}=\left(x+\dfrac{15}{2}\right)^2+\dfrac{39}{4}\ge\dfrac{39}{4}>0\)
=>x+15=0
\(\Leftrightarrow x=-15\)

mk lưu nhầm ảnh ở bài dưới của câu

\(\dfrac{8}{x}-8+\dfrac{11}{x}-11=\dfrac{9}{x}-9+\dfrac{10}{x}-10\)\(\Leftrightarrow\dfrac{8}{x}+\dfrac{11}{x}-\dfrac{9}{x}-\dfrac{10}{x}=8+11-9-10\)

\(\Leftrightarrow\dfrac{8+11-9-10}{x}=0\)

\(\Leftrightarrow\dfrac{0}{x}=0\)

\(\Leftrightarrow x=0\)

S=\(\left\{0\right\}\)

Có : \(2x^2+3x+2\)

\(\Leftrightarrow\) \(\left(x^2+2x+1^2\right)+\left(x^2+x+1^2\right)\)

\(\Leftrightarrow\) \(\left(x^2+2.x.1+1^2\right)\) + \(\left(x^2+2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}\right)\)

\(\Leftrightarrow\) \(\left(x+1\right)^2+\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(x+1\right)^2\ge0và\left(x+\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow\) \(\left(x+1\right)^2+\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

Vậy \(2x^2+3x+2>0\left(\forall_x\right)\)

x² - 8x - 9  ≥ 0

<=> (x+1)(x-9)\(\ge\)0

<=> \(\hept{\begin{cases}x+1\ge0\\x-9\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x+1\le0\\x-9\le0\end{cases}}\)

<=> \(\orbr{\begin{cases}x\ge9\\x\le-1\end{cases}}\)