\(1.đặt:\sqrt[3]{x-9}=y-3\)
\(\Rightarrow y^3-9y^2+27y-27=x-9\Leftrightarrow y^3-9y^2+27y-x=18\left(1\right)\)
\(x^3-9x^2+27x-21=y-3\Leftrightarrow x^3-9x^2+27x-y=18\left(2\right)\)
\(\left(2\right)-\left(1\right)\Rightarrow\left(x-y\right)\left(x^2+xy-9x+y^2-9y+28\right)=0\Leftrightarrow\left[{}\begin{matrix}x=y\left(3\right)\\x^2+xy-9x+y^2-9y+28=0\left(4\right)\end{matrix}\right.\)
\(\left(3\right)thay\left(1\right)\Rightarrow x=1\)
\(\left(4\right)\Leftrightarrow x^2+x\left(y-9\right)+y^2-9y+28=0\Rightarrow\Delta=\left(y-9\right)^2-4\left(y^2-9y+28\right)=-\left(3y^2-18y+31\right)< 0\Rightarrow x;y=\varnothing\)
\(vậy:x=1\)
\(2.\left(đk:x\ge-1\right)\) \(x^3-3x^2-8x+40=8\sqrt[4]{4x+4}=\sqrt{16.16.16.\left(4x+4\right)}\le\dfrac{16+16+16+4x+4}{4}=\dfrac{4x+52}{4}\Rightarrow4\left(x^3-3x^2-8x+40\right)-\left(4x+52\right)\le0\Leftrightarrow4\left(x-3\right)^2\left(x+3\right)\le0\Leftrightarrow\left(x-3\right)^2\left(x+3\right)\le0\left(1\right)\)
\(xét\) \(đk:x\ge-1\Rightarrow x+3\ge2>0\Rightarrow\left(x-3\right)^2\left(x+3\right)\le0\Leftrightarrow\left(x-3\right)^2\le0\Leftrightarrow x=3\)
\(thử:x=3\) \(vào\) \(pt\) \(thấy\) \(thỏa\)
3. ĐKXĐ : \(x\ge-2\)
Đặt \(\sqrt{x+5}=a;\sqrt{x+2}=b\left(a;b\ge0\right)\)
Khi đó phương trình tương đương
\(\left(a-b\right)\left(ab+1\right)=a^2-b^2\)
\(\Leftrightarrow\left(a-b\right)\left(a-1\right)\left(b-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=1\\b=1\end{matrix}\right.\)
Khi a = 1 => x = -4 (loại)
Khi b = 1 => x = -1 (tm)
Khi a = b => \(\sqrt{x+5}=\sqrt{x+2}\Leftrightarrow\left\{{}\begin{matrix}x+5=x+2\\x\ge-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}∄x\\x\ge-2\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
Tập nghiệm phương trình \(S=\left\{-1\right\}\)
