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 :))

1) Ta có: 3x-12=5x(x-4)

\(\Leftrightarrow3x-12-5x\left(x-4\right)=0\)

\(\Leftrightarrow3x-12-5x^2+20x=0\)

\(\Leftrightarrow-5x^2+23x-12=0\)

\(\Leftrightarrow-5x^2+20x+3x-12=0\)

\(\Leftrightarrow\left(-5x^2+20x\right)+\left(3x-12\right)=0\)

\(\Leftrightarrow5x\left(-x+4\right)+3\left(x-4\right)=0\)

\(\Leftrightarrow5x\left(4-x\right)-3\left(4-x\right)=0\)

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

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

Vậy: \(x\in\left\{4;\frac{3}{5}\right\}\)

2) Ta có: 3x-15=2x(x-5)

\(\Leftrightarrow3x-15-2x\left(x-5\right)=0\)

\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)

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

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

Vậy: \(x\in\left\{5;\frac{3}{2}\right\}\)

3) Ta có: 3x(2x-3)+2(2x-3)=0

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

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

Vậy: \(x\in\left\{\frac{3}{2};-\frac{2}{3}\right\}\)

4) Ta có: (4x-6)(3-3x)=0

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

Vậy: \(x\in\left\{\frac{3}{2};1\right\}\)

4) (4x - 6 ) ( 3 - 3x ) = 0

<=> \(\left[{}\begin{matrix}4x-6=0\\3-3x=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}4x=6\\3x=3\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\frac{3}{2}\\x=1\end{matrix}\right.\)

Bài 1 :

a, Ta có : \(3x-12=5x\left(x-4\right)\)

=> \(3x-12=5x^2-20x\)

=> \(3x-12-5x^2+20x=0\)

=> \(5x^2-23x+12=0\)

=> \(5x^2-20x-3x+12=0\)

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

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

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

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

Vậy phương trình có nghiệm là x = \(\frac{3}{5}\) và x = 4 .

b, Ta có : \(3x-15=2x\left(x-5\right)\)

=> \(3x-15-2x\left(x-5\right)=0\)

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

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

=> \(\left[{}\begin{matrix}3-2x=0\\x-5=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{3}{2}\\x=5\end{matrix}\right.\)

Vậy phương trình có nghiệm là x = \(\frac{3}{2}\) và x = 5 .

c, Ta có : \(3x\left(2x-3\right)+2\left(2x-3\right)=0\)

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

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

=> \(\left[{}\begin{matrix}3x=-2\\2x=3\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-\frac{2}{3}\\x=\frac{3}{2}\end{matrix}\right.\)

Vậy phương trình có nghiệm là x = \(-\frac{2}{3}\) và x = \(\frac{3}{2}\) .

d, Ta có : \(\left(4x-6\right)\left(3-3x\right)=0\)

=> \(\left[{}\begin{matrix}4x-6=0\\3-3x=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}4x=6\\-3x=-3\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{6}{4}\\x=1\end{matrix}\right.\)

Vậy phương trình có nghiệm là x = 1 và x = \(\frac{6}{4}\) .

<=>4x-8=0 

<=>4x=8 

=.x=2(nhan)

cái bài này tìm nghiệm là ra mà bạn

câu trả lời của thu hương rất hay!

Mình làm được khổ nỗi lại chưa biết nghiệm là gì? @ thu hương có thể giải thích cho minh không

 hiihhi  

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

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

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

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

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

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

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

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

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

Vậy tập nghiệm của pt là \(S=\left\{-\frac{1}{3};2\right\}\)

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

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

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

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

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

Vậy phương trình có tập nghiệm \(S=\left\{\frac{-1}{3};2\right\}\)

\(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)

\(\Leftrightarrow\frac{24\left(x-3\right)}{24}-\frac{4\left(x-3\right)\left(2x-5\right)}{24}=-\frac{6\left(x-3\right)\left(x-3\right)}{24}\)

\(\Leftrightarrow24\left(x-3\right)-4\left(x-3\right)\left(2x-5\right)+6\left(x-3\right)^2=0\)

\(\Leftrightarrow\left(x-3\right)\left[24-4\left(2x-5\right)+6\left(x-3\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(24-8x+20+6x-18\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(26-2x\right)=0\)

\(\Leftrightarrow2\left(x-3\right)\left(13-x\right)=0\)

\(\Leftrightarrow\left[\begin{matrix}x-3=0\\13-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}x=3\\x=13\end{matrix}\right.\)

Vậy tập nghiệm của pt là \(S=\left\{3;13\right\}\)

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

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

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+1=0\\-x-2=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-\frac{1}{3}\\x=-2\end{cases}}\)

Vậy...

a) \(2x^3+3x^2-8x-12=0\)

\(\Leftrightarrow\left(2x^3-8x\right)+\left(3x^2-12\right)=0\)

\(\Leftrightarrow2x\left(x^2-4\right)+3\left(x^2-4\right)=0\)

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

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

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

hoặc \(x+2=0\)

hoặc \(2x+3=0\)

\(\Leftrightarrow\)\(x=2\)

hoặc \(x=-2\)

hoặc \(x=-\frac{3}{2}\)

Vậy tập nghiệm của phương trình là \(S=\left\{2;-2;-\frac{3}{2}\right\}\)

b) \(x^3-4x^2-x+4=0\)

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

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

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

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

hoặc \(x-1=0\)

hoặc \(x+1=0\)

\(\Leftrightarrow\)\(x=4\)

hoặc \(x=1\)

hoặc \(x=-1\)

Vậy tập nghiệm của phương trình là \(S=\left\{4;1;-1\right\}\)

c) \(x^3-x^2-x-2=0\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x^2+x+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\\left(x+\frac{1}{2}\right)^2+\frac{3}{4}=0\left(ktm\right)\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{2\right\}\)

d) \(x^4-3x^3+3x^2-x=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{0;1\right\}\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2-2x+3=x^2-x+1\end{cases}}\)

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

Vậy tập nghiệm của phương trình là \(S=\left\{-1;2\right\}\)

g) \(x^3+3x^2+3x+1=4x+4\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\\left(x+1\right)^2=4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x+1=\pm2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\)  hoặc   \(x=1\)

Vậy tập nghiệm của phương trình là \(S=\left\{-1;1;-3\right\}\)

b) \(x^3-4x^2-x+4=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x^2-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=\pm1\end{cases}}\)

c) \(x^3-x^2-x-2=0\)

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

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

\(\Leftrightarrow x=2\) ( Do \(x^2+x+1>0\) )

a) \(2x^3+3x^2-8x-12=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}2x+3=0\\x^2-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=\pm2\end{cases}}\)

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: ...

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