sai đề rồi

dung ma

chỗ cuối là -1 chứ

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

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

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

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

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

vì 1/2016+1/2015+1/2014+1/6 khác 0

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

vậy...................

chúc bạn học tốt ^ ^

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

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

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

Do 2 thừa số ở VT đều > 0

\(\Rightarrow\) PTVN

\(x^4+x^3+3x^2+2x+2=0\\ \Leftrightarrow x^4+x^3+x^2+2x^2+2x+2=0\\ \Leftrightarrow x^2\left(x^2+x+1\right)+2\left(x^2+x+1\right)=0\\ \Leftrightarrow\left(x^2+x+1\right)\left(x^2+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+x+1=0\left(VN\right)\\x^2+2=0\left(VN\right)\end{matrix}\right.\)

Vậy phương trình vô nghiệm

x^3 + x^2 + 4 = -2^3 + 2^2 + 4

                     = 0

\(x^3+x^2+4=0\Leftrightarrow x^3+2x^2-x^2-2x+2x+4=0\Leftrightarrow x^2\left(x+2\right)-x\left(x+2\right)+2\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-x+2\right)=0\)
vì x^2 -x +2 >0 nên \(x+2=0\Rightarrow x=-2\)
Vậy nghiệm phương trình là x=-2

x^3+x^2+4=0

x^3+2x^2-x^2-2x+2x+4=0

x^2(x+2)-x(x+2)+2(x+2)=0

(x+2)(x^2-x+2)=0

\(\Rightarrow\orbr{\begin{cases}x+2=0\\x^2-x+2=0\left(voli\right)\end{cases}\Rightarrow\hept{x=-2}}\)

Vay pt co tap hop nghiem la : \(S=\left\{-2\right\}\)

Vì \(\left(x+1\right)^4\ge0\forall x\); \(\left(x-3\right)^4\ge0\forall x\)

\(\Rightarrow\left(x+1\right)^4+\left(x-3\right)^4\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^4=0\\\left(x-3\right)^4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}\left(ktm\right)}\)

=> Pt vô nghiệm

a)   ( x + 1 ) ⁴  +  ( x - 3 ) ⁴   = 0

Vì \(\left(x+1\right)^4\ge0\forall x\inℤ\)

     \(\left(x-3\right)^4\ge0\forall x\inℤ\)

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

\(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)^4=0\\\left(x-3\right)^4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+1=0\\x-3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-1\\x=3\end{cases}}}\)

Vậy .....

( x + 1 )⁴ + ( x - 3 )⁴ = 0 

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

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x+1=0\\x-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\x=3\end{cases}}\)( mâu thuẫn )

=> Pt vô nghiệm

a, Đặt \(x^2=t\left(t\ge0\right)\)

Khi đó \(PT< =>t^1+4t-5=0\)

\(< =>t^2-1+4t-4=0\)

\(< =>\left(t-1\right)\left(t+1\right)+4\left(t-1\right)=0\)

\(< =>\left(t-1\right)\left(t+5\right)=0\)

\(< =>\orbr{\begin{cases}t=1\left(tm\right)\\t=-5\left(loai\right)\end{cases}}\)

\(< =>x^2=1< =>\orbr{\begin{cases}x=-1\\x=1\end{cases}}\)

Vậy ...

Thay m = 2 vào , ta có :

\(PT< =>x^2-2\left(2+1\right)x+2^2+3.2-4=0\)

\(< =>x^2-6x+6=0\)

\(< =>\left(x^2-6x+9\right)-\sqrt{3}^2=0\)

\(< =>\left(x-3-\sqrt{3}\right)\left(x-3+\sqrt{3}\right)=0\)

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

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

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

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

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

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

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

- Khi x - 1 = 0 thì x = 1

- Khi x + 1 = 0 thì x = -1

- Khi \(x^2-4x+1=0\Leftrightarrow\left(x-2\right)^2=3\Leftrightarrow\orbr{\begin{cases}x=\sqrt{3}+2\\x=-\sqrt{3}+2\end{cases}}\)

Pt có tậo nghiệm là: \(S=\left\{1;-1;\sqrt{3}+2;-\sqrt{3}+2\right\}\)

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

hay \(x\in\left\{\dfrac{1+\sqrt{13}}{2};\dfrac{1-\sqrt{13}}{2};\dfrac{-1+\sqrt{5}}{2};\dfrac{-1-\sqrt{5}}{2}\right\}\)