Giải:

\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-3\right)\left(x-2\right)}=\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-1\right)\left(x-4\right)}\)

ĐKXĐ: \(x\ne\left\{1;2;3;4\right\}\)

\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-3\right)\left(x-2\right)}=\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-1\right)\left(x-4\right)}\)

\(\Rightarrow\left(x-3\right)\left(x-4\right)+\left(x-1\right)\left(x-4\right)=\left(x-1\right)\left(x-2\right)+\left(x-2\right)\left(x-3\right)\)

\(\Leftrightarrow\left(x-4\right)\left[\left(x-3\right)+\left(x-1\right)\right]=\left(x-2\right)\left[\left(x-1\right)+\left(x-3\right)\right]\)

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

\(\Leftrightarrow0x=2\)

Vậy ...

\(48x\left(x+1\right)\left(x^3-4\right)=\left(x^4+8x+12\right)^2\)

\(\Leftrightarrow4\left(12x+12\right)\left(x^4-4x\right)=\left(x^4+8x+12\right)^2\)

Đặt \(\left\{{}\begin{matrix}x^4-4x=a\\12x+12=b\end{matrix}\right.\)

\(\Rightarrow4ab=\left(a+b\right)^2\)

\(\Leftrightarrow4ab=a^2+a^2+2ab\)

\(\Leftrightarrow\left(a-b\right)^2=0\)

\(\Leftrightarrow a-b=0\)

\(\Leftrightarrow x^4-16x-12=0\)

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

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

\(\Rightarrow x=1\pm\sqrt{3}\)

\(\left(x+4\right)^2\)nhấn lộn.mn giúp đỡ

\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x^2+\frac{1}{x^2}+2\right)=\left(x+4\right)^2\)

\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2-8\left(x^2+\frac{1}{x^2}\right)=\left(x+2\right)^2\)

\(\Leftrightarrow8\left(x^2+\frac{1}{x^2}+2\right)-8\left(x^2+\frac{1}{x^2}\right)=\left(x+2\right)^2\)

\(\Leftrightarrow\left(x+2\right)^2=16\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

ĐK: \(x\in R\backslash\left\{-4,-3,-2,-1\right\}\)

PT ban đầu

\(\Leftrightarrow\frac{x+2-x-1}{\left(x+1\right)\left(x+2\right)}+\frac{x+3-x-2}{\left(x+2\right)\left(x+3\right)}+\frac{x+4-x-3}{\left(x+3\right)\left(x+4\right)}+\frac{x+5-x-4}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+1}-403\\ \Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}=\frac{1}{x+1}-403\\ \Leftrightarrow\frac{1}{x+5}=403\\ \Leftrightarrow x+5=\frac{1}{403}\Leftrightarrow x=\frac{-2014}{403}\)

Chúc bạn học tốt nha.

a.

\(\left\{{}\begin{matrix}x^4+y^4=34\\y=2-x\end{matrix}\right.\)

\(\Rightarrow x^4+\left(x-2\right)^4=34\)

Đặt \(x-1=t\)

\(\Rightarrow\left(t+1\right)^4+\left(t-1\right)^4=34\)

\(\Leftrightarrow t^4+6t^2-16=0\Rightarrow\left[{}\begin{matrix}t^2=2\\t^2=-8\left(loại\right)\end{matrix}\right.\)

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

b.

\(\left\{{}\begin{matrix}xy^2-x^2y+6x-y^2-y-6=0\\x^2y-xy^2+6y-x^2-x-6=0\end{matrix}\right.\) (1)

Lần lượt cộng 2 vế và trừ 2 vế ta được:

\(\left\{{}\begin{matrix}-x^2-y^2+5x+5y-12=0\\2xy\left(y-x\right)+7\left(x-y\right)+\left(x-y\right)\left(x+y\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2-5\left(x+y\right)+12=0\\\left(y-x\right)\left(2xy-x-y-7\right)=0\end{matrix}\right.\)

Th1: \(\left\{{}\begin{matrix}x=y\\x^2+y^2-5\left(x+y\right)+12=0\end{matrix}\right.\)

\(\Rightarrow2x^2-10x+12=0\Rightarrow...\)

TH2: \(\left\{{}\begin{matrix}2xy-\left(x+y\right)-7=0\\x^2+y^2-5\left(x+y\right)+12=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2xy-\left(x+y\right)-7=0\\\left(x+y\right)^2-2xy-5\left(x+y\right)+12=0\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2v-u-7=0\\u^2-2v-5u+12=0\end{matrix}\right.\)

\(\Rightarrow u^2-6u+5=0\)

\(\Leftrightarrow...\)

Chú ý:

\(\left(x^2+2x\right)^2+4\left(x+1\right)^2=\left(x^2+2x\right)^2+4\left(x^2+2x+1\right)=\left(x^2+2x\right)^2+4\left(x^2+2x\right)+4\)

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

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

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

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

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

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

èo =))

ĐKXĐ: \(x\ne0\)

Ta có \(\left(x+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2\)

Đặt \(x^2+\frac{1}{x^2}=a\Rightarrow\left(x+\frac{1}{x}\right)^2=a+2\) pt trở thành:

\(8\left(a+2\right)+4a^2-4a\left(a+2\right)=\left(x+4\right)^2\)

\(\Leftrightarrow8a+16+4a^2-4a^2-8a=\left(x+4\right)^2\)

\(\Leftrightarrow\left(x+4\right)^2=16\)

\(\Rightarrow\left[{}\begin{matrix}x+4=4\\x+4=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=-8\end{matrix}\right.\)

\(\left(x^2+x+1\right)^2=3\left(x^4+x^2+1\right)\)

\(\Leftrightarrow\left(x^2+x+1\right)^2=3\left[\left(x^4+2x^2+1\right)-x^2\right]\)

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

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

Ta thấy \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\) nên phương trình đã cho tương đương:

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

\(\Leftrightarrow2x^2-4x+2=0\)

\(\Leftrightarrow2\left(x-1\right)^2=0\Leftrightarrow x=1\)

Vậy phương trình đã cho có nghiệm duy nhất là x=1.

=>(x^2+x+1)^2=3[x^4+2x^2+1-x^2]

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

=>(x^2+x+1)(3x^2-3x+3-x^2-x-1)=0

=>(x^2+x+1)(2x^2-4x+2)=0

=>x=1