Cảm ơn diễn quỳnh

Mình là diễm quỳnh chứ không phải diễn quỳnh nha bạn

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

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

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

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

\(\Leftrightarrow-3=x\)

\(\Leftrightarrow x=-3\)

Vậy: nghiệm phương trình là -3

\(3,\text{ }\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16=0\)

\(\Rightarrow\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)=0-16\)

\(\Rightarrow\text{ Có lẻ thừa số âm }\)

Mà \(\left(x+8\right)>\left(x+6\right)>\left(x+4\right)>\left(x+2\right)\)

Ta có hai trường hợp : 

\(TH\text{ }1\text{ :}\) Có một thừa số âm

\(\Rightarrow\text{ }\left(x+2\right)< 0\)

\(\Rightarrow\text{ }x< -2\)

\(TH\text{ }2\text{ : }\) Có 3 thừa số âm

\(\Rightarrow\text{ }\hept{\begin{cases}\left(x+2\right)< 0\\\left(x+4\right)< 0\\\left(x+6\right)< 0\end{cases}}\)                \(\Rightarrow\text{ }\left(x+2\right)< 0\text{ }\Rightarrow\text{ }x< -2\)

Si thì thôi nha ! Mong bạn thông cảm !

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

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

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

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

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

a) \(x^3-5x^2+8x-4=0\)

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

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

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

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

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

Vậy nghiệm của phương trình là: \(x=\left\{1;2\right\}\)

b: =>2x^3+2x^2-3x^2-3x+6x+6=0

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

=>x+1=0

=>x=-1

c: =>(x^2+x)^2+(x^2+x)-6=0

=>(x^2+x-2)=0

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

=>x=1 hoặc x=-2

d: =>(x^2-4x-3)(x^2-4x-5)=0

=>(x-5)(x+1)(x^2-4x-3)=0

hay \(x\in\left\{2+\sqrt{7};2-\sqrt{7};5;-1\right\}\)

\(\left(4-3x\right)\left(10x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)

\(\left(7-2x\right)\left(4+8x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)

rồi thực hiện đến hết ... 

Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>

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

\(2x^2-7x+3=4x^2+4x-3\)

\(2x^2-7x+3-4x^2-4x+3=0\)

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

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

\(2x^2+12x-x-6=0\)

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

\(\left(x+6\right)\left(2x-1\right)=0\)

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

\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)

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

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

\(x=0\)

\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)

Tự lm tiếp nha 

2)

a) \(x^3-5x^2+8x-4=0\)

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

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

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

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

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

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

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

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

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

Vậy x=1 ; x=2

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

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

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

\(\Leftrightarrow2x\left(x^2-1\right)-x\left(x+1\right)+6\left(x+1\right)=0\)

\(\Leftrightarrow2x\left(x-1\right)\left(x+1\right)-x\left(x+1\right)+6\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x^2-2x-x+6\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x^2-3x+6\right)=0\)

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

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

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

Vậy x=-1