2)  \(x^3-6x^2+11x-6=0\)

\(\Leftrightarrow\)\(x^3-3x^2-3x^2+9x+2x-6=0\)

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

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

bn giải tiếp nha

3)   \(x^3-4x^2+x+6=0\)

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

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

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

lm tiếp nha

4)  \(x^3-3x^2+4=0\)

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

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

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

lm tiếp nha

Mk làm mẫu 1 bài cho nha !

1. <=> (x^3-x^2)+(5x^2-5x)+(6x-6) = 0

<=> (x-1).(x^2+5x+6) = 0

<=> (x-1).[(x^2+2x)+(3x+6)] = 0

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

<=> x-1=0 hoặc x+2=0 hoặc x+3=0

<=> x=1 hoặc x=-2 hoặc x=-3

Vậy ..............

Tk mk nha

2.  x3−6x2+11x−6=0

⇔x3−3x2−3x2+9x+2x−6=0

⇔(x−3)(x2−3x+2)=0

⇔(x−3)(x−2)(x−1)=0

bn giải tiếp nha

3)   x3−4x2+x+6=0

⇔x3−3x2−x2+3x−2x+6=0

⇔(x−3)(x2−x−2)=0

⇔(x−3)(x−2)(x+1)=0

lm tiếp nha

4)  x3−3x2+4=0

⇔x3+x2−4x2−4x+4x+4=0

⇔(x+1)(x2−4x+4)=0

⇔(x+1)(x−2)2=0

lm tiếp nha

\(1,\)

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

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

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

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

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

\(2,\)

\(3x\left(x+5\right)-6\left(x+5\right)=0\)

\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)

\(3,\)

\(x^4-x^2=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

\(4,\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(5,\)

\(x\left(x+6\right)-10\left(x-6\right)=0\)

\(\Leftrightarrow x^2+6x-10x+60=0\)

\(\Leftrightarrow x^2-4x+60=0\)

\(\Leftrightarrow x^2-4x+4+56=0\)

\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)

=> Phương trình vô nghiệm

Answer:

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

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

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)

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

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

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

\(x^2-5x+6=0\)

\(\Rightarrow x^2-2x-3x+6=0\)

\(\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)

\(\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

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

\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Rightarrow\left(5x-1\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)

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

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

\(\Rightarrow\left(x-1\right)^2=-4\) (Vô lý)

Vậy không có giá trị \(x\) thoả mãn

\(x^2+x-6=0\)

\(\Rightarrow x^2+3x-2x-6=0\)

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

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

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

1) (x+6)(3x-1)+x+6=0

⇔(x+6)(3x-1)+(x+6)=0

⇔(x+6)(3x-1+1)=0

⇔3x(x+6)=0

2) (x+4)(5x+9)-x-4=0

⇔(x+4)(5x+9)-(x+4)=0

⇔(x+4)(5x+9-1)=0

⇔(x+4)(5x+8)=0

3)(1-x)(5x+3)÷(3x-7)(x-1)

=\(\frac{\left(1-x\right)\left(5x+3\right)}{\left(3x-7\right)\left(x-1\right)}=\frac{\left(1-x\right)\left(5x+3\right)}{\left(7-3x\right)\left(1-x\right)}=\frac{\left(5x+3\right)}{\left(7-3x\right)}\)

1)x=1;2)x=1;3)x=3;4)x=2;5)x=2;6)x=4.

moi hoc lop 6 thoi

1) \(x^2+x-2=0\)

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

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

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

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

2) \(x^2+2x-3=0\)

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

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

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

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

3) \(x^2-x-6=0\)

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

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

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

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

4) \(x^2+x-6=0\)

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

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

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

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

5) \(2x^2-x-6=0\)

\(\Leftrightarrow2x^2-4x+3x-6=0\)

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

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

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

Quên , còn ý 6 

\(3x^3+10x-8=0\)

\(\Leftrightarrow3x^2+12x-2x-8=0\)

\(\Leftrightarrow3x\left(x+4\right)-2\left(x+4\right)=0\)

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

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

1: =>(x+3)(x-5)=0

=>x=5 hoặc x=-3

2: =>(x-1)(5x-1)=0

=>x=1/5 hoặc x=1

5: =>(x-4)*x=0

=>x=0 hoặc x=4

10: =>(x+5)(x-3)=0

=>x=3 hoặc x=-5

9: =>(x-2)(x-4)=0

=>x=2 hoặc x=4

7: =>(x-6)(2x-1)=0

=>x=1/2 hoặc x=6

8: =>(2x-1)(3x-12)=0

=>x=4 hoặc x=1/2

1)\(2x+6=0\)

\(\Leftrightarrow2x=-6\)

\(\Leftrightarrow x=-3\)

Vậy : x=3 là nghiệm PT

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

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

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

Vậy:....

3)\(\frac{x-2}{x+2}+\frac{3}{x-2}=\frac{x^2-11}{x^2-4}\)

\(\Rightarrow\left(x-2\right)^2+3\left(x+2\right)=x^2-11\)

\(\Leftrightarrow x^2-4x+4+3x+6-x^2+11=0\)

\(\Leftrightarrow-x+21=0\)

\(\Leftrightarrow-x=-21\)

\(\Leftrightarrow x=21\)

Vậy:......

4) \(x\left(x^2-1\right)=0\)

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

Vậy:........

5)\(4x+20=0\)

\(\Leftrightarrow4x=-20\)

\(\Leftrightarrow x=-5\)

Vậy:...

6)\(\frac{x+3}{x+1}+\frac{x-2}{x}=2\)

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

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

\(\Leftrightarrow-2=0\)(vô lí)

Vậy : PT vô nghiệm

7)\(\frac{1+2x-5}{6}=\frac{3-x}{4}\)

\(\Leftrightarrow\frac{-4+2x}{6}=\frac{3-x}{4}\)

\(\Rightarrow2\left(-4+2x\right)=3\left(3-x\right)\)

\(\Leftrightarrow-8+4x-9+3x=0\)

\(\Leftrightarrow-17+7x=0\)

\(\Leftrightarrow7x=17\)

\(\Leftrightarrow x=\frac{17}{7}\)

8) Làm tương tự

9) \(2\left(x+1\right)=5x-7\)

\(\Leftrightarrow2x+2-5x+7=0\)

\(\Leftrightarrow-3x+9=0\)

\(\Leftrightarrow-3x=-9\)

\(\Leftrightarrow x=3\)

#H

1.\(2x+6=0\)

\(\Leftrightarrow2\left(x+3\right)=0\)

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=3\)

Vậy tập nghiệm của PT là \(S=\left\{3\right\}\)

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

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

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

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

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

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

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

3.\(\frac{x-2}{x+2}+\frac{3}{x-2}=\frac{x^2-11}{x^2-4}\)

ĐKXĐ :\(x\ne\pm2\)

Ta có ; \(\frac{x-2}{x+2}+\frac{3}{x-2}=\frac{x^2-11}{x^2-4}\)

\(\Leftrightarrow\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{x^2-11}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\frac{x^2-4x+4+3x+6}{\left(x-2\right)\left(x+2\right)}=\frac{x^2-11}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\frac{x^2-x+10}{\left(x-2\right)\left(x+2\right)}=\frac{x^2-11}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow x^2-x+10=x^2-11\)

\(\Leftrightarrow21-x=0\)

\(\Leftrightarrow x=21\)(Thỏa mãn ĐKXĐ)

Vậy tập nghiệm của PT là \(S=\left\{21\right\}\)

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

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

\(\Leftrightarrow x=0\)

hoặc \(x-1=0\)

hoặc \(x+1=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

Vậy tập nghiệm của PT là \(S=\left\{0;\pm1\right\}\)

5.\(4x+20=0\)

\(\Leftrightarrow4\left(x+5\right)=0\)

\(\Leftrightarrow x+5=0\)

\(\Leftrightarrow x=-5\)

Vậy tập nghiệm của PT là \(S=\left\{-5\right\}\)

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

ĐKXĐ : \(x\notin\left\{-1;0\right\}\)

Ta có : \(\frac{x+3}{x+1}+\frac{x-2}{x}=2\)

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

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

\(\Leftrightarrow\frac{x^2+2x-2}{x\left(x+1\right)}=\frac{2x^2+2x}{x\left(x+1\right)}\)

\(\Rightarrow2x^2+2x-2=2x^2+2x\)

\(\Leftrightarrow0x=2\)(Vô lí)

Vậy PT vô nghiệm 

7.\(1+\frac{2x-5}{6}=\frac{3-x}{4}\)

\(\Leftrightarrow\frac{12}{12}+\frac{2\left(2x-5\right)}{12}=\frac{3\left(3-x\right)}{12}\)

\(\Leftrightarrow\frac{12+4x-10}{12}=\frac{9-3x}{12}\)

\(\Leftrightarrow\frac{4x+2}{12}=\frac{9-3x}{12}\)

\(\Rightarrow4x+2=9-3x\)

\(\Leftrightarrow7x=7\)

\(\Leftrightarrow x=1\)

Vậy tập nghiệm của PT là \(S=\left\{1\right\}\)

8.\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\)

ĐKXĐ : \(x\notin\left\{0;2\right\}\)

Ta có : \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\)

\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)

\(\Leftrightarrow\frac{x^2+2x-x+2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)

\(\Leftrightarrow\frac{x^2+x+2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)

\(\Rightarrow x^2+x+2=2\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)(Không thỏa mãn ĐKXĐ)_(Thỏa mãn ĐKXĐ)

Vậy tập nghiệm của PT là \(S=\left\{-1\right\}\)

9.\(2\left(x+1\right)=5x-7\)

\(\Leftrightarrow2x+2=5x-7\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\)

Vậy tập nghiệm của PT là \(S=\left\{3\right\}\)

1)3x(x-2)=7(x-2)

<=>3x(x-2)-7(x-2)=0

<=>(x-2)(3x-7)=0

x-2=0=>x=2

3x-7=0=>x=7/3

cn lại lm tg tự

10)\(x^2-9x+20=0\)

\(\Leftrightarrow x^2-4x-5x+20=0\)

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

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

\(\Leftrightarrow\hept{\begin{cases}x=4\\x=5\end{cases}}\)

16) \(\left(x^2+x\right)\left(x^2+x+1\right)=6\)

\(\Leftrightarrow x^4+x^3+x^2+x^3+x^2+x=6\)

\(\Leftrightarrow x^4+2x^3+2x^2+x-6=0\)

\(\Leftrightarrow x^4+2x^3+2x^2+4x-3x-6=0\)

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

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

\(\Leftrightarrow\left(x+2\right)\left(x^3+\frac{1}{4}x-x+\frac{11}{4}x-\frac{11}{4}-\frac{1}{4}+x^2-x^2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left[\left(x^3-x^2\right)+\left(x^2-x\right)+\left(\frac{1}{4}x-\frac{1}{4}\right)+\left(\frac{11}{4}x-\frac{11}{4}\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right)\left[x^2\left(x-1\right)+x\left(x-1\right)+\frac{1}{4}\left(x-1\right)+\frac{11}{4}\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x^2+x+\frac{1}{4}+\frac{11}{4}\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{11}{4}\right]=0\)

\(\Leftrightarrow\hept{\begin{cases}x+2=0\\x-1=0\\\left(x+\frac{1}{2}\right)^2+\frac{11}{4}=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=-2\\x=1\\\left(x+\frac{1}{2}\right)^2+\frac{11}{4}=0->ktm\end{cases}}\)

\(\left(x+\frac{1}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}>0\)=>ko thỏa mãn(đây là giải thích cho phần trên)

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

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

\(\Leftrightarrow x^2-4x-26=0\)

đến đây nếu phân tích tam thức bậc hai này thì tìm đc x là số thập phân vô hạn ko tuần hoàn nên mk nghĩ là đề bài câu này sai

\(a,x=3x^2\Rightarrow x-3x^2=0\Rightarrow x\left(1-3x\right)=0\Rightarrow\orbr{\begin{cases}x=0\\1-3x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{3}\end{cases}}\)

\(b,\left(2x-6\right)\left(x+4\right)+2\left(2x-6\right)=0\)

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

\(\Rightarrow\left(2x-6\right)\left(x+6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-6=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)

\(c,\left(2x-5\right)\left(x+9\right)+6x-15=0\)

\(\Rightarrow\left(2x-5\right)\left(x+9\right)+3\left(2x-5\right)=0\)

\(\Rightarrow\left(2x-5\right)\left(x+9+3\right)=0\)

\(\Rightarrow\left(2x-5\right)\left(x+12\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-5=0\\x+12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-12\end{cases}}\)