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\(ĐKXĐ:x\ge-1,5\)
\(=>\left(2\sqrt{2x^3+5x^2+9x+9}\right)^2=\left(x^2+3x+6\right)^2\)
=>\(8x^3+20x^2=x^4+6x^3+21x^2\) ( Đã đc rút gọn )
=> \(x^4+6x^3+21x^2-\left(8x^3+20x^2\right)=0\)
=> \(x^4-2x^3+x^2=0\)
=> \(x^2\left(x-1\right)^2=0\)
=> \(\left[{}\begin{matrix}x^2=0\\\left(x-1\right)^2=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left|x\right|=\sqrt{0}\\\left|x-1\right|=\sqrt{0}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy....
ĐKXĐ:...
\(\sqrt{3x^2-5x-1}-\sqrt{3x^2-7x+9}+\sqrt{x^2-2}-\sqrt{x^2-3x+13}=0\)
\(\Leftrightarrow\frac{2\left(x-5\right)}{\sqrt{3x^2-5x-1}+\sqrt{3x^2-7x+9}}+\frac{3\left(x-5\right)}{\sqrt{x^2-2}+\sqrt{x^2-3x+13}}=0\)
\(\Leftrightarrow\left(x-5\right)\left(\frac{2}{\sqrt{3x^2-5x-1}+\sqrt{3x^2-7x+9}}+\frac{3}{\sqrt{x^2-2}+\sqrt{x^2-3x+13}}\right)=0\)
\(\Leftrightarrow x-5=0\) (ngoặc to phía sau luôn dương)
\(\Rightarrow x=5\)
ĐK: \(x^3+3x^2-3x+1\ge0\)
\(pt\Leftrightarrow\sqrt[3]{9x^2-15x+9}-\left(2-x\right)+\sqrt{x^3+3x^2-3x+1}=0\)
\(\Leftrightarrow\frac{9x^2-15x+9-\left(2-x\right)^3}{A^2+AB+B^2}+\sqrt{x^3+3x^2-3x+1}=0\)
\(\left(A=\sqrt[3]{9x^2-15x+9};\text{ }B=2-x\right)\)\(\text{(}A^2+AB+B^2=\left(A+\frac{B}{2}\right)^2+\frac{3B^2}{4}>0\text{)}\)
\(\Leftrightarrow\frac{x^3+3x^2-3x+1}{A^2+AB+B^2}+\sqrt{x^3+3x^2-3x+1}=0\)
\(\Leftrightarrow\sqrt{x^3+3x^2-3x+1}\left(\frac{\sqrt{x^3+3x^2-3x+1}}{A^2+AB+B^2}+1\right)=0\)
\(\Leftrightarrow x^3+3x^2-3x+1=0\text{ (do }\frac{\sqrt{x^3+3x^2-3x+1}}{A^2+AB+B^2}+1>0\text{)}\)
\(\Leftrightarrow\left(x+1+\sqrt[3]{2}+\sqrt[3]{4}\right)\left[x^2+\left(2-\sqrt[3]{2}-\sqrt[3]{4}\right)x+\sqrt[3]{2}-1\right]=0\)
\(\Leftrightarrow x+1+\sqrt[3]{2}+\sqrt[3]{4}=0\text{ (}pt\text{ }x^2+\left(2-\sqrt[3]{2}-\sqrt[3]{4}\right)x+\sqrt[3]{2}-1=0\text{ vô nghiệm do }\Delta
\(\left(x^2-3x+9\right)\left(x^2+5x+9\right)=9x^2\)
\(\Leftrightarrow x^4+5x^3+9x^2-3x^3-15x^2-27x+9x^2+45x+81=9x^2\)
\(\Leftrightarrow x^4+2x^3+3x^2+18x+81=9x^2\)
\(\Leftrightarrow x^4+2x^3+3x^2+18x+81-9x^2=0\)
\(\Leftrightarrow x^4+2x^2-6x^2+18x+81=0\)
\(\Leftrightarrow\left(x^3-x^2-3x+27\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x^2-4x+9\right)\left(x+3\right)\left(x+3\right)=0\)
Vì \(x^2-4x+9\ne0\) nên:
\(\Rightarrow x+3=0\)
\(x=-3\)
Vậy: nghiệm phương trình là: {-3}