pt<=>\(\sqrt{\left(x+6\right)^3}+\sqrt{x+6}=\left(x^2+4x\right)^3+x^2+4x\)

đặt\(\sqrt{x+6}=a;x^2+4x=b\)

a) Ta có: \(x^3-9x^2+19x-11=0\)

\(\Leftrightarrow x^3-x^2-8x^2+8x+11x-11=0\)

\(\Leftrightarrow x^2\left(x-1\right)-8x\left(x-1\right)+11\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-8x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2-8x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\sqrt{5}+4\\x=-\sqrt{5}+4\end{matrix}\right.\)

Vậy: \(S=\left\{1;\sqrt{5}+4;-\sqrt{5}+4\right\}\)

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

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

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)

\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

\(a,9\left(2x+1\right)=4\left(x-5\right)^2\)

\(4x^2-40x+100=18x+9\)

\(4x^2-58x+91=0\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{29+3\sqrt{53}}{4}\\x=\frac{29-3\sqrt{53}}{4}\end{cases}}\)

\(b,x^3-4x^2-12x+27=0\)

\(\left(x+3\right)\left(x^2-7x+9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\x^2-7x+9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{7\pm\sqrt{13}}{2}\end{cases}}}\)

\(c,x^3+3x^2-6x-8=0\)

\(\left(x+4\right)\left(x-2\right)\left(x+1\right)=0\)

\(Th1:x+4=0\Leftrightarrow x=-4\)

\(Th2:x-2=0\Leftrightarrow x=2\)

\(Th3:x+1=0\Leftrightarrow x=-1\)

\(a,9.\left(2x+1\right)=4.\left(x-5\right)^2\)

\(< =>4x^2-40x+100=18x+9\)

\(< =>4x^2+58x+91=0\)

\(< =>\orbr{\begin{cases}x=\frac{29-3\sqrt{53}}{4}\\x=\frac{29+3\sqrt{53}}{4}\end{cases}}\)

\(b,x^3-4x^2-12x+27=0\)

\(< =>\left(x+3\right)\left(x^2-7x+9\right)=0\)

\(< =>\orbr{\begin{cases}x+3=0\\x^2-7x+9=0\end{cases}}\)

\(< =>\orbr{\begin{cases}x=-3\\x=\frac{7\pm\sqrt{13}}{2}\end{cases}}\)

\(x^4-9x^2+24x-16=\)\(0\)

\(\Leftrightarrow x^4-\left(9x^2-24x+16\right)=0\)

\(\Leftrightarrow x^4-\left(3x-4\right)^2=0\)

\(\Leftrightarrow\left(x^2+3x-4\right)\left(x^2-3x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-1\right)\left[\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\right]=0\)

Vì \(\left(x-\frac{3}{2}\right)^2+\frac{7}{4}>0\forall x\)nên:

\(\left(x+4\right)\left(x-1\right)=0:\left[\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\right]\)

\(\Leftrightarrow\left(x+4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-4\\x=1\end{cases}}\)

Vậy phương trình có tập nghiệm \(S=\left\{1;-4\right\}\)

\(x^4=6x^2+12x+\)\(8\)

\(\Leftrightarrow x^4-2x^2+1=4x^2+12x+9\)

\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)

\(\Leftrightarrow|x^2-1|=|2x+3|\)\(|\)

xét các trường hợp:

- Trường hợp 1:

\(x^2-1=2x+3\)

\(\Leftrightarrow x^2-1-2x-3=0\)

\(\Leftrightarrow x^2-2x-4=0\)

\(\Leftrightarrow\left(x-1\right)^2-5=0\Leftrightarrow\left(x-1\right)^2=5\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=\sqrt{5}\\x-1=-\sqrt{5}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1+\sqrt{5}\\x=1-\sqrt{5}\end{cases}}}\)

-Trường hợp 2:

\(x^2-1=-2x-3\)

\(\Leftrightarrow x^2-1+2x+3=0\)

\(\Leftrightarrow x^2+2x+2=0\)

\(\Leftrightarrow\left(x+1\right)^2+1=0\)

\(\Leftrightarrow\left(x+1\right)^2=-1\left(vn\right)\)(vô nghiệm)

Vậy phương trình có tập nghiệm: \(S=\left\{1\pm\sqrt{5}\right\}\)

\(x^4=4x+1\)

\(\Leftrightarrow x^4+2x^2+1=2x^2+4x+2\)

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

\(\Leftrightarrow|x^2+1|=|x\sqrt{2}+\sqrt{2}|\)

Xét các trường hợp sau:

-Trường hợp 1:

\(x^2+1=x\sqrt{2}+\sqrt{2}\)

\(\Leftrightarrow x^2+1-x\sqrt{2}-\sqrt{2}=0\)

\(\Leftrightarrow\left(x^2-2x.\frac{\sqrt{2}}{2}+\frac{1}{2}\right)-\frac{2\sqrt{2}-1}{2}=0\)

\(\Leftrightarrow\left(x-\frac{1}{\sqrt{2}}\right)^2=\frac{2\sqrt{2}-1}{2}\)

Vì \(\frac{2\sqrt{2}-1}{2}>0\)nên:

\(\left|x-\frac{1}{\sqrt{2}}\right|=\left|\sqrt{\frac{2\sqrt{2}-1}{2}}\right|\)

Lại xét các trường hợp:

+Trường hợp 1.1:

\(x-\frac{1}{\sqrt{2}}=\frac{\sqrt{2\sqrt{2}-1}}{\sqrt{2}}\)\(\Leftrightarrow x=\frac{\sqrt{2\sqrt{2}-1}+1}{\sqrt{2}}\)

+Trường hợp 1.2:

\(x-\frac{1}{\sqrt{2}}=\frac{\sqrt{2\sqrt{2}-1}}{\sqrt{2}}\Leftrightarrow x=\frac{1-\sqrt{2\sqrt{2}-1}}{\sqrt{2}}\)

-Trường hợp 2:

\(x^2+1=-x\sqrt{2}-\sqrt{2}\)(2)

\(\Leftrightarrow x^2+1+x\sqrt{2}+\sqrt{2}=0\)

\(\Leftrightarrow\left(x^2+2x.\frac{\sqrt{2}}{2}+\frac{1}{2}\right)+\frac{1+2\sqrt{2}}{2}=0\)

\(\Leftrightarrow\left(x+\frac{1}{\sqrt{2}}\right)^2=\frac{-1-2\sqrt{2}}{2}\)(vô nghiệm)

Do đó phương trình (2) vô nghiệm.

Vậy phương trình có tập nghiệm : \(S=\left\{\frac{1\pm\sqrt{2\sqrt{2}-1}}{\sqrt{2}}\right\}\)

\(6x^4+25x^3+12x^2-25x+6=0.\)

\(\Leftrightarrow\left(2x^2+x-2\right)\left(3x^2+8x-3\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x+2\right)\left(x+3\right)\left(3x-1\right)=0\)

x²  - 1 + 3(6x + 6)=0