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

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

\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+1=0\\2x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=-1\\x=-\frac{3}{2}\end{matrix}\right.\)

b) Có \(\left|2x+1\right|\ge0;\left|4x^2-1\right|\ge0\forall x\)

\(\Rightarrow\left|2x+1\right|+\left|4x^2-1\right|\ge0\forall x\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}2x+1=0\\4x^2-1=0\end{matrix}\right.\Leftrightarrow x=-\dfrac{1}{2}\)

c) \(\left|2x-1\right|=\left|x+5\right|\) 

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x+5\\2x-1=-x-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-\dfrac{4}{3}\end{matrix}\right.\)

2.5+4.5567895

a, ĐKXĐ: ...

\(\sqrt{3x^2-2x+6}+3-2x=0\)

\(\Leftrightarrow\sqrt{3x^2-2x+6}=2x-3\)

\(\Leftrightarrow3x^2-2x+6=4x^2-12x+9\)

\(\Leftrightarrow4x^2-10x+3=0\)

.....

b, ĐKXĐ: ...

\(\sqrt{x+1}+\sqrt{x-1}=4\\ \Leftrightarrow x+1+x-1+2\sqrt{\left(x+1\right)\left(x-1\right)}=16\\ \Leftrightarrow2\sqrt{x^2-1}=16-2x\\ \Leftrightarrow\sqrt{x^2-1}=8-x\\ \Leftrightarrow x^2-1=64-16x+x^2\\ \Leftrightarrow65-16x=0\\ \Leftrightarrow x=\dfrac{65}{16}\)

a, \(5\left|2x-1\right|-3=7\Leftrightarrow5\left|2x-1\right|=10\Leftrightarrow\left|2x-1\right|=2\)

TH1 : \(2x-1=2\Leftrightarrow x=\frac{3}{2}\)

TH2 : \(2x-1=-2\Leftrightarrow x=-\frac{1}{2}\)

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

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

c, \(\frac{2x-3}{2}< \frac{1-3x}{-5}\Leftrightarrow\frac{2x-3}{2}+\frac{1-3x}{5}< 0\)

\(\Leftrightarrow\frac{10x-15+2-6x}{10}< 0\Rightarrow4x-13< 0\Leftrightarrow x< \frac{13}{4}\)

cái này dễ đợi tí mình giải cho, gõ đáp số mất khá nhiều thời gian

\(\frac{x}{2x-6}+\frac{x}{2x+2}-\frac{2x}{\left(x+1\right).\left(x-3\right)}=0\)

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

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

tự làm tiếp nha bạn :)))

Lời giải:

1. 

PT $\Leftrightarrow (x^2+5x)^2+2(x^2+5x)-24=0$

$\Leftrightarrow t^2+2t-24=0$ (đặt $x^2+5x=t$)

$\Leftrightarrow (t-4)(t+6)=0$

$\Rightarrow t-4=0$ hoặc $t+6=0$

Nếu $t-4=0\Leftrightarrow x^2+5x-4=0$

$\Leftrightarrow x=\frac{-5\pm \sqrt{41}}{2}$

Nếu $t+6=0$

$\Leftrightarrow x^2+5x+6=0$

$\Leftrightarrow (x+2)(x+3)=0\Rightarrow x=-2$ hoặc $x=-3$

2.

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

$\Leftrightarrow t^2+2t-3=0$ (đặt $x^2-4x+1=t$)

$\Leftrightarrow (t-1)(t+3)=0$

$\Rightarrow t-1=0$ hoặc $t+3=0$

Nếu $t-1=0\Leftrightarrow x^2-4x=0\Leftrightarrow x(x-4)=0$

$\Rightarrow x=0$ hoặc $x=4$

Nếu $t+3=0\Leftrightarrow x^2-4x+4=0$

$\Leftrightarrow (x-2)^2=0\Leftrightarrow x=2$