(x-1)^3-(x+3)(x^2-3x+9)+3(x^2-4)=2

=>x^3-3x^2+3x-1-x^3-27+3x^2-12=2

=>3x-40=2

=>x=42/3=14

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

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

$\Leftrightarrow 65x=-100$

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

\(\left|x-1\right|+2\left|x-2\right|+3\left|x-3\right|+4\left|x-4\right|+5\left|x-5\right|+20x=0\left(1\right)\)

TH1: x<1

(1) trở thành 1-x+2(2-x)+3(3-x)+4(4-x)+5(5-x)+20x=0

=>\(1-x+4-2x+9-3x+16-4x+25-5x+20x=0\)

=>\(5x+55=0\)

=>x=-11(nhận)

TH2: 1<=x<2

Phương trình (1) sẽ trở thành:

\(x-1+2\left(2-x\right)+3\left(3-x\right)+4\left(4-x\right)+5\left(5-x\right)+20x=0\)

=>\(x-1+4-2x+9-3x+16-4x+25-5x+20x=0\)

=>\(7x+53=0\)

=>\(x=-\dfrac{53}{7}\left(loại\right)\)

TH3: 2<=x<3

Phương trình (1) sẽ trở thành:

\(x-1+2\left(x-2\right)+3\left(3-x\right)+4\left(4-x\right)+5\left(5-x\right)+20x=0\)

=>\(x-1+2x-4+9-3x+16-4x+25-5x+20x=0\)

=>\(11x+45=0\)

=>\(x=-\dfrac{45}{11}\left(loại\right)\)

TH4: 3<=x<4

Phương trình (1) sẽ trở thành:

\(x-1+2\left(x-2\right)+3\left(x-3\right)+4\left(4-x\right)+5\left(5-x\right)+20x=0\)

=>\(x-1+2x-4+3x-9+16-4x+25-5x+20x=0\)

=>\(-3x+27=0\)

=>x=9(loại)

TH5: 4<=x<5

Phương trình (1) sẽ trở thành:

\(\left(x-1\right)+2\left(x-2\right)+3\left(x-3\right)+4\left(x-4\right)+5\left(5-x\right)+20x=0\)

=>\(x-1+2x-4+3x-9+4x-16+25-5x+20x=0\)

=>\(25x-5=0\)

=>x=1/5(loại)

TH6: x>=5

Phương trình (1) sẽ trở thành:

\(\left(x-1\right)+2\left(x-2\right)+3\left(x-3\right)+4\left(x-4\right)+5\left(x-5\right)+20x=0\)

=>\(x-1+2x-4+3x-9+4x-16+5x-25+20x=0\)

=>35x-55=0

=>x=55/35(loại)

Ta có: \(\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+...+\left(x+99\right)=0\)

\(\dfrac{\left[\left(x+1\right)+\left(x+99\right)\right].50}{2}=0\)

\(\left(x+50\right).50=0\)

\(x+50=0\)

\(x=-50\)

\(\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+...+\left(x+99\right)=0\)

Có tất cả số hạng là

\(\left(99-1\right):2+1=50số\)

Ta có: \(\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+...+\left(x+99\right)=0\)

hay: \(\left(x+50\right).50=0\)

\(x+50=0\)

\(=>x=-50\)

`(x+3)^2014 = (x+3)^2012`

`(x+3)^2014 -(x+3)^2012 =0`

`(x+3)^2012 [(x+3)^2 -1]=0`

TH1 :`(x+3)^2012 =0 => x+3 =0 => x=-3`

TH2 :`(x+3)^2 -1 =0 => (x+3)^2 =1 => [(x+3=1),(x+3=-1):}`

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

`(x+3)^2014 = (x+3)^2012`

`=> (x+3)^2014 - (x+3)^2012 = 0`

`=> (x+3)^2 * (x+3)^2012 - (x+3)^2012 = 0`

`=> (x+3)^2012 * [ (x+3)^2 - 1] =0`

`=>`\(\left[{}\begin{matrix}\left(x+3\right)^{2012}=0\\\left(x+3\right)^2-1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x+3=0\\\left(x+3\right)^2=1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=-3\\x+3=1\\x+3=-1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=-3\\x=-2\\x=-4\end{matrix}\right.\)

Vậy, `x = {-3; -2; -4}.`

Từ gt =>X² = lXl²

=>X TÙY Ý

mình nghĩ zậy(ko chắc)

Đặt \(t=\left|x\right|\) , \(t\ge0\)

PT đã cho trở thành :

\(t^2\left(2t-3\right)=t^2.\left(2t-3\right)\)

Vì \(x^2=\left|x\right|^2\Rightarrow t^2=t^2\) nên từ đó suy ra pt trên vô số nghiệm.

a: \(\left(\sqrt{3}\right)^x=243\)

=>\(3^{\dfrac{1}{2}\cdot x}=3^5\)

=>\(\dfrac{1}{2}\cdot x=5\)

=>x=10

b: \(0,1^x=1000\)

=>\(\left(\dfrac{1}{10}\right)^x=1000\)

=>\(10^{-x}=10^3\)

=>-x=3

=>x=-3

c: \(\left(0,2\right)^{x+3}< \dfrac{1}{5}\)

=>\(\left(0,2\right)^{x+3}< 0,2\)

=>x+3>1

=>x>-2

d: \(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{5}{3}\right)^2\)

=>\(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{3}{5}\right)^{-2}\)

=>2x+1<-2

=>2x<-3

=>\(x< -\dfrac{3}{2}\)

e: \(5^{x-1}+5^{x+2}=3\)

=>\(5^x\cdot\dfrac{1}{5}+5^x\cdot25=3\)

=>\(5^x=\dfrac{3}{25,2}=\dfrac{1}{8,4}=\dfrac{10}{84}=\dfrac{5}{42}\)

=>\(x=log_5\left(\dfrac{5}{42}\right)=1-log_542\)

\(\left(x-1\right)^3+\left(x-3\right)^3=\left(2x-4\right)^3\)

\(\Leftrightarrow\left(x-1\right)^3+\left(x-3\right)^3-\left(2x-4\right)^3=0\)

\(\Leftrightarrow\left(x-1\right)^3+\left(x-3\right)^3+\left(4-2x\right)^3=0\)

Đặt \(\left(x-1\right)=a;\left(x-3\right)=b;\left(4-2x\right)=c\)ta có:

\(a^3+b^3+c^3=0\)

\(\Leftrightarrow\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)^3=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

Thay lại, ta được:

\(\left(a+b+c\right)^3=\left(x-1+x-3+4-2x\right)^3=0\)

\(\Rightarrow3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Leftrightarrow\left(x-1+x-3\right)\left(x-3+4-2x\right)\left(4-2x+x-1\right)=0\)

\(\Leftrightarrow\left(2x-4\right)\left(1-x\right)\left(3-x\right)=0\)

\(\Leftrightarrow2x-4=0\Leftrightarrow2x=4\Rightarrow x=2\)

hoặc \(1-x=0\Leftrightarrow x=1\)

hay \(3-x=0\Leftrightarrow x=3\)

Vậy \(x\in\left\{1;2;3\right\}\). Xong :))