(x-5)^x+1.(1-(x-5)

=>\(\left(x-5\right)^{x+1}\left[1-\left(x-5\right)^{12}\right]=0\)

=>\(\orbr{\begin{cases}\left(x-5\right)^{x+1}=0\\1-\left(x-5\right)^{12}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-5=0\\\left(x-5\right)^{12}=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x-5=\pm1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x\in\left\{6;4\right\}\end{cases}}}\)

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

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

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

\(\Leftrightarrow7x=1\)

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

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

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

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

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

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

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

`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):}`

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

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

1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)

=>-13x=0

=>x=0

2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

=>3x=13

=>x=13/3

3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)

=>-2x^2=0

=>x=0

4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

=>-8x=6-14=-8

=>x=1

`1)2x(x-5)-(3x+2x^2)=0`

`<=>2x^2-10x-3x-2x^2=0`

`<=>-13x=0`

`<=>x=0`

___________________________________________________

`2)x(5-2x)+2x(x-1)=13`

`<=>5x-2x^2+2x^2-2x=13`

`<=>3x=13<=>x=13/3`

___________________________________________________

`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`

`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`

`<=>x=0`

___________________________________________________

`4)5x(x-1)-(x+2)(5x-7)=0`

`<=>5x^2-5x-5x^2+7x-10x+14=0`

`<=>-8x=-14`

`<=>x=7/4`

___________________________________________________

`5)6x^2-(2x-3)(3x+2)=1`

`<=>6x^2-6x^2-4x+9x+6=1`

`<=>5x=-5<=>x=-1`

___________________________________________________

`6)2x(1-x)+5=9-2x^2`

`<=>2x-2x^2+5=9-2x^2`

`<=>2x=4<=>x=2`

a) \(\Rightarrow\dfrac{1}{3}x\left(x-2\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

b) \(\Rightarrow\left(x+5\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)

c) \(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

e) \(\Rightarrow\left(x+2\right)\left(x+2-x+2\right)=0\Rightarrow\left(x+2\right).4=0\Rightarrow x=-2\)

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

g) \(\Rightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left(3x-2\right)\left(3x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)

h) \(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)

i) \(\Rightarrow4x\left(x+1\right)+5\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(4x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{4}\end{matrix}\right.\)

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

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: \(\left(x+50\right).50=0\)

\(x+50=0\)

\(=>x=-50\)

\(\Leftrightarrow\left(x-5\right)^{x+1}\left[1-\left(x-5\right)^{12}\right]=0\)

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

=>(x-5)(6-x)(-4+x)=0

hay \(x\in\left\{5;6;4\right\}\)

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

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

$\Leftrightarrow 65x=-100$

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