minh biet

ta có : 

\(\left|x+1\right|+\left|x-1\right|=1+\left|\left(x-1\right)\left(x+1\right)\right|\)

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

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

\(\Leftrightarrow x\in\left\{-2,0,2\right\}\)

1. a = 3 thì phương trình trở thành:

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

\(\Leftrightarrow\frac{\left(x+3\right)^2+\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}=\frac{-3\left[-9+1\right]}{9}-x^2\)

\(\Leftrightarrow\frac{x^2+6x+9+x^2-6x+9}{\left(3-x\right)\left(3+x\right)}=\frac{-3.\left(-8\right)}{9}-x^2\)

\(\Leftrightarrow\frac{2x^2+18}{9-x^2}=\frac{24}{9}-x^2\)

\(\Leftrightarrow\frac{2x^2+18}{9-x^2}+x^2=\frac{24}{9}\)

\(\Leftrightarrow\frac{2x^2+18+9x^2-x^4}{9-x^2}=\frac{24}{9}\)

\(\Leftrightarrow\frac{11x^2+18-x^4}{9-x^2}=\frac{24}{9}\)

\(\Leftrightarrow99x^2+18-9x^4=216-24x^2\)

\(\Leftrightarrow9x^4-123x^2+198=0\)

Đặt \(x^2=t\left(t\ge0\right)\)

Phương trình trở thành \(9t^2-123t+198=0\)

Ta có \(\Delta=123^2-4.9.198=8001,\sqrt{\Delta}=3\sqrt{889}\)

\(\Rightarrow\orbr{\begin{cases}t=\frac{123+3\sqrt{889}}{18}=\frac{41+\sqrt{889}}{6}\\t=\frac{123-3\sqrt{889}}{18}=\frac{41-\sqrt{889}}{6}\end{cases}}\)

Lúc đó \(\orbr{\begin{cases}x^2=\frac{41+\sqrt{889}}{6}\\x^2=\frac{41-\sqrt{889}}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm\sqrt{\frac{41+\sqrt{889}}{6}}\\x=\pm\sqrt{\frac{41-\sqrt{889}}{6}}\end{cases}}\)

Vậy pt có 4 nghiệm \(S=\left\{\pm\sqrt{\frac{41+\sqrt{889}}{6}};\pm\sqrt{\frac{41-\sqrt{889}}{6}}\right\}\)

Sửa)):

a = -3 mà ghi lôn a = 3.giải tương tự như 3

\(\Leftrightarrow\frac{y+x}{xy}=\frac{1}{2}\)

=>\(\frac{x+y}{xy}-\frac{1}{2}=0\)

\(\Rightarrow\frac{-\left(x-2\right)y-2x}{2xy}=0\)

=>(x-2)y-2x=0

=>x-2=0( vì x-2=0 thì nhân y-2x ms =0 )

=>x=2

=>y-2=0

=>y=2

vậy x=y=2

ĐKXĐ: \(x\notin\left\{0;1\right\}\)

a) Thay m=1 vào phương trình, ta được:

\(\dfrac{2x+1}{x}=1+\dfrac{x+1}{x-1}\)

\(\Leftrightarrow\dfrac{2x+1}{x}=\dfrac{x-1+x+1}{x-1}\)

\(\Leftrightarrow\dfrac{2x+1}{x}=\dfrac{2x}{x-1}\)

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

\(\Leftrightarrow2x^2=2x^2-2x+x-1\)

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

\(\Leftrightarrow x-1=0\)

hay x=1(loại)

Vậy: Khi m=1 thì \(S=\varnothing\)

`a,m=1`

`=>(2x+1)/x=(2x)/(x-1)`

`<=>2x^2-x-1=2x^2`

`<=>-x-1=0`

`<=>x=-1`

`b,(2x+m)/x=(2x)/(x-1)`

`<=>2x^2=2x^2-2x+mx-m`

`<=>mx-2x=m`

`<=>x(m-2)=m`

PT có nghiệm duy nhất

`<=>m-2 ne 0<=>m ne 2`

PT vô nghiệm

`<=>m-2=0,m ne 0`

`<=>m=2`

PT có vô số nghiệm

`<=>m=2,m=2` vô lý.

Ta có : 17 - 14(x + 1) = 13 - 4(x + 1) - 5(x - 3)

<=> 17 - 14x - 14 = 13 - 4x - 4 - 5x + 15

<=> -14x + 3 = -9x + 24

<=> -14x + 9x = 24 - 3

<=> -5x = 21

=> x = -4,2

Ta có :  5x + 3,5 + (3x - 4) = 7x - 3(x - 0,5)

<=>  5x + 3,5 + 3x - 4 = 7x - 3x + 1,5 

<=> 8x - 0,5 = 4x + 1,5

=> 8x - 4x = 1,5 + 0,5

=> 4x = 2

=> x = \(\frac{1}{2}\)

17-14(x+1)=13-4(x+1)-5(x-3)