A = \(\frac{1}{2}.24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

A = \(\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

A = \(\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

A = \(\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

A = \(\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

A = \(\frac{1}{2}\left(5^{32}-1\right)\left(5^{32}+1\right)\)

A = \(\frac{1}{2}\left(5^{64}-1\right)\)

\(2A=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

        \(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

       \(=5^{64}-1\)

=> \(A=\frac{5^{64}-1}{2}\)

Ta có; \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\Rightarrow\widehat{A}+\widehat{B}=220^o\)

Lại có: \(\left(\widehat{A}+\widehat{B}\right)+\left(\widehat{A}-\widehat{B}\right)=220^o+10^o\)

\(\Rightarrow2\widehat{A}=230^o\Rightarrow\widehat{A}=115^o\)

\(\Rightarrow B=115^o-10^o=105^o\)

Vậy..

a) x² - 2x + 5

= x² - x - x + 1 + 4

= (x² - x) - (x - 1) + 4

= x.(x-1) - (x-1) + 4

= (x-1)^2 + 4

Có: (x-1)^2 \(\ge\)0 => (x-1)^2 + 4\(\ge4\)

Dấu ''='' xảy ra khi x-1=0 => x = 1.

Vậy Min của x^2 - 2x + 5 bằng 4 khi x = 1

\(\left(a+b\right)^0=1\)

\(\left(a+b\right)^1=a+b\)

\(\left(a+b\right)^2=a^2+2ab+b^2\)

\(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3\)

\(\left(a+b\right)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)

\(\left(a+b\right)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5\)

Tổng quát:

\(\left(a+b\right)^n=C_0a^n+C_1a^{n-1}b+...+C_nb^n\)

Trong đó : C₀, C₁, ..., C_n là các hệ số trong tam giác cân Paxcan:

(a + b)^0 1 (a + b)^1 1 1 (a + b)^2 1 2 1 (a + b)^3 1 3 3 1 (a + b)^4 1 4 6 4 1 (a + b)^5 1 5 10 10 5 1 (a + b)^6 1 6 15 20 15 6 1 ........... ...........

Chúc bn học tốt <3

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

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

\(\Rightarrow\orbr{\begin{cases}3x-5=0\\2x+4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-2\end{cases}}\)

Đây ạ bạn

Đây nhé bạn

đâu bạn

\(\left(2m^2-\frac{10}{2}\right)^2\)

\(=\left(2m^2\right)^2-2.2m^2.\frac{10}{2}+\left(\frac{10}{2}\right)^2\)

\(=4m^4-20m^2+25\)

Không rõ đề bài ....

a) Ta có \(a\left(b+1\right)+b\left(a+1\right)=\left(a+1\right)\left(b+1\right)\Rightarrow2ab+a+b=a+b+ab+1\)

=> ab=1

b) Ta có \(2\left(a+1\right)\left(b+1\right)=\left(a+b\right)\left(a+b+2\right)\Leftrightarrow2ab+2a+2b+2=a^2+ab+2a+b^2+ab+2b\)

=> a^2+b^2=2

^_^

vũ tiền châu bạn làm rõ đc ko ạk

a ) x^1 - 2223. x^2 + 223x + 2223 tại x = 222

     = 222 - 2223 . 222^2 + 223.222 + 2223

     = 222( 1 + 223 ) - 2223( 222^2 - 1 ) 

     = 222 . 224 - 2223 . 49283

     = -109506381

b ) Sửa đề , - 2009.x chứ ko phải + 2009.x

     x^14 - 2009.x^13 + 2009.x^12 - 2009.x^11 + ... - 2009.x + 2009 tại x = 2009

     = 2009^14 - 2009^14 + 2009^13 - 2009^12 + 2009^11 - 2009^10 + 2009^9 - 2009^8 + 2009^7 - 2009^6 + 2009^5 - 2009^4 + 2009^3 - 2009^2 + 2009

     = 2009^12( 2009 - 1 ) + 2009^10( 2009 - 1 ) + 2009^8( 2009 - 1 ) + 2009^6( 2009 - 1 )+ 2009^4( 2009 - 1 ) + 2009^2( 2009 - 1 ) + 2009 

     = ( 2009 - 1 )( 2009^12 + 2009^10 + 2009^8 + 2009^6 + 2009^4 + 2009^2 ) + 2009

     = 2008( 2009^12 + 2009^10 + 2009^8 + 2009^6 + 2009^4 + 2009^2 ) + 2009

P/s : ko chắc 

\(a^2+b^2-2a\left(b+2\right)=0\Rightarrow a^2-2ab+b^2-4a=0\Rightarrow\left(a-b\right)^2-4a=0\Rightarrow\left(a-b\right)^2=4a\)

\(\Rightarrow a=\frac{\left(a+b\right)^2}{4}=\left(\frac{a+b}{2}\right)^2\)là số chính phương