= x ^1 +100 =  x^101

\(x\times x^2\times x^3\times x^4\times...\times x^{100}=x^{1+2+3+4+...+100}=x^{101\times500}=x^{5050}\)

\(x\times x^2\times x^3\times x^4\times...\times x^{100}\)

\(=x^{\left(1+100\right)\times100\div2}\)

\(=x^{5050}\)

nha

a) -12.(x-5)+7.(3.x)=5

<=> -12x+60+21+7x=5

<=>-5x+81=5

<=>-5x=5-81=-76

<=>x=-76/-5=76/5=15,2

b) 30.(x+2)-6.(x-5)-24.x=100

<=> 30x+60-6x+30-24x=100

<=> 0x=100-60-30=10

=> không có giá trị nào của x để 0x=10

c) \(|5.x-2|< 13\)

Khi 5x-2 < 13

<=> 5x<15 <=> x<3

Khi 5x-2 <-13

<=> 5x<-11 <=> x<-11/5 <=> x<-2,2

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

\(\left(x-3\right)^2=9+16\)

\(\left(x-3\right)^2=25\)

\(\left(x-3\right)^2=5^2\)

\(\Rightarrow x-3=5\)

\(\Rightarrow x=5+3=8\)

Vậy x = 8

b) -12 . (x - 5) +7 . (3 - x) = 5

-12x + 60 + 21 - 7x = 5

-12x - 7x = 5 - 60 - 21

-19x = -76

x = -76 : (-19) =4

Vậy x = 4

c) 30 . (x + 2) - 6 . ( x - 5) - 24x = 100

30x + 60 - 6x + 30 - 24x =100

30x - 6x - 24x = 100 - 30 - 60

0x = 10

x=10 : 0

\(\Rightarrow x\in\varnothing\)

1.

a) \(5x.5x.5x=\left(5x\right)^3.\)

b) \(x^1.x^2.....x^{2006}=x^{\frac{\left(2006+1\right).2006}{2}=}x^{2013021}.\)

c) \(x^1.x^4.x^7.....x^{100}=x^{\frac{\left(100+1\right).\left(\frac{100-1}{3}+1\right)}{2}}=x^{1717}.\)

d) \(x^2.x^5.x^8.....x^{2003}=x^{\frac{\left(2003+2\right).\left(\frac{2003-2}{3}+1\right)}{2}}=x^{669670}.\)

2.

\(2^x+80=3^y\)

Với \(x>0\Rightarrow2^x\) chẵn

Và 80 chẵn

\(\Rightarrow2^x+80\) chẵn.

Mà \(3^y\) lẻ

\(\Rightarrow x< 0.\)

Mà \(x\in N\)

\(\Rightarrow x=0.\)

\(\Rightarrow2^0+80=3^y\)

\(\Rightarrow1+80=3^y\)

\(\Rightarrow3^y=81\)

\(\Rightarrow3^y=3^4\)

\(\Rightarrow y=4.\)

Vậy \(\left(x;y\right)=\left(0;4\right).\)

Chúc bạn học tốt!

1.viết tích dưới dạng lũy thừa

a.5x.5x.5x

=(5x)\(^3\)

b.x\(^1\) . x \(^2\).......x \(^{2006}\)

=x \(^{2013021}\)

c.x\(^1\).x \(^4\) .x \(^7\)......x \(^{100}\)

=x \(^{1717}\)

d.x \(^2\) .x \(^5\).x \(^8\).......x\(^{2003}\)

=x \(^{669670}\)

2 \(^x\)+80=3\(^y\)

ta có 3\(^y\)lẻ nên 2 \(^x\)+80 lẻ .

Với x \(\ge\)1

thì 2 \(^x\)+80 chẵn không thỏa mãn

nên x=0

=>1+80=3\(^y\)

=>81=3\(^y\)

=>3\(^4\)=3\(^y\)

=>y=4

a: \(=12x^2-9x-12x^2-10x+6x+5=-13x+5\)

b: \(=3x\left(x^2-2x+1\right)-2x\left(x^2-9\right)+4x^2-16x\)

\(=3x^3-6x^2+3x-2x^3+18x+4x^2-16x\)

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

c: \(=x^3-3x^2+3x-1+x^3+8+3\left(x^2-16\right)\)

\(=2x^3-3x^2+3x+7+3x^2-48=2x^3+3x-41\)

d: \(=\left(x^3+1\right)\left(x^3-1\right)=x^6-1\)

?? .......fmdnfhxvfvhzyv fhddddgnfhgmjvhnjghytdthtanththnghng nhng 

a: =>2x-2/3-3x+3/2=1/2x

=>-3/2x=-5/6

=>x=5/6:3/2=5/6x2/3=10/18=5/9

b: =>-3x+3/4-1/3x-1/6=x

=>-13/3x=-7/12

=>x=7/12:13/3=7/12x3/13=21/156=7/52

\(A\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\\\Leftrightarrow \left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\\ \Leftrightarrow\left(x-1\right)^{x+2}\left(\left(x-1\right)^{x+2}+1\right)=0\\ \Leftrightarrow\left(x-1\right)^{x+2}=0hoac\left(x-1\right)^{x+2}+1=0\)

Giả tiếp đc x=1