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

\(\left(2x+1\right)\left(2x+3\right)\left(x+1\right)^2=18\)

\(\Leftrightarrow\left(2x+2-1\right)\left(2x+2+1\right)\left(x+1\right)^2=18\)

\(\Leftrightarrow\left(\left(2x+2\right)^2-1\right)\left(x+1\right)^2=18\)

\(\Leftrightarrow4\left(x+1\right)^4-\left(x+1\right)^2-18=0\)

Đặt t = \(\left(x+1\right)^2\) \(\left(t\ge0\right)\)

pt \(\Leftrightarrow4t^2-t-18=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{9}{4}\left(nh\right)\\t=-2\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\left(x+1\right)^2-\dfrac{9}{4}=0\)

\(\Leftrightarrow\left(x+1-\dfrac{3}{2}\right)\left(x+1+\dfrac{3}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

\(\left(2x+1\right)\left(2x+3\right)\left(x+1\right)^2-18=0\)

\(\Leftrightarrow\left(x^2+2x+1\right)\left(4x^2+8x+3\right)-18=0\)

Đặt \(x^2+2x+1=a\ge0\)

\(\Rightarrow a\left(4a-1\right)-18=0\)

\(\Leftrightarrow4a^2-a-18=0\)

\(\Leftrightarrow\left(4a^2+8a\right)+\left(-9a-18\right)=0\)

\(\Leftrightarrow\left(a+2\right)\left(4a-9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=-2\left(l\right)\\a=\frac{9}{4}\end{cases}}\)

\(\Rightarrow x^2+2x+1=\frac{9}{4}\)

\(\Leftrightarrow4x^2+8x-5=0\)

\(\Leftrightarrow\left(4x^2-2x\right)+\left(10x-5\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{5}{2}\end{cases}}\)

(2x+1)(x+1)²(2x+3)=18 

<=> (2x+2-1)(x+1)²(2x+2+1)=18

Đặt y=x+1, ta có: 

(2y-1)y²(2y+1)=18

Ta có 

(2x+1)(x+1)²(2x+3)=18

=> (x+1)²(4x²+8x+3)-18=0

=> (x²+2x+1)(4x²+8x+3)-18=0

Đặt x²+2x+1=a ta có 

a.(4a-1)-18=0

=> 4a²-a-18=0

=> 4a² +8a-9a-18=0

=> 4a(a+2)-9(a+2)=0

=> (a+2)(4a-9)=0

Với a=x²+2x+1biểu thức trên trở thành

(x²+2x+3)(4x²+8x-5)=0

=> x²+2x+3=0 hoặc 4x²+8x-5=0

• x²+2x+3=0 => phương trình vô nghiệm

• 4x²+8x-5=0 => x=1/2 hoặc x=-5/2

Vậy x=1/2 và x=-5/2 là nghiệm của phương trình

\(\frac{2x}{x+1}+\frac{18}{x^2+2x-3}=\frac{2x-5}{x+3}ĐKXĐ:x\ne-1;-3\)

\(\frac{2x}{x+1}+\frac{18}{\left(x-1\right)\left(x+3\right)}=\frac{2x-5}{x+3}\)

\(2x\left(x-1\right)\left(x+3\right)+18\left(x+1\right)=\left(2x-5\right)\left(x+1\right)\left(x-1\right)\)

\(4x^2+12x+18=-2x-5x^2+5\)

\(4x^2+12x+18+2x+5x^2-5=0\)

\(9x^2-14x+13=0\)

=> vô nghiệm