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

bai 1

1 thay k=0 vao pt ta co 4x^2-25+0^2+4*0*x=0

<=>(2x)^2-5^2=0

<=>(2x+5)*(2x-5)=0

<=>2x+5=0 hoăc 2x-5 =0 tiếp tục giải ý 2 tương tự

\(\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)

đặt \(x^2+5x+5=t\)

\(\Leftrightarrow t^2-25=0\Rightarrow\left\{{}\begin{matrix}t=5\\t=-5\end{matrix}\right.\)

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

\(\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-4}-3\left(\frac{2x-4}{x-4}\right)^2=0\)

<=> \(\left(x+1\right)^2.\left(x-2\right)^2.\left(x-4\right)^2+\frac{x+1}{x-4}.\left(x-2\right)^2.\left(x-4\right)^2-\frac{3\left(2x-4\right)^2}{\left(x-4\right)^2}.\left(x-2\right)^2.\left(x-4\right)^2\)\(=0.\left(x-2\right)^2.\left(x-4\right)^2\)

<=> \(\left(x+1\right)^2.\left(x-4\right)^2+\left(x+1\right).\left(x-2\right)^2.\left(x-4\right)^2-3\left(2x-4\right)^2.\left(x-2\right)^2=0\)

<=> \(-\left(x-3\right)\left(5x-4\right)\left(2x^2-9x+16\right)=0\)

<=> \(\orbr{\begin{cases}x-3=0\\5x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{4}{5}\end{cases}}\)

Mà vì: \(2x^2-9x+16\ne0\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{4}{5}\end{cases}}\)

a.\(\left(3x\right)^2-4\left(x-3\right)^2=0\)

<=> \(9x^2-4\left(x^2-6x+9\right)=0\)

<=> \(9x^2-4x^2+24x-36=0\)

<=>\(5x^2+24x-36=0\)

giải pt bậc hai thì pt có hai nghiệm x={1,2;-6}

a) (3x)² - 4(x- 3)² = 0

\(\Leftrightarrow\) (3x - 2x + 6)(3x + 2x - 6) = 0

\(\Leftrightarrow\) (x+ 6)(5x - 6) = 0

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x+6=0\\5x-6=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-6\\x=\dfrac{6}{5}\end{matrix}\right.\)

Vậy phượng trình có tập nghiệm là: S = {-6;\(\dfrac{6}{5}\)}

b) x³ + x² + 4 = 0

\(\Leftrightarrow\) x³ + 2x² - x² + 4 = 0

\(\Leftrightarrow\) (x³ + 2x²) - (x² - 4) = 0

\(\Leftrightarrow\) x²(x + 2) - (x + 2)(x - 2) = 0

\(\Leftrightarrow\) (x² - x + 2)(x + 2) = 0

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x^2-x+2=0\left(vôli\right)\\x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\) x = -2

Vậy phương trình có tập nghiệm là: S={-2}

c) (x - 1)²(x - 3) + (1 - x)²(x + 3) = 72

\(\Leftrightarrow\) (x - 1)²(x - 3) + (x - 1)²(x + 3) = 72

\(\Leftrightarrow\) (x - 1)²(x - 3 + x + 3) = 72

\(\Leftrightarrow\) 2x(x² - 2x + 1) = 72

\(\Leftrightarrow\) 2x³ - 4x² + 2x - 72 = 0

\(\Leftrightarrow\) 2(x³ - 2x² + x - 36) = 0

\(\Leftrightarrow\) x³ - 2x² + x - 36 = 0

\(\Leftrightarrow\) x³ - 4x² + 2x² - 8x + 9x - 36 = 0

\(\Leftrightarrow\) (x³ - 4x²) + (2x² - 8x) + (9x - 36) = 0

\(\Leftrightarrow\) x²(x - 4) + 2x(x - 4) + 9(x - 4)= 0

\(\Leftrightarrow\) (x² + 2x + 9)(x - 4) = 0

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x^2+2x+9=0\left(vôli\right)\\x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\) x = 4

Vậy phương trình có tập nghiệm là: S={4}

(4x - 3)² - (2x + 1)² = 0

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

\(\Leftrightarrow\) (2x - 4)(6x - 2) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\)

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

Vậy ...

3x - 12 - 5x(x - 4) = 0

\(\Leftrightarrow\) 3x - 12 - 5x² + 20x = 0

\(\Leftrightarrow\) -5x² + 23x - 12 = 0

\(\Leftrightarrow\) 5x² - 23x + 12 = 0

\(\Leftrightarrow\) 5x² - 20x - 3x + 12 = 0

\(\Leftrightarrow\) 5x(x - 4) - 3(x - 4) = 0

\(\Leftrightarrow\) (x - 4)(5x - 3) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-4=0\\5x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy ...

(8x + 2)(x² + 5)(x² - 4) = 0

\(\Leftrightarrow\) (8x + 2)(x² + 5)(x - 2)(x + 2) = 0

Vì x² \(\ge\) 0 \(\forall\) x nên x² + 5 > 0 \(\forall\) x

\(\Rightarrow\) (8x + 2)(x - 2)(x + 2) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}8x+2=0\\x-2=0\\x+2=0\end{matrix}\right.\)

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

Vậy ...

Chúc bn học tốt!

a) Ta có: \(\left(4x-3\right)^2-\left(2x+1\right)^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{2;\dfrac{1}{3}\right\}\)

b) Ta có: \(3x-12-5x\left(x-4\right)=0\)

\(\Leftrightarrow3\left(x-4\right)-5x\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(3-5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{4;\dfrac{3}{5}\right\}\)

c) Ta có: \(\left(8x+2\right)\left(x^2+5\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow2\left(4x+1\right)\left(x^2+5\right)\left(x-2\right)\left(x+2\right)=0\)

mà \(2>0\)

và \(x^2+5>0\forall x\)

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

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

Vậy: \(S=\left\{-\dfrac{1}{4};2;-2\right\}\)

\(\left(x+\dfrac{1}{x}\right)^2-4\left(x+\dfrac{1}{x}\right)^2+3=0\\\Leftrightarrow3\left(x+\dfrac{1}{x}\right)^2=3\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=1\\x+\dfrac{1}{x}=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2-x+1=0\\x^2+x+1=0\end{matrix}\right.\\ \Leftrightarrow x\in\varnothing \)