Bài làm:

Ta có: \(x=-9\Leftrightarrow-10=x-1\Rightarrow10=1-x\)nên thay vào ta tính:

\(P\left(-9\right)=1+\left(1-x\right)x+\left(1-x\right)x^2+\left(1-x\right)x^3+...+\left(1-x\right)x^{19}+\left(1-x\right)x^{20}\)

\(P\left(-9\right)=1+x-x^2+x^2-x^3+x^3-x^4+...+x^{20}-x^{21}\)

\(P\left(-9\right)=1+x-x^{21}\)

\(P\left(-9\right)=1-9+9^{21}\)

\(P\left(-9\right)=9^{21}-8\)

Vậy khi \(x=-9\)thì \(P\left(x\right)=9^{21}-8\)

Học tốt!!!!

(3-x)^10x:(3-x)²⁰=1

=> (3-x)^10x=(3-x)²⁰

=> 10x=20

=> x=2

\(\frac{\left(3-x\right)^{10x}}{\left(3-x\right)^{20}}=1\Leftrightarrow\left(3-x\right)^{10x-20}=\left(3-x\right)^{10\left(x-2\right)}=1\\ \)

\(\orbr{\begin{cases}x-2=0=>x=2\\3-x=+-1\orbr{\begin{cases}x=2\\x=4\end{cases}}\end{cases}}\)\(\orbr{\begin{cases}x-2=0\Rightarrow x=2\\3-x=!1!\end{cases}}\)\(\orbr{\begin{cases}x=2\\x=4\end{cases}}\)

(3-x)^10x-20=1

Vì a^0=1

=>10x-20=0

10x=20

x=2

a) (3x-15)⁷ = 0

3x-15 = 0

3x = 0+15

3x = 15

x = 15:3

x = 5

b) 4^2x-6 = 1

 2x-6 = 0

2x = 0+6

2x = 6

x = 6:2

x = 3

c) Tớ ko bít 

d) (x - 6)³ = (x - 6)²

Th₁:

x - 6 = 1

x = 1 + 6

x = 7

Th₂:

x - 6 = 0

x = 6

Vậy x = 7

      x = 6

--thodagbun--

a, (3x-15)^7=0 <=> 3x-15=0 <=> x=5

b, 4²x+6=1 <=> 16x=-5 <=>x=-5/16

c, \(\dfrac{\left(3-x\right)^{10x}}{\left(3-x\right)^{20}}=1\Leftrightarrow\left(3-x\right)^{10x-20}=1\)

TH1: 10x-20 = 0 <=> x=2

TH2: 3-x=1 <=> x=2

Vậy x=2

d, (x-6)^3 = (x-6)^2

<=> (x-6)^2.[(x-6)-1]=0

<=> (x-6)^2=0 hoặc (x-6)-1=0

<=> x=6 hoặc x=7