\(\left(x-2\right)\left(x^2+6x-11\right)^2=\left(5x^2-10x+1\right)^2\) \(\Rightarrow x>2\)

\(\Rightarrow x^2+6x-11>0\)

\(pt\Leftrightarrow x-2=\left(\dfrac{5x^2-10x+1}{x^2+6x-11}\right)^2\Leftrightarrow\sqrt{x-2}=\dfrac{5x^2-10x+1}{x^2+6x-11}\)

\(\Leftrightarrow\sqrt{x-2}-1=\dfrac{5x^2-10x+1}{x^2+6x-11}-1=\dfrac{4x^2-16x+12}{x^2+6x+12}\)

\(\Leftrightarrow\dfrac{x-3}{\sqrt{x-2}+1}=\dfrac{4\left(x-1\right)\left(x-3\right)}{x^2+6x-11}\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\\\dfrac{1}{\sqrt{x-2}+1}=\dfrac{4\left(x-1\right)}{x^2+6x-11}\left(1\right)\end{matrix}\right.\)

Xét (1):

\(x^2+6x-11=4\left(x-1\right)+4\left(x-1\right)\sqrt{x-2}\)

\(\Leftrightarrow x^2+2x-7-4\left(x-1\right)\sqrt{x-2}=0\)

\(\Leftrightarrow x^2-2x+1-2\left(x-1\right)\sqrt{4x-8}+4x-8=0\)

\(\Leftrightarrow\left(x-1\right)^2-2\left(x-1\right)\sqrt{4x-8}+\left(\sqrt{4x-8}\right)^2=0\)

\(\Leftrightarrow\left(x-1-\sqrt{4x-8}\right)^2=0\)

\(\Leftrightarrow x-1=\sqrt{4x-8}\)

\(\Leftrightarrow x^2-2x+1=4x-8\)

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

Vậy pt đã cho có nghiệm duy nhất \(x=3\)

Đặt \(y=x-2\), phương trình đã cho trở thành:

\( y{\left[ {{{\left( {y + 2} \right)}^2} + 6\left( {y + 2} \right) - 11} \right]^2} = {\left[ {5{{\left( {y + 2} \right)}^2} - 10\left( {y + 2} \right) + 1} \right]^2}\\ \Leftrightarrow y{\left( {{y^2} + 10y + 5} \right)^2} = {\left( {5{y^2} + 10y + 1} \right)^2}\\ \Leftrightarrow {y^5} - 5{y^4} + 10{y^3} - 10{y^2} + 5y - 1 = 0 \Leftrightarrow {\left( {y - 1} \right)^5} = 0 \Leftrightarrow y = 1 \)

Với \(y=1\) ta có \(x-2=1\) \(\Rightarrow x=3\)

Vậy \(x = 3 \)

a: =>(x^2+4x-5)(x^2+4x-21)=297

=>(x^2+4x)^2-26(x^2+4x)+105-297=0

=>x^2+4x=32 hoặc x^2+4x=-6(loại)

=>x^2+4x-32=0

=>(x+8)(x-4)=0

=>x=4 hoặc x=-8

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

hay \(x\in\left\{\dfrac{1+\sqrt{13}}{2};\dfrac{1-\sqrt{13}}{2};\dfrac{-1+\sqrt{17}}{2};\dfrac{-1-\sqrt{17}}{2}\right\}\)

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

hay \(x\in\left\{1;-2;3+\sqrt{11};3-\sqrt{11}\right\}\)

\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)

= \(\frac{3x\left(x-y\right)}{5.2.\left(x+y\right)\left(x-y\right)}-\frac{x\left(x+y\right)}{10\left(x^2-y^2\right)}\)

= \(\frac{3x^2-3xy-x^2-xy}{10\left(x^2-y^2\right)}\)

= \(\frac{3x\left(x-y\right)}{10\left(x^2-y^2\right)}\)

= \(\frac{3x}{10\left(x+y\right)}\)

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

`<=>[(2x^2-5x+3=0),(x^2-4x+3=0):}<=>[(x=3/2),(x=1),(x=3):}`

  `=>A={3/2;1;3}`

`b)(x^2-10x+21)(x^3-x)=0`

`<=>[(x^2-10x+21=0),(x^3-x=0):}<=>[(x=7),(x=3),(x=0),(x=+-1):}`

   `=>B={0;+-1;3;7}`

`c)(6x^2-7x+1)(x^2-5x+6)=0`

`<=>[(6x^2-7x+1=0),(x^2-5x+6=0):}<=>[(x=1),(x=1/6),(x=2),(x=3):}`

    `=>C={1;1/6;2;3}`

`d)2x^2-5x+3=0<=>[(x=1),(x=3/2):}`   Mà `x in Z`

    `=>D={1}`

`e){(x+3 < 4+2x),(5x-3 < 4x-1):}<=>{(x > -1),(x < 2):}<=>-1 < x < 2`

    Mà `x in N`

   `=>E={0;1}`

`f)|x+2| <= 1<=>-1 <= x+2 <= 1<=>-3 <= x <= -1`

      Mà `x in Z`

  `=>F={-3;-2;-1}`

`g)x < 5`  Mà `x in N`

   `=>G={0;1;2;3;4}`

`h)x^2+x+3=0` (Vô nghiệm)

   `=>H=\emptyset`.

\(\left(x-2\right)^3-\left(x+5\right)\left(x^2-5x+25\right)+6x^2=11\)

=>\(x^3-6x^2+12x-8-\left(x^3+125\right)+6x^2=11\)

=>\(x^3+12x-8-x^3-125=11\)

=>12x-133=11

=>12x=144

=>\(x=\dfrac{144}{12}=12\)

một lượt tối đa 2 câu làm vậy có thánh nào dmas beensg tới