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Lâm Khánh Linh

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Ukm The Moon đề sai đổi \(-3^{100}\) thành \(+3^{100}\)

                                                                Bài giải

\(S=1-3+3^2-3^3+...+3^{99}-3^{100}\)

\(S=\left(1-3\right)+\left(3^2-3^3\right)+...+\left(3^{98}-3^{99}\right)+3^{100}\)

\(S=\left(1-3\right)+3^2\left(1-3\right)+...+3^{98}\left(1-3\right)+3^{100}\)

\(S=\left(1-3\right)\left[1+3^2+3^4+...+3^{98}\right]+3^{100}\)

Đặt \(A=1+3^2+3^4+...+3^{98}\)

\(3^2A=3^2+3^4+...+3^{100}\)

\(3^2A-A=9A-A=8A=3^{100}-1\)

\(A=\frac{3^{100}-1}{8}\)

Thay \(A=\frac{3^{100}-1}{8}\) vào biểu thức ta được :

\(S=\left(1-3\right)\cdot\frac{3^{100}-1}{8}+3^{100}\)

\(S=-2\cdot\frac{3^{100}-1}{8}+3^{100}\)

\(S=-\frac{\left(3^{100}-1\right)}{4}+3^{100}\)

                                                        Bài giải

a, \(3^{450}=\left(3^3\right)^{150}=9^{150}\)

\(5^{300}=\left(5^2\right)^{150}=25^{150}\)

\(\text{Vì }9^{150}< 25^{150}\) \(\Rightarrow\text{ }3^{450}< 5^{300}\)

b, \(333^{444}=\left(333^4\right)^{111}=12296370321^{111}\)

\(444^{333}=\left(444^3\right)^{111}=87528384^{111}\)

Vì  \(12296370321^{111}>87528384^{111}\) \(\Rightarrow\text{ }333^{444}>444^{333}\)

                                                        Bài giải

a, \(3^{450}=\left(3^3\right)^{150}=9^{150}\)

\(5^{300}=\left(5^2\right)^{150}=25^{150}\)

\(\text{Vì }9^{150}< 25^{150}\) \(\Rightarrow\text{ }3^{450}< 5^{300}\)

b, \(333^{444}=\left(333^4\right)^{111}=12296370321^{111}\)

\(444^{333}=\left(444^3\right)^{111}=87528384^{111}\)

Vì  \(12296370321^{111}>87528384^{111}\) \(\Rightarrow\text{ }333^{444}>444^{333}\)

Ta có: \(\left(x-5\right)^4=\left(x-5\right)^6\)

\(\Leftrightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)

\(\Leftrightarrow\left(x-5\right)^4\left[\left(x-5\right)^2-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-5=\pm1\end{cases}}\)

Làm nốt nha 

\(\left(x-5\right)^6=\left(x-5\right)^4\)

\(\Leftrightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)

\(\Leftrightarrow\left(x-5\right)^4\left[\left(x-5\right)^2-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-5=0\\\left(x-5\right)^2=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=6;x=4\end{cases}}\)

Vậy \(x=5;x=6;x=4\)

=7^3.(1+7^2).5^4.(1+5^2).(3^4-3^4)

=7^3.(1+7^2).5^4.(1+5^2).0

=0

\((7^3+7^5)\times(5^4+5^6)\times(3^3\times3-9^2)\)

\(=(7^3+7^5)\times(5^4+5^6)\times0\)

\(=0\)

\(\left(x+1\right)+\left(x+2\right)+...\left(x+20\right)=930\)

\(\Rightarrow\left(x+x+...+x\right)+\left(1+2+...+20\right)=930\)

\(\Leftrightarrow20x+\left(\frac{\left(20+1\right).20}{2}\right)=930\)

\(\Leftrightarrow20x+210=930\)

\(\Leftrightarrow20x=930-210=720\)

\(\Leftrightarrow x=\frac{720}{20}=36\)

Vậy : \(x=36\)

\(\left(x+1\right)+\left(x+2\right)+...+\left(x+20\right)=930\)

\(\Rightarrow\left(x+x+..+x\right)+\left(1+2+...+20\right)=930\)

\(\Rightarrow20x+210=930\)

\(\Rightarrow20x=720\Rightarrow x=36\)

\(n\inℕ;n^2+n=n\left(n+1\right)\)

\(+;n⋮2\Rightarrow n\left(n+1\right)⋮2\)

\(+;n⋮̸2\Rightarrow n+1⋮2\Rightarrow n\left(n+1\right)⋮2\)

\(\Rightarrow n^2+n⋮2\)

\(\Rightarrow n^2+n+2⋮2\)

\(\left(n+1\right)-n=1;\left(n+1\right)n\ne\cdot\cdot\cdot8\)

\(\Rightarrow n^2+n+2\ne\cdot\cdot\cdot0\)

\(\Rightarrow n^2+n+2⋮̸5\)

                    Bài giải

\(n\left(n^2+5\right)=n^3+5n\)

Nếu n lẻ thì : \(n^3+5n=\text{ lẻ }+\text{ lẻ }=\text{ chẵn }⋮\text{ }2\)

Nếu n chẵn thì : \(n^3+5n=\text{ chẵn }+\text{ chẵn }=\text{ chẵn }⋮\text{ }2\)

\(\Rightarrow\text{ }n\left(n^2+5\right)\text{ }⋮\text{ }2\)

a) \(37.12=37.3.4=\left(37.3\right).4=111.4=444\)

\(37.27=37.3.9=\left(37.3\right).9=111.9=999\)

b) \(15873.28=15873.7.4=\left(15873.7\right).4=111111.4=444444\)

\(15873.63=15873.7.9=\left(15873.7\right).9=111111.9=999999\)

a) Do \(x,y\inℤ\Rightarrow\hept{\begin{cases}x+2\inℤ\\y-1\inℤ\end{cases}}\)

\(\Rightarrow x+2,y-1\)là các cặp ước của 3.

Ta có bảng sau :

Vậy : \(\left(x,y\right)\in\left\{\left(-1,4\right);\left(-3,-2\right);\left(1,2\right);\left(-5,0\right)\right\}\)

a) ( x + 2 ) ( y - 1 ) = 3

Mà x,y  Z

=>( x + 2 ) và ( y - 1 )  Ư(3)={±1;±3}

Ta có bảng 

Vậy (x,y) thuộc {(-1;4);(-3;-2);(1;2);(-5;0)}

b) ( 3 -x ) ( xy + 5 ) = -1

Vì x,y thuộc Z

=>( 3 -x ) và ( xy + 5 ) thuộc Ư(-1)={ ±1}

Ta có bảng

Vậy x,y thuộc {(2;-3);(4;-1)}