a: =>5x-5+17x=1-12x-4

=>22x-5=-12x-3

=>34x=2

hay x=1/17

b: =>\(\left(x-3\right)^2-4x\left(x-3\right)=0\)

=>(x-3)(-3x-3)=0

=>x=3 hoặc x=-1

c: =>(x-4)(x-6)=0

=>x=4 hoặc x=6

a) \(\left(x+1+\dfrac{1}{x}\right)^2=\left(x-1-\dfrac{1}{x}\right)^2\\ \Leftrightarrow\left(x+1+\dfrac{1}{x}\right)^2-\left(x-1-\dfrac{1}{x}\right)^2=0\\ \Leftrightarrow\left(x+1+\dfrac{1}{x}-x+1+\dfrac{1}{x}\right)\left(x+1+\dfrac{1}{x}+x-1-\dfrac{1}{x}\right)=0\\ \Leftrightarrow2\left(1+\dfrac{1}{x}\right)\cdot2x=0\\ \Leftrightarrow4x\left(1+\dfrac{1}{x}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

\(S=\left\{-1;0\right\}\) là tập nghiệm của pt.

b) Ta có: \(\left(x-1\right)^2+3x^2=0\)

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

\(\Leftrightarrow4x^2-2x+1=0\)

\(\text{Δ}=\left(-2\right)^2-4\cdot4\cdot1=4-16=-12< 0\)

=> Phương trình vô nghiệm

Vậy: \(S=\varnothing\)

`a)(x+6)(3x-1)=(x-6)(x+6)`

`<=>(x+6)(3x-1+6-x)=0`

`<=>(x+6)(2x+5)=0`

`<=>[(x=-6),(x=-5/2):}`

`b)(x+1)^2=(2x+3)^2`

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

`<=>(x+1-2x-3)(x+1+2x+3)=0`

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

`<=>[(x=-2),(x=-4/3):}`

a)

`(x+6)(3x-1)=(x-6)(x+6)`

`<=> (x+6)(3x-1)-(x-6)(x+6)=0`

`<=> (x+6)(3x-1-x+6)=0`

`<=> (x+6)(2x+5)=0`

\(< =>\left[{}\begin{matrix}x+6=0\\2x+5=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=-6\\x=-\dfrac{5}{2}\end{matrix}\right.\)

b)

`(x+1)^2 =(2x+3)^2`

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

`<=> (x+1-2x-3)(x+1+2x+3)=0`

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

\(< =>\left[{}\begin{matrix}-x-2=0\\3x+4=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=-2\\x=-\dfrac{4}{3}\end{matrix}\right.\)

`a)(x+6)(3x-1)=(x-6)(x+6)`

`<=> (x+6)(3x-1-x +6) =0`

`<=> (x+6)(2x+5)=0`

`<=> [(x+6=0),(2x=-5):}`

`<=>[(x=-6),(x=-5/2):}`

`b)(x+1)^2=(2x+3)^2`

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

`<=> (x+1 -2x -3)(x+1+2x +3) =0`

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

`<=> (x+2)(3x +4) =0`

`<=> [(x =-2),(x =-4/3):}`

Ta có (\(^{x^{2^{ }}^{ }+3x}\)) (\(^{x^{2^{ }}+3x+4}\))

Đặt \(x^{2^{ }^{ }}+3x\) là a ta có

a.(a+4)=-4

4a+\(a^2\) -4=0

\(^{ }\left(a-2\right)^2\)=0

Suy ra a=2

hay \(x^{2^{ }^{ }^{ }}+3x=2\)

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

𝑥=−3±17√/2

a.\(\left(x^2+2x+5\right)\left(x^2+4x\right)=0\)

Ta có: \(x^2+2x+5=x^2+2x+1+4=\left(x+1\right)^2+4\ge4>0;\forall x\)

 \(\Rightarrow x^2+4x=0\)

\(\Leftrightarrow x\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x=3\end{matrix}\right.\)

c.\(1,2x^3-x^2-0,2x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-\dfrac{1}{6}\end{matrix}\right.\)

\(a)PT\Leftrightarrow4x^2-9-4x^2+20x+3x=0.\\ \Leftrightarrow23x=9.\\ \Leftrightarrow x=\dfrac{9}{23}.\\ b)PT\Leftrightarrow\left(2x+1\right)\left(4x-3\right)-\left(2x+1\right)\left(2x-1\right)=0.\\\Leftrightarrow\left(2x+1\right)\left(4x-3-2x+1\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)=0. \)

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

a) ĐKXĐ: \(x\ne3\)

Ta có: \(\dfrac{x^2-x-6}{x-3}=0\)

\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x-3\right)}{x-3}=0\)

Suy ra: x+2=0

hay x=-2(thỏa ĐK)

Vậy: S={-2}

d)

ĐKXĐ: \(x\notin\left\{1;3\right\}\)

Ta có: \(\dfrac{x+5}{x-1}=\dfrac{x+1}{x-3}-\dfrac{8}{x^2-4x+3}\)

\(\Leftrightarrow\dfrac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}-\dfrac{8}{\left(x-1\right)\left(x-3\right)}\)

Suy ra: \(x^2-3x+5x-15=x^2-1-8\)

\(\Leftrightarrow2x-15+9=0\)

\(\Leftrightarrow2x-6=0\)

hay x=3(loại)

Vậy: \(S=\varnothing\)

a,\(x\in\left\{5;1,5;\dfrac{-4}{3}\right\}\)