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

<=>4x-8=0 

<=>4x=8 

=.x=2(nhan)

a: Đặt x-3=a; x+1=b

Theo đề, ta có: \(a^3+b^3=\left(a+b\right)^3\)

\(\Leftrightarrow3ab\left(a+b\right)=0\)

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

hay \(x\in\left\{3;-1;1\right\}\)

b: \(\Leftrightarrow\left(2x^2+1\right)^2+2x\left(2x^2+1\right)-15x^2-9x^2=0\)

\(\Leftrightarrow\left(2x^2+1\right)^2+2x\left(2x^2+1\right)-24x^2=0\)

\(\Leftrightarrow\left(2x^2+1\right)^2+6x\left(2x^2+1\right)-4x\left(2x^2+1\right)-24x^2=0\)

\(\Leftrightarrow\left(2x^2+1\right)\left(2x^2+6x+1\right)-4x\left(2x^2+6x+1\right)=0\)

\(\Leftrightarrow\left(2x^2-4x+1\right)\left(2x^2+6x+1\right)=0\)

\(\Leftrightarrow x^2+3x+\dfrac{1}{2}=0\)

\(\Leftrightarrow x^2+3x+\dfrac{9}{4}=\dfrac{7}{4}\)

\(\Leftrightarrow\left(x+\dfrac{3}{2}\right)^2=\dfrac{7}{4}\)

hay \(x\in\left\{\dfrac{\sqrt{7}-3}{2};\dfrac{-\sqrt{7}-3}{2}\right\}\)

\(\Delta'=\left(-2\right)^2-3.\left(-8\right)=4+24=28>0.\)

\(\Rightarrow\) Pt có 2 nghiệm phân biệt \(x_1;x_2.\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{2+2\sqrt{7}}{3}.\\x_2=\dfrac{2-2\sqrt{7}}{3}.\end{matrix}\right.\)

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

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

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

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

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

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

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

Dễ thấy \(x^2+x+1>0\forall x;x^2+1>0\forall x\)

\(\Rightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy....

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

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

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

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

...

\(\Leftrightarrow x=1\)

p/s: có bác nào giải đc pt \(x^3+4x^2+2x+3=0\)thì giúp nhé :))

\(a.-3x^2+15x=0\)

\(\Leftrightarrow3x\left(-x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\-x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

\(b.2x^2-32=0\)

\(\Leftrightarrow2x^2=32\)

\(\Leftrightarrow x^2=16\)

\(\Leftrightarrow\left|x\right|=4\)

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

\(c.2x^2-5x+1=0\)

\(a=2;b=-5;c=1\)

\(\Delta=\left(-5\right)^2-4.2.1=17>0\)

Do \(\Delta>0\) nên phương trình có 2 nghiệm phân biệt:

\(x_1=\dfrac{5+\sqrt{17}}{4}\)

\(x_2=\dfrac{5-\sqrt{17}}{4}\)

\(a,-3x^2+15x=0\\ -3x\left(x-5\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=5\end{matrix}\right.\) 

\(b,\\ 2\left(x^2-16\right)=0\\ \Leftrightarrow x^2-16=0\\ \Leftrightarrow\left(x-4\right)\left(x+4\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\) 

\(c,\\ \Delta=5^2-4.2=17\\ \Rightarrow x_1,x_2=\dfrac{\Delta\pm b}{2ac}\\ =\dfrac{5\pm\sqrt{17}}{4}\)

a)    (2x + 1)(3x - 2) = (5x - 8)(2x + 1)

 <=> 6x² - x - 2 = 10x² - 11x - 8

<=>  6x² - 10x² - x + 11x -2 + 8 = 0

<=>  -4x² + 10x + 6  = 0

<=> -2 (2x² - 5x - 3) = 0

<=> 2x² - 5x - 3 = 0 

<=> 2x² - 6x + x - 3 = 0

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

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

* x - 3 = 0  => x = 3

* 2x + 1 = 0 => x = -1/2 

S = {-1/2; 3}

b) 4x² – 1 = (2x +1)(3x -5)

<=> 4x² – 1 - (2x +1)(3x -5) = 0

<=> (2x - 1) (2x + 1) - (2x + 1)(3x - 5) = 0

<=>  (2x + 1) (2x - 1 - 3x + 5) = 0

<=>  (2x + 1) (-x + 4) = 0

* 2x + 1 = 0  <=> x = -1/2

* -x + 4 = 0 <=> x = 4

S = {-1/2; 4}

c) (x + 1)² = 4(x² – 2x + 1)

<=> (x + 1)² - 4(x² – 2x + 1) = 0

<=> (x + 1)² - 4(x² – 1)² = 0

* (x + 1)² = 0   <=> x = -1

* 4(x² - 1)² = 0  <=> x = 1 và x = -1

S = {-1;  1}

d) 2x³ + 5x² – 3x = 0

<=> x (2x² + 5x - 3) = 0

<=> x (2x² + 6x - x - 3) = 0

<=> x [x(2x - 1) + 3 (2x - 1)] = 0

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

* x = 0

* 2x - 1 = 0  <=> x = 1/2

* x + 3 = 0  <=> x = -3

S = { -3; 0; 1/2}