Hùng còn lại số tiền là:

     45 000x 4/5 = 36 000 ( đồng)

Hùng có tất cả số tiền là:

             45 000 + 36000 = 81 000 ( đồng)

                                     Đáp số:..............

Số tiền Hùng còn lại là :

`45 000xx 4/5 = 36 000 ( đồng)`

Số tiền mà lúc đầu Hùng có :

`45 000 + 36000 = 81 000 ( đồng)`

Đs

a, \(2A=2+2^3+2^4+...+2^{64}\Rightarrow2A-A=2+2^3+...+2^{64}-1-2^2-2^3-...-2^{63}=1-2^2+2^{64}=-3+2^{64}\)

b, \(3B=3+3^2+...+3^{2001}\Rightarrow3B-B=3+...+3^{2001}-1-3-...-3^{2000}=-1+3^{2001}\Rightarrow B=\dfrac{-1+3^{2001}}{2}\)

c, tương tự 

\(A(x)+B(x)=3x^4-\dfrac{3}{4}x^3+2x^2-3+8x^4+\dfrac{1}{5}x^3-9x+\dfrac{2}{5}\)

         \(=11x^4-\dfrac{11}{20}x^3+2x^2-9x-\dfrac{13}{5}\)

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\(A(x)-B(x)=3x^4-\dfrac{3}{4}x^3+2x^2-3-8x^4-\dfrac{1}{5}x^3+9x-\dfrac{2}{5}\)

         \(=-5x^4-\dfrac{19}{20}x^3+2x^2+9x-\dfrac{17}{5}\)

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\(B(x)-A(x)=8x^4+\dfrac{1}{5}x^3-9x+\dfrac{2}{5}-3x^4+\dfrac{3}{4}x^3-2x^2+3\)

         \(=5x^4+\dfrac{19}{20}x^3-2x^2-9x+\dfrac{17}{5}\)

Bài 1 

\(A\left(x\right)+B\left(x\right)=3x^4-\dfrac{3}{4}x^3+2x^2-3+8x^4+\dfrac{1}{5}x^3-9x+\dfrac{2}{5}=11x^4-\dfrac{11}{20}x^3+2x^2-9x-\dfrac{13}{5}\)

\(A\left(x\right)-B\left(x\right)=-5x^4-\dfrac{19}{20}x^3+2x^2+9x-\dfrac{17}{5}\)

\(B\left(x\right)-A\left(x\right)=5x^4+\dfrac{19}{20}x^3-9x-2x^2+\dfrac{17}{5}\)

a,3/4+2/5+1/2

= 23/20 + 1/2

= 43/20

b,4/9×3/8+1/4

= 1/6 x 1/4 

= 1/24

c,3/10 +2/5:4/15

= 3/10 + 3/2

= 9/5

d,4/7×3+2/3

= 12/7 + 2/3

= 50/21

`3/4 + 2/5 + 1/2 = 15/20 + 8/20 + 10/20 =(15+8+10)/20 = 43/20`

`4/9 . 3/8 + 1/4 = 1/6 + 1/4 =2/12 + 3/12= (2+3)/12= 5/12`

`3/10 + 2/5 : 4/15 = 3/10 + 2/5 xx 15/4 = 3/10 + 3/2 =3/10 + 15/10 = (3+15)/10= 18/10 = 9/5`

`4/7 . 3 + 2/3 = 12/7 + 2/3 = 50/21`

Hộp 1 : Thêm 3

Hộp 2 : thêm 3 

 -> 30 gấp đôi 15 =))

Hộp 1 :  3

Hộp 2 :  3 

 30 gấp đôi 15 

b, \(5^{n-2}=5^4\Rightarrow n-2=4\Leftrightarrow n=6\)

c, \(N^{50}-N=0\Leftrightarrow N\left(N^{49}-1\right)=0\Leftrightarrow N=0;N=1\)

`b)5^{n}:5^{2}=625`

`5^{n-2}=625`

`5^{n-2}=5^{4}`

`n-2=4`

`x=4+2=6`

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`c)n^{50}=n`

`n^{50}-n=0`

\(n \times (n^{49}-1)=0\)

\(@TH1:n=0\)

\(@TH2:n^{49}-1=0\)

       \(=>n^{49}=1\)

      \(=>n^{49}=1^{49}\)

      \(=>n=1\)

Vậy `n=0` hoặc `n=1`

B( 2) = { 0;2;4;6;8;10;12:.........}

B₍₂₎ = {0;2;4;6;8;10;12;14;...}

`(5^{x})^{2}=125^{3}:5^{2}`

`5^{2x}=(5^{2})^{3}:5^{2}`

`5^{2x}=5^{6}:5^{2}`

`5^{2x}=5^{4}`

`2x=4`

`x=4:2=2`

\(D=\dfrac{1}{1 \times 3}+\dfrac{1}{3 \times 5}+\dfrac{1}{5 \times 7}+...+\dfrac{1}{2007 \times 2009}+\dfrac{1}{2009 \times 2011}\)

\(D=\dfrac{1}{2} \times (\dfrac{2}{1 \times 3}+\dfrac{2}{3 \times 5}+\dfrac{2}{5 \times 7}+...+\dfrac{2}{2007 \times 2009}+\dfrac{2}{2009 \times 2011})\)

\(D=\dfrac{1}{2} \times (1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2009}-\dfrac{1}{2011})\)

\(D=\dfrac{1}{2} \times (1-\dfrac{1}{2011})\)

\(D=\dfrac{1}{2} \times \dfrac{2010}{2011}\)

\(D=\dfrac{1005}{2011}\)

\(D=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+.....+\dfrac{1}{2007.2009}+\dfrac{1}{2009.2011}\)

`=>` \(D=\dfrac{1}{2}.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+....+\dfrac{1}{2007.2009}+\dfrac{1}{2009.2011}\right)\)

\(D=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)

`C=1/2 . (1-1/2011)`

`C=1/2 . (2011/2011- 1/101)`

`C=1/2 . 2010/2011`

`C=1005/2011`