Lời giải:

Đặt $(x-3)^2=a$. Khi đó pt đã cho tương đương với:

$(x^2-6x+9-9)^2+13(x-3)^2-77=0$

$\Leftrightarrow [(x-3)^2-9]^2+13(x-3)^2-77=0$

$\Leftrightarrow (a-9)^2+13a-77=0$

$\Leftrightarrow a^2-5a+4=0$

$\Leftrightarrow (a-1)(a-4)=0$

$\Leftrightarroe a=1$ hoặc $a=4$

Đến đây thì đơn giản rồi.

Với \(x=0\) không phải nghiệm

Với \(x\ne0\) chia 2 vế cho \(x^2\) ta được:

\(x^2+x-4+\dfrac{1}{x}+\dfrac{1}{x^2}=0\)

\(\Leftrightarrow\left(x^2+\dfrac{1}{x^2}+2\right)+x+\dfrac{1}{x}-6=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2+x+\dfrac{1}{x}-6=0\)

Đặt \(x+\dfrac{1}{x}=t\)

\(\Rightarrow t^2+t-6=0\)

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

\(\Rightarrow\left[{}\begin{matrix}t=2\\t=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=2\\x+\dfrac{1}{x}=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2+1=2x\\x^2+1=-3x\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-2x+1=0\\x^2+3x+1=0\end{matrix}\right.\) (bấm máy)

    x⁴ - 3x² +1

= x⁴ - 2x² + 1 - x²

= ( x² - 1 )² - x²

= ( x²- 1 +x ) ( x² - 1 - x ) 

umk, mình biết rồi. Tại đọc nhầm đề ấy mà

\(2x^3+7x^2+7x+2=0\)

\(\Leftrightarrow\left(2x^3+4x^2\right)+\left(3x^2+6x\right)+\left(x+2\right)=0\)

\(\Leftrightarrow2x^2\left(x+2\right)+3x\left(x+2\right)+\left(x+2\right)=0\)

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

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

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

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\(x^3-8x^2-8x+1=0\)

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

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cảm ơn nha 

Lời giải:
Đặt $\sqrt[3]{x+1}=a;\sqrt[3]{x-1}=b$ thì pt trở thành:

\(\left\{\begin{matrix} a^2+b^2+ab=1\\ a^3-b^3=2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a^2+ab+b^2=1\\ (a-b)(a^2+ab+b^2)=2\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} a^2+ab+b^2=1\\ a-b=2\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} (a-b)^2+3ab=1\\ a-b=2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a(-b)=1\\ a+(-b)=2\end{matrix}\right.\)

Theo đl Viet đảo thì $a,-b$ là nghiệm của pt $X^2-2X+1=0$

$\Rightarrow a=-b=1$

$\Leftrightarrow \sqrt[3]{x+1}=1; \sqrt[3]{x-1}=-1$

$\Rightarrow x=0$

Vậy.........

1.

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y+x^3y+xy^2+xy=-\dfrac{5}{4}\\x^4+y^2+xy\left(1+2x\right)=-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+y\right)+xy+xy\left(x^2+y\right)=-\dfrac{5}{4}\\\left(x^2+y\right)^2+xy=-\dfrac{5}{4}\end{matrix}\right.\left(1\right)\)

Đặt \(\left\{{}\begin{matrix}x^2+y=a\\xy=b\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}a+b+ab=-\dfrac{5}{4}\\a^2+b=-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-a^2-\dfrac{5}{4}-a\left(a^2+\dfrac{5}{4}\right)=-\dfrac{5}{4}\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^2-a^3-\dfrac{1}{4}a=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-a\left(a^2-a+\dfrac{1}{4}\right)=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a\left(a-\dfrac{1}{2}\right)^2=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=0\\b=-\dfrac{5}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=-\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}a=0\\b=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y=0\\xy=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{\sqrt[3]{10}}{2}\\y=-\dfrac{5}{2\sqrt[3]{10}}\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y=\dfrac{1}{2}\\xy=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-\dfrac{3}{2}\end{matrix}\right.\)

Kết luận: Phương trình đã cho có nghiệm \(\left(x;y\right)\in\left\{\left(\dfrac{\sqrt[3]{10}}{2};-\dfrac{5}{2\sqrt[3]{10}}\right);\left(1;-\dfrac{3}{2}\right)\right\}\)

2.

\(\left\{{}\begin{matrix}\left(x+1\right)^3-16\left(x+1\right)=\left(\dfrac{2}{y}\right)^3-4\left(\dfrac{2}{y}\right)\\1+\left(\dfrac{2}{y}\right)^2=5\left(x+1\right)^2+5\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+1=u\\\dfrac{2}{y}=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u^3-16u=v^3-4v\\v^2=5u^2+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u^3-v^3=16u-4v\\4=v^2-5u^2\end{matrix}\right.\)

\(\Rightarrow4\left(u^3-v^3\right)=\left(16u-4v\right)\left(v^2-5u^2\right)\)

\(\Leftrightarrow21u^3-5u^2v-4uv^2=0\)

\(\Leftrightarrow u\left(7u-4v\right)\left(3u+v\right)=0\Rightarrow\left[{}\begin{matrix}u=0\Rightarrow v^2=4\\u=\dfrac{4v}{7}\Rightarrow4=v^2-5\left(\dfrac{4v}{7}\right)^2\\v=-3u\Rightarrow4=\left(-3u\right)^2-5u^2\end{matrix}\right.\) 

\(\Rightarrow...\)

Mình ko biết đặt biến phụ nên mình sẽ giải bừa :>

\(x^4+4x^3+6x^2+4x+1=0\)

\(\Leftrightarrow x^4+2x^3+x^2+2x^3+4x^2+2x+x^2+2x+1=0\)

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

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

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

Thấy ngay x= 0 không phải là nghiệm của pt. Chia 2 vế của pt cho x² ta được:

\(x^2+4x+6+4.\frac{1}{x}+\frac{1}{x^2}=0\)

\(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)+4\left(x+\frac{1}{x}\right)+6=0\left(1\right)\)

Đặt \(x+\frac{1}{x}=t\Rightarrow\left(x+\frac{1}{x}\right)^2=t^2\Rightarrow x^2+\frac{1}{x^2}=t^2-2\) Khi đó ta có:

\(\left(1\right)\Leftrightarrow t^2-2+4t+6=0\)

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

Vậy pt có 1 nghiệm x = -1