ta có : x^5+2x^4+3x^3+3x^2+2x+1=0

\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0

\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0

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

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

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

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

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

VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)

\(\Rightarrow\)x+1=0

\(\Rightarrow\)x=-1

CÒN CÂU B TỰ LÀM (02042006)

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

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

=>(x-1)(x^3+4x^2+2x+3)=0

=>x-1=0

=>x=1

a, đk : x >= 1

\(\left[{}\begin{matrix}3x+5=2x-2\\3x+5=2-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\left(ktm\right)\)

vậy pt vô nghiệm 

b, đk >= 0 

\(\left[{}\begin{matrix}x^2+1=2x\\x^2+1=-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)

c, \(\left[{}\begin{matrix}2x^2+2x=0\\2x^2+4x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x\left(x+1\right)=0\\x^2+2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0;x=-1\\x=-1\end{matrix}\right.\)

\(\left|2x+1\right|=4.\\ \Leftrightarrow\left[{}\begin{matrix}2x+1=-4.\\2x+1=4.\end{matrix}\right.\) 

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}.\\x=\dfrac{3}{2}.\end{matrix}\right.\)

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

\(\Leftrightarrow\left|3x-2\right|=-1\) (vô lý).

\(\Rightarrow x\in\phi.\)

\(\left|2x-6\right|=3x+1\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-6=3x+1\\6-2x=3x+1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=7\\-5x=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=1\end{matrix}\right.\)

Vậy \(S=\left\{1;-7\right\}\)

Ta có: \(\left|2x-6\right|=3x+1\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-6=3x+1\left(x\ge3\right)\\2x-6=-3x-1\left(x< 3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=7\\5x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)

đk: \(-x^4+3x-1\ge0\)

Có \(-\left(x^4+1\right)\le-2x^2\)

 \(\Rightarrow\sqrt{-x^4+3x-1}+\sqrt{2x^2-3x+2}\le\sqrt{3x-2x^2}+\sqrt{2x^2-3x+2}\) 

Áp dụng bunhia có: \(\sqrt{3x-2x^2}+\sqrt{2x^2-3x+2}\le\sqrt{\left(1+1\right)\left(3x-2x^{^2}+2x^2-3x+2\right)}=2\)

\(\Rightarrow\sqrt{-x^4+3x-1}+\sqrt{2x^2-3x+2}\le2\)  (*)

Có: \(x^4-x^2-2x+4=\left(x^4+1\right)-x^2-2x+3\ge2x^2-x^2-2x+3=\left(x-1\right)^2+2\ge2\) (2*)

Từ (*) (2*) dấu = xảy ra khi x=1 (TM)

Vậy x=1

Bạn tham khảo nhé

Giải Phương Trình: (x+1)\(\sqrt{2x^2-2x}\) = 2x2-3x-2 - Hoc24

TK: https://hoc24.vn/cau-hoi/giai-phuong-trinh-x1sqrt2x2-2x-2x2-3x-2.261337627197

\(\Leftrightarrow6x^2-3x+4x-2-5=0\)

\(\Leftrightarrow6x^2+x-7=0\)

\(\Leftrightarrow6x^2+7x-6x-7=0\)

=>(6x+7)(x-1)=0

=>x=-7/6 hoặc x=1