I do not know what to do.

a) Ta có: \(\left(x+1\right)^4+\left(x-3\right)^4=0\)

Nhận thấy: \(\hept{\begin{cases}\left(x+1\right)^4\ge0\left(\forall x\right)\\\left(x-3\right)^4\ge0\left(\forall x\right)\end{cases}\Rightarrow}\left(x+1\right)^4+\left(x-3\right)^4\ge0\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x+1\right)^4=0\\\left(x-3\right)^4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\x=3\end{cases}}\) (mâu thuẫn)

=> pt vô nghiệm

b) \(x^4+2x^3-4x^2-5x-6=0\)

\(\Leftrightarrow\left(x^4-2x^3\right)+\left(4x^3-8x^2\right)+\left(4x^2-8x\right)+\left(3x-6\right)=0\)

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

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

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

Mà \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\left(\forall x\right)\)

=> \(\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

a,\(\left(x+1\right)^4+\left(x-3\right)^4=0\)

\(x^4-1+x^4-81=0\)

\(2x^4-82=0\)

\(2x^4=82\)

\(x^4=41\)

\(x=\sqrt[4]{41}\)

\(\Rightarrow\)vô nghiệm

\(x^2+2>0\Rightarrow4x+6=0\Leftrightarrow x=-\frac{3}{2}\)

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

\(\iff 4x+6=0 \) hoặc \(x^{2}+2=0\)

\(\iff 4x=6\) hoặc \(x^{2}\) =-2 (loại, vì \(x^{2}>0\) )

\(\iff\) x=\(\dfrac{3}{2}\)

x² - 3x - 2x +6 = x(x - 3) - 2(x - 3)

                  =(x - 3)(x - 2)

suy ra ta tìm được nghiệm của pt là x= 3 hoặc x=2

x=3;-0,5;-2

1. \(\left(2x+5\right)^2=\left(x+2\right)^2\Leftrightarrow\left(2x+5+x+2\right)\left(2x+5-x-2\right)=0\)\(\Leftrightarrow\left(3x+7\right)\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=-\frac{7}{3}\\x=-3\end{cases}}\)
2. \(x^2-5x+6=0\Leftrightarrow x^2-6x+x-6=0\Leftrightarrow x\left(x-6\right)+\left(x-6\right)=0\)\(\left(x+1\right)\left(x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=6\\x=-1\end{cases}}\)
3. \(2x^3+6x^2=x^2+3x\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)\(\Leftrightarrow\left(x+3\right)\left(2x^2-x\right)=0\Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\Leftrightarrow\)\(x=0\)hoặc \(x=\frac{1}{2}\)hoặc \(x=-3\)

\(\dfrac{x+2}{2016}+\dfrac{x+3}{2015}+\dfrac{x+4}{2014}+\dfrac{x+2036}{6}=0\)

<=>\(\dfrac{x+2}{2016}+1+\dfrac{x+3}{2015}+1+\dfrac{x+4}{2014}+1+\dfrac{x+2036}{6}-3=0\)

<=>\(\dfrac{x+2018}{2016}+\dfrac{x+2018}{2015}+\dfrac{x+2018}{2014}+\dfrac{x+2018}{6}=0\)

<=>\(\left(x+2018\right)\left(\dfrac{1}{2016}+\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{6}\right)=0\)

vì 1/2016+1/2015+1/2014+1/6 khác 0

=>x+2018=0<=>x=-2018

vậy...................

chúc bạn học tốt ^ ^