a/ \(x=\dfrac{-5}{12}\)

b/ \(x\approx-1,9526\)

c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)

d/ \(x=\dfrac{-20}{13}\)

a) (x-2)³+6(x+1)²-x³+12=0

⇒ x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

⇒ x³-6x²+12x-8+6x²+12x+6-x³+12=0

⇒ 24x+10=0

⇒ 24x=-10

⇒ x=-5/12

a.

PT \(\Leftrightarrow x^3-6x^2+12x-8+6(x^2+2x+1)-x^3+12=0\)

\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x^3+12=0\)

\(\Leftrightarrow 24x+10=0\Leftrightarrow x=\frac{-5}{12}\)

b. Bạn xem lại đề, nghiệm khá xấu không phù hợp với mức độ tổng thể của bài.

c.

PT $\Leftrightarrow (4x^2+12x+9)+(x^2-1)=5(x^2+4x+4)+(x^2-4x-5)+9(x^2+6x+9)$
$\Leftrightarrow 10x^2+42x+64=0$

$\Leftrightarrow x^2+(3x+7)^2=-15< 0$ (vô lý) 

Do đó pt vô nghiệm.

d.

PT $\Leftrightarrow (1-6x+9x^2)-(9x^2-17x-2)=(9x^2-16)-9(x^2+6x+9)$

$\Leftrightarrow 11x+3=-54x-97$

$\Leftrightarrow 65x=-100$

$\Leftrightarrow x=\frac{-20}{13}$

a: Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)

\(\Leftrightarrow10x=20\)

hay x=2

b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow x^3-1-x^3+4x=5\)

\(\Leftrightarrow4x=6\)

hay \(x=\dfrac{3}{2}\)

\(\Leftrightarrow x^2-6x+9-4x^2-4x-1-2\left(x^2+x-2\right)=3\left(x-3\right)-\left(4x^2+8x-x-2\right)\)

\(\Leftrightarrow-3x^2-10x+8-2x^2-2x+4=3\left(x-3\right)-4x^2-7x+2\)

\(\Leftrightarrow-5x^2-12x+12=3x-9-4x^2-7x+2\)

\(\Leftrightarrow-5x^2-12x+12=-4x^2-4x-7\)

\(\Leftrightarrow-4x^2-4x-7+5x^2+12x-12=0\)

\(\Leftrightarrow x^2+8x-19=0\)

\(\text{Δ}=8^2-4\cdot1\cdot\left(-19\right)=76+64=140\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-8-2\sqrt{35}}{2}=-4-\sqrt{35}\\x_2=-4+\sqrt{35}\end{matrix}\right.\)

\(\Leftrightarrow x^3-3x^2+3x-1-2x^2+8x-8=\left(3x+2\right)^3-3x^2-6x-3-x^2+3x-2\)

\(\Leftrightarrow x^3-5x^2+11x-9=\left(3x+2\right)^3-4x^2-3x-5\)

\(\Leftrightarrow x^3-5x^2+11x-9+4x^2+3x+5-27x^3-54x^2-36x-8=0\)

\(\Leftrightarrow-26x^3-55x^2-22x-12=0\)

Bạn xem lại đề nhé, nghiệm rất xấu

\(8,1-\left(x-6\right)=4\left(2-2x\right)\)

\(\Leftrightarrow1-x+6=8-8x\)

\(\Leftrightarrow-x+8x=8-1-6\)

\(\Leftrightarrow7x=1\)

\(\Leftrightarrow x=\dfrac{1}{7}\)

\(9,\left(3x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)

\(10,\left(x+3\right)\left(x^2+2\right)=0\)

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

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

`8)1-(x-5)=4(2-2x)`

`<=>1-x+5=8-6x`

`<=>5x=2<=>x=2/5`

`9)(3x-2)(x+5)=0`

`<=>[(x=2/3),(x=-5):}`

`10)(x+3)(x^2+2)=0`

  Mà `x^2+2 > 0 AA x`

 `=>x+3=0`

`<=>x=-3`

`11)(5x-1)(x^2-9)=0`

`<=>(5x-1)(x-3)(x+3)=0`

`<=>[(x=1/5),(x=3),(x=-3):}`

`12)x(x-3)+3(x-3)=0`

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

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

`13)x(x-5)-4x+20=0`

`<=>x(x-5)-4(x-5)=0`

`<=>(x-5)(x-4)=0`

`<=>[(x=5),(x=4):}`

`14)x^2+4x-5=0`

`<=>x^2+5x-x-5=0`

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

`<=>[(x=-5),(x=1):}`

\(11,=>\left[{}\begin{matrix}5x-1=0\\x^2-9=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\\x=-3\end{matrix}\right.\\ 12,=>\left(x+3\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ 13,=>x\left(x-5\right)-4\left(x-5\right)=0\\ =>\left(x-4\right)\left(x-5\right)=0\\ =>\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

\(14,=>x^2+5x-x-5=0\\ =>x\left(x+5\right)-\left(x+5\right)=0\\ =>\left(x-1\right)\left(x+5\right)=0\\ =>\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

\(\left(x^3-3x^2+3x-1\right)-\left(2x^2+x-2\right)=\left(x^3+6x^2+12x+8\right)-\left(3x^2-6x+3\right)+\left(2x^2+8x+8\right)\)

\(-10x^2-24x-12=0\)

\(5x^2+12x+6=0\)

....

\(a,\left(x+2\right)^2-9=0\\ \Leftrightarrow\left(x+2-3\right)\left(x+2+3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{1;-5\right\}\)

\(b,x^2-2x+1=25\\ \Leftrightarrow\left(x-1\right)^2=25\\ \Leftrightarrow\left(x-1\right)^2-25=0\\ \Leftrightarrow\left(x-1-5\right)\left(x-1+5\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{6;-4\right\}\)

\(c,\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\\ \Leftrightarrow25x^2+10x+1-25x^2+9=30\\ \Leftrightarrow25x^2+10x-25x^2=30-1-9\\ \Leftrightarrow10x=20\\ \Leftrightarrow x=2\\ Vậy\dfrac{ }{ }S=\left\{2\right\}\)

\(d,\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\\ \Leftrightarrow x^3-1-x\left(x^2-4\right)=5\\ \Leftrightarrow x^3-1-x^3+4x=5\\ \Leftrightarrow x^3-x^3+4x=5+1\\ \Leftrightarrow4x=6\\ \Leftrightarrow x=\dfrac{3}{2}\\ Vậy\dfrac{ }{ }S=\left\{\dfrac{3}{2}\right\}\)

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

=>(x-1)(x+5)=0

=>x=1 hoặc x=-5

b: =>(x-1)^2=25

=>x-1=5 hoặc x-1=-5

=>x=-4 hoặc x=6

c: =>25x^2+10x+1-25x^2+9=30

=>10x+10=30

=>x+1=3

=>x=2

d: =>x^3-1-x(x^2-4)=5

=>x^3-1-x^3+4x=5

=>4x=6

=>x=3/2

\(P\ge\dfrac{1}{2}\left(\dfrac{1}{x-2}+\dfrac{1}{3-x}\right)^2+\dfrac{4}{\left(x-2+3-x\right)^2}=\dfrac{1}{2}\left(\dfrac{1}{x-2}+\dfrac{1}{3-x}\right)^2+4\)

\(P\ge\dfrac{1}{2}\left(\dfrac{4}{x-4+3-x}\right)^2+4=12\)

Dấu "=" xảy ra khi \(x-2=3-x\Rightarrow x=\dfrac{5}{2}\)

a) \(x-2=\left(x-2\right)^2\)

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

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

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

\(\Rightarrow x-2=0\) hoặc \(x-3=0\)

*) \(x-2=0\)

\(x=2\)

*) \(x-3=0\)

\(x=3\)

Vậy \(x=2;x=3\)

b) \(x+5=2\left(x+5\right)^2\)

\(2\left(x+5\right)^2-\left(x+5\right)=0\)

\(\left(x+5\right)\left[2\left(x+5\right)-1\right]=0\)

\(\left(x+5\right)\left(2x+10-1\right)=0\)

\(\left(x+5\right)\left(2x+9\right)=0\)

\(\Rightarrow x+5=0\) hoặc \(2x+9=0\)

*) \(x+5=0\)

\(x=-5\)

*) \(2x+9=0\)

\(2x=-9\)

\(x=-\dfrac{9}{2}\)

Vậy \(x=-5;x=-\dfrac{9}{2}\)

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

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

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

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

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

\(\Rightarrow2x-1=0\) hoặc \(x^2+2=0\)

*) \(2x-1=0\)

\(2x=1\)

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

*) \(x^2+2=0\) 

\(x^2=-2\) (vô lí)

Vậy \(x=\dfrac{1}{2}\)

d) Sửa đề:

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

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

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

\(\left(x+1\right)\left(x^2+4\right)=0\)

\(\Rightarrow x+1=0\) hoặc \(x^2+4=0\)

*) \(x+1=0\)

\(x=-1\)

*) \(x^2+4=0\)

\(x^2=-4\) (vô lí)

Vậy \(x=-1\)