Ta có : \(\sqrt{3x^2-18x+28}=\sqrt{3\left(x^2-6x+9\right)-27+28}=\sqrt{3\left(x-3\right)^2+1}\ge1\)

\(\sqrt{4x^2-24x+45}=\sqrt{4\left(x^2-6x+9\right)-36+45}=\sqrt{4\left(x-3\right)^2+9}\ge\sqrt{9}=3\)

=> VT >= 1 + 3 = 4 

VP = \(6x-x^2-5=-\left(x^2-6x+9\right)+9-5=-\left(x-3\right)^2+4\le4\)

Vậy VT = VP = 4 

Dấu = xảy ra khi x = 3 

Vậy x = 3 là n* của pt 

\(ĐK:x\ge0;x\ne4\\ P=\dfrac{5x+10\sqrt{x}-\left(3-\sqrt{x}\right)\left(\sqrt{x}-2\right)-6x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{5x+10\sqrt{x}-5\sqrt{x}+6+x-6x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{5\sqrt{x}+6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(P=\dfrac{5\sqrt{x}}{\sqrt{x}-2}-\dfrac{3-\sqrt{x}}{\sqrt{x}+2}+\dfrac{6x}{4-x}\left(đk:x\ge0,x\ne4\right)\)

\(=\dfrac{5\sqrt{x}\left(\sqrt{x}+2\right)-\left(3-\sqrt{x}\right)\left(\sqrt{x}-2\right)-6x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{5x+10\sqrt{x}+x-5\sqrt{x}+6-6x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{5\sqrt{x}+6}{x-4}\)

\(\Leftrightarrow\sqrt{2\left(x^2-1\right)^2+9}+\sqrt{3\left(x-1\right)^2+25}=-3\left(x-1\right)^2+8\)

Ta có:

\(\left\{{}\begin{matrix}\sqrt{2\left(x^2-1\right)^2+9}+\sqrt{3\left(x-1\right)^2+25}\ge\sqrt{9}+\sqrt{25}=8\\-3\left(x-1\right)^2+8\le8\end{matrix}\right.\)

\(\Rightarrow\sqrt{2\left(x^2-1\right)^2+9}+\sqrt{3\left(x-1\right)^2+25}\ge-3\left(x-1\right)^2+8\)

Đẳng thức xảy ra khi và chỉ khi \(x=1\)

\(3x+3+\sqrt{x^3-x+1}-1=0\)

\(3\left(x+1\right)+\frac{x\left(x+1\right)\left(x-1\right)}{\sqrt{x^3-x+1}}=0\)

\(\left(x+1\right)\left(3+\frac{x\left(x-1\right)}{\sqrt{x^3-x+1}}\right)=0\)

Đk :\(-1\le x\le0,x\ge1\)

Kết hợp điều kiện ta được \(\left(3+\frac{x\left(x-1\right)}{\sqrt{x^3-x+1}}\right)\ge0\)

vậy  x  =  - 1

x,y là số  nguyên tố đúng ko? bn có nhiueeuf câu hỏi nên mik trả lời nhầm.(ko phait thì thui nhé)

\(\left(3x^2+6x+3\right)+\left(3y^2+3y+1\right)+y^2-8=0\)

\(\Leftrightarrow3\left(x+1\right)^2+3\left(y+\frac{1}{2}\right)^2-\frac{9}{4}-8=0\)

\(\Leftrightarrow12\left(x+1\right)^2+3\left(y+1\right)^2=41\)

\(\Rightarrow12\left(x+1\right)^2\le41\Rightarrow\left(x+1\right)^2\le3\Rightarrow x+1\in\left\{1;0;-1\right\}\Rightarrow x\in\left\{0;-1;-2\right\}\)

Bạn làm nốt

bạn chỉ cần cố gắng là làm được