x⁴+2x³-2x²+2x-3=0

=>  (x⁴ - 1) + (2x³-2x² )+ (2x-2)=0

=> (x - 1).(x+1).(x² + 1) + 2x².(x - 1) + 2.(x -1) = 0

=> (x -1). [(x+1).(x² + 1) + 2x² + 2] = 0

<=> (x - 1). (x³ + x + x² + 1 + 2x² + 2)= 0

<=> (x - 1). (x³ + x + 3x² + 3)= 0

<=> x - 1 = 0 hoặc x³ + x + 3x² + 3 = 0

+) x - 1 = 0 => x  =1 

+) x³ + x + 3x² + 3 = 0 <=> x. (x² + 1) + 3.(x² + 1) = 0

<=> (x+3). (x² +1) = 0 <=> x + 3 = 0 (vì x² + 1 > 0 với mọi x)

<=> x = -3

Vậy pt có 2 nghiệm x = 1 ; x = -3

X^4+2X^3-X^2+2X+1=0 LAM TN

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

Ta có: \(\dfrac{5}{-x^2+5x-6}+\dfrac{x+3}{2-x}=0\)

\(\Leftrightarrow\dfrac{-5}{\left(x-2\right)\left(x-3\right)}-\dfrac{x^2-9}{\left(x-2\right)\left(x-3\right)}=0\)

\(\Leftrightarrow-5-x^2+9=0\)

\(\Leftrightarrow x=-2\)

\(a,PT\Leftrightarrow\left(x+2\right)\left(3x+5\right)-\left(2x-4\right)\left(x+1\right)=0\)

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

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

<=> \(\left(x+2\right)\left(2x-2\right)=0\)

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

Vậy: ...

\(b,PT\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-4\right)\left(x+5\right)=0\)

<=> \(\left(x-4\right)\left(2x+4+x+5\right)=0\)

<=> \(\left(x-4\right)\left(3x+9\right)=0\)

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

Vậy: ...

a) \(\left(8x+5\right)^2\left(4x+3\right)\left(2x+1\right)=9\)

\(\Leftrightarrow\left(64x^2+8x+25\right)\left(8x^2+10x+3\right)-9=0\)

Đặt a = \(8x^2+10x+3\)

\(\left(8a+1\right)a-9=0\)

\(\Leftrightarrow8a^2+a-9=0\)

\(\Leftrightarrow\left(a-1\right)\left(8a+9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=1\\a=-\frac{9}{8}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}8x^2+10x+3=1\\8x^2+10x+3=-\frac{9}{8}\end{cases}}\)

mà \(8x^2+10x+3=1\Rightarrow8x^2+10x+2=0\)

\(\Rightarrow2\left(x+1\right)\left(4x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\x=-0,25\end{cases}}\)

cảm ơn bạn còn mấy phần còn lại ạ

\(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-5}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\\ \Leftrightarrow\dfrac{5\left(x+3\right)+4\left(x-3\right)}{x^2-9}=\dfrac{x-5}{x^2-9}\\ \Leftrightarrow5x+15+4x-12=x-5\\ \Leftrightarrow5x+4x-x=-5-15+12\\ \Leftrightarrow8x=-8\\ \Leftrightarrow x=-1\left(TM\right)\\ Vậy:S=\left\{-1\right\}\)

Lời giải:

$\Delta'=(\sqrt{3}-1)^2+4\sqrt{3}=(\sqrt{3}+1)^2$
Do đó pt có 2 nghiệm:
\(x_1=\frac{-b'+\sqrt{\Delta'}}{a}=\frac{1-\sqrt{3}+\sqrt{3}+1}{\sqrt{3}}=\frac{2}{\sqrt{3}}\)

\(x_2=\frac{-b'-\sqrt{\Delta'}}{a}=\frac{1-\sqrt{3}-\sqrt{3}-1}{\sqrt{3}}=-2\)

\(a,\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x-6=0\Leftrightarrow x=6\\2x-5=0\Leftrightarrow x=\dfrac{5}{2}\\3x+9=0\Leftrightarrow x=-3\end{matrix}\right.\)

\(b,2x\left(x-3\right)+5\left(x-3\right)=0\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\2x+5=0\Leftrightarrow x=-\dfrac{5}{2}\end{matrix}\right.\)

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

\(x=-7\left(2m-5\right)x-2m^2+8\Leftrightarrow x+7\left(2m-5\right)=8-2m^2\Leftrightarrow x\left(14m-34\right)=8-2m^2\)

\(ycđb\Leftrightarrow14m-34\ne0\Leftrightarrow m\ne\dfrac{34}{14}\)\(\Rightarrow x=\dfrac{8-2m^2}{14m-34}\)

\(3.17\Leftrightarrow4x^2-4x+1-2x-1=0\Leftrightarrow4x^2-6x=0\Leftrightarrow x\left(4x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

3.15:

a, \(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\2x-5=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=\dfrac{5}{2}\\x=-\dfrac{9}{3}=-3\end{matrix}\right.\)

b, \(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)

c, \(\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\)

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

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

3.16

\(\Leftrightarrow\left(2m-5\right).-7-2m^2+8=0\)

\(\Leftrightarrow-14m+35-2m^2+8=0\)

\(\Leftrightarrow-14m-2m^2+43=0\)

\(\Leftrightarrow-2\left(7m+m^2\right)=-43\)

\(\Leftrightarrow m\left(7-m\right)=\dfrac{43}{2}\)

\(\Leftrightarrow\dfrac{m\left(7-m\right)}{1}-\dfrac{43}{2}=0\)

\(\Leftrightarrow\dfrac{14m-2m^2}{2}-\dfrac{43}{2}=0\)

pt vô nghiệm