\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

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

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

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

a: Sửa đề: \(x^2+\left(x-1\right)^2=\left(2x-1\right)^2+2\)

Đặt x=a; x-1=b

=>\(a^2+b^2=\left(a+b\right)^2+2\)

=>2ab+2=0

=>ab+1=0

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

=>x²-x+1=0

hay \(x\in\varnothing\)

b: Đặt x-2=a; 3x+2=b

=>\(a^3+b^3=\left(a+b\right)^3\)

\(\Leftrightarrow3ab\left(a+b\right)=0\)

=>4x(x-2)(3x+2)=0

hay \(x\in\left\{0;2;-\dfrac{2}{3}\right\}\)

pt<=>\(\sqrt{\left(x+6\right)^3}+\sqrt{x+6}=\left(x^2+4x\right)^3+x^2+4x\)

đặt\(\sqrt{x+6}=a;x^2+4x=b\)

=>\(\dfrac{3x^3-9x^2+9x-2x^3+2x^2-6x}{\left(x^2-3x+3\right)\left(x^2-x+3\right)}=-1\)

=>x^3-7x^2+3x=-[(x^2+3)^2-4x(x^2+3)+3x^2]

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

=>x^3-4x^2+3x+x^4+6x^2+9-4x^3-12x=0

=>x^4-3x^3+2x^2-9x+9=0

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

=>x=3;x=1

`|5x| = - 3x + 2`

Nếu `5x>=0<=> x>=0` thì phương trình trên trở thành :

`5x =-3x+2`

`<=> 5x +3x=2`

`<=> 8x=2`

`<=> x= 2/8=1/4` ( thỏa mãn )

Nếu `5x<0<=>x<0` thì phương trình trên trở thành :

`-5x = -3x+2`

`<=>-5x+3x=2`

`<=> 2x=2`

`<=>x=1` ( không thỏa mãn ) 

Vậy pt đã cho có nghiệm `x=1/4`

__

`6x-2<5x+3`

`<=> 6x-5x<3+2`

`<=>x<5`

Vậy bpt đã cho có tập nghiệm `x<5`

1/ ( x-3) ²=16

\(\Rightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)

2/ (3x-1)³=8

\(\Rightarrow3x-1=2\\ \Rightarrow3x=3\\ \Rightarrow x=1\)

3/ (x-11)³=-27

\(\Rightarrow x-11=-3\\ \Rightarrow x=8\)

phần 4 mình ko rõ đề

4) \(x^3-3x^2+3x-1=-64\)

\(\Rightarrow x^3-3x^2+3x+63=0\\ \Rightarrow\left(x^3+3x^2\right)-\left(6x^2+18x\right)+\left(21x+63\right)=0\\ \Rightarrow x^2\left(x+3\right)+6x\left(x+3\right)+21\left(x+3\right)=0\\ \Rightarrow\left(x+3\right)\left(x^2+6x+21\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+3=0\\x^2+6x+21=0\end{matrix}\right.\)

\(x+3=0\\ \Rightarrow x=-3\)

\(x^2+6x+21=0\\ \Rightarrow\left(x^2+6x+9\right)+12=0\\ \Rightarrow\left(x+3\right)^2+12=0\)

Vì \(\left(x+3\right)^2\ge0;12>0\Rightarrow\left(x+3\right)^2+12>0\Rightarrow x^2+6x+21vônghiệm\)

Vậy \(x=-3\)

1:

a: =>3x=6

=>x=2

b: =>4x=16

=>x=4

c: =>4x-6=9-x

=>5x=15

=>x=3

d: =>7x-12=x+6

=>6x=18

=>x=3

2:

a: =>2x<=-8

=>x<=-4

b: =>x+5<0

=>x<-5

c: =>2x>8

=>x>4