Toán 9 Giải đề

3.1
[tex]a^3+b^3+c^3-3abc=0\Leftrightarrow (a+b+c)\left [ (a-b)^2+(b-c)^2+(c-a)^2 \right ]=0\Leftrightarrow a=b=c\Rightarrow P=2018^3[/tex]
3.2
[tex]\sqrt{3x+y}-2\sqrt{y}+y(x-y)=0\Leftrightarrow \frac{3(x-y)}{\sqrt{3x+y}+2\sqrt{y}}+y(x-y)=0\Leftrightarrow x=y[/tex]
Thay xuống dưới:
[tex]x^3-5x^2+8x+2=\sqrt{5x-6}+\sqrt{9x-2}\Leftrightarrow x^3-5x^2+6x+(x-\sqrt{5x-6})+(x+2-\sqrt{9x-2})=0[/tex]
Đến đây bạn tự giải nốt
5.
[tex]GT\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{2}[/tex]
[tex]P=\sum \frac{2a^2}{2\sqrt{(a+1)(a^2-a+1)}+2}\geq \sum \frac{2a^2}{a^2+4}=6-8\sum \frac{1}{a^2+4}\geq 6-8\left ( \frac{1}{4a}+\frac{1}{4b}+\frac{1}{4c} \right )=3[/tex]
Dấu "=" xảy ra khi a=b=c=2

3.1.
[tex]a^3+b^3+c^3-3abc=0\Leftrightarrow (a+b+c).\frac{1}{2}[(a-b)^2+(b-c)^2+(c-a)^2]=0[/tex]
Mà [tex]a+b+c\neq 0\Rightarrow (a-b)^2+(b-c)^2+(c-a)^2=0\Rightarrow a=b=c[/tex]
3.2.
[tex]\sqrt{3x+y}+xy=y^2+2\sqrt{y}\Rightarrow \sqrt{3x+y}-2\sqrt{y}+xy-y^2=0\Rightarrow \frac{3(x-y)}{\sqrt{3x+y}+2\sqrt{y}}+y(x-y)=0\Rightarrow (x-y)(\frac{3}{\sqrt{3x+y}+2\sqrt{y}}+y)=0 \Rightarrow x=y[/tex]
Từ phương trình 2 ta có: [tex]x^3-5x^2+8x+2=\sqrt{5x-6}+\sqrt{9x-2}\Rightarrow (x-2)(x^2-3x)+x-\sqrt{5x-6}+x+2-\sqrt{9x-2}=0\Rightarrow x(x^2-5x+6)+\frac{x^2-5x+6}{x+\sqrt{5x-6}}+\frac{x^2-5x+6}{x+2+\sqrt{9x-2}}=0\Rightarrow (x^2-5x+6)(x+\frac{1}{x+\sqrt{5x-6}}+\frac{1}{x+2+\sqrt{9x-2}})=0[/tex]