\(ĐK:x\ne\pm1\)

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

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

\(\Leftrightarrow\hept{\begin{cases}x\ne\pm1\\a+b-1=0\\-b-1=a\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ne\pm1\\a+b-1=0\\-b-1=a\end{cases}}\)

Giải ra :D

sửa đề : 

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

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

b: \(\Delta=\left\lbrack2\left(m+1\right)\right\rbrack^2-4\cdot1\cdot\left(2m+1\right)\)

\(=4m^2+8m+4-8m-4=4m^2\) >=0∀m

=>Phương trình luôn có hai nghiệm

Theo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=2\left(m+1\right)=2m+2;x_1x_2=\frac{c}{a}=2m+1\)

\(x_1-x_2=8\)

=>\(\left(x_1-x_2\right)^2=8^2=64\)

=>\(\left(x_1+x_2\right)^2-4x_1x_2=64\)

=>\(\left(2m+2\right)^2-4\left(2m+1\right)=64\)

=>\(4m^2+8m+4-8m-4=64\)

=>\(4m^2=64\)

=>\(m^2=16\)

=>m=4 hoặc m=-4

\(x_1x_2-2\left(x_1+x_2\right)\le5\)

=>2m+1-2(2m+2)<=5

=>2m+1-4m-2<=5

=>-2m-1<=5

=>-2m<=6

=>m>=-3

`a)(2x-1)^2-0,25=0`

`<=>(2x-1-0,5)(2x-1+0,5)=0`

`<=>(2x-1,5)(2x-0,5)=0`

`<=>[(x=0,75)(x=0,25):}`

`b)x^2+9=6x`

`<=>(x-3)^2=0`

`<=>x-3=0`

`<=>x=3`

`c)(x^2-4)-3x-6=0`

`<=>(x-2)(x+2)-3(x+2)=0`

`<=>(x+2)(x-2-3)=0`

`<=>(x+2)(x-5)=0`

`<=>[(x=-2),(x=5):}`

a: =>(2x-1-0,5)(2x-1+0,5)=0

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

=>x=0,25 hoặc x=0,75

b: =>x^2-6x+9=0

=>(x-3)^2=0

=>x-3=0

=>x=3

c: =>(x-2)(x+2)-3(x+2)=0

=>(x+2)(x-5)=0

=>x=5 hoặc x=-2

a,\(\left(x-4-5\right)\left(x-4+5\right)=0\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\Leftrightarrow x=9;x=-1\)

b, \(\left(x-3-x-1\right)\left(x-3+x+1\right)=0\Leftrightarrow2x-2=0\Leftrightarrow x=1\)

c, \(\left(x^2-4\right)\left(2x-3\right)-\left(x^2-4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(2x-3-x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-2\right)=0\Leftrightarrow x=-2;x=2\)

d, \(\left(3x-7\right)^2-\left(2x+2\right)^2=0\Leftrightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\)

\(\Leftrightarrow\left(x-9\right)\left(5x-5\right)=0\Leftrightarrow x=1;x=9\)

a) Ta có: 4x-20=0

$\iff 4x=20$

hay x=5

Vậy: S={5}

b) Ta có: $2x+x+12=0$

$\iff 3x+12=0$

$\iff 3x=-12$

hay x=-4

Xét phương trình:

ax²=3x-1

\(\Leftrightarrow\)ax²-3x+1=0 (1)

Xét phương trình (1) có \(\Delta\)=(-3)² - 4.a.1

= 9-4a

- (p) và (d) có 2 giao điểm phân biệt khi \(\Delta\)>0 \(\Leftrightarrow\) 9-4a>0 \(\Leftrightarrow\)a<\(\dfrac{9}{4}\)

- (p) và (d) có 1 điểm chung duy nhất khi \(\Delta\)=0 \(\Leftrightarrow\) 9-4a=0 \(\Leftrightarrow\) a=\(\dfrac{9}{4}\)

- (p) và (d) không có giao điểm khi \(\Delta\)<0 \(\Leftrightarrow\) 9-4a<0 \(\Leftrightarrow\) a>\(\dfrac{9}{4}\)

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

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

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

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

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