\(A=3\left(x^2+y^2\right)-2\left(x^3+y^3\right)\)

\(=3x^2+3y^2-2\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(=3x^2+3y^2-2.1\left(x^2-xy+y^2\right)\)

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

\(=x^2+2xy+y^2=\left(x+y\right)^2=1^2=1\)

\(B=x^3+y^3+3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)\)

\(=x^3+y^3+3xy\left[\left(x+y\right)^2-2xy\right]+6x^2y^2.1\)

\(=x^3+y^3+3xy\left(x+y\right)^2-6x^2y^2+6x^2y^2\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\)

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

\(=x^2+2xy+y^2=\left(x+y\right)^2=1^2=1\)

a) \(A=x^2-4x+9=x^2-4x+4+5=\left(x-2\right)^2+5\ge5\)

Vậy \(A_{min}=5\Leftrightarrow x=2\)

b) \(B=x^2-x+1=x^2-x+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Vậy \(B_{min}=\frac{3}{4}\Leftrightarrow x=\frac{1}{2}\)

Giả sử trong hai số a, b không đồng thời chia hết cho 3 

=> a+b không chia hết cho 3 => m+2n+n+2m=3(m+n) không chia hết cho 3 ( vô lí ) 

=> điều giả sử sai => đpcm 

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

\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Ta có: \(a^2+b^2+c^2+3=2.\left(a+b+c\right)\)

\(\Rightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)

\(\Rightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\\\left(c-1\right)^2=0\end{cases}}\text{vì }\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\)

\(\Leftrightarrow a=b=c=1\)

Vậy a=b=c=1

=(50²-49²)+...+(2²-1²)

=(50-49)(50+49)+(48-47)(48+47)+...+(2-1)(2+1)

=1.99+1.95+1.91+...+1.3

=99+95+91+...+3

=(99+3)+(95+7)+...+

         \(50^2-49^2+48^2-47^2+...+2^2-1^2\)

\(=\left(50^2-49^2\right)+\left(48^2-47^2\right)+...+\left(2^2-1^2\right)\)

\(=\left(50+49\right)\left(50-49\right)+\left(48+47\right)\left(48-47\right)+...+\left(2+1\right)\left(2-1\right)\)

\(=99.1+95.1+...+3.1\)

\(=99+95+...+3\)

\(=3+...+95+99\)

Từ 3 đến 99 có: \(\left(99-3\right):4+1=25\left(\text{số hạng}\right)\)

Tổng là: \(\frac{\left(99+3\right)\times25}{2}=1275\)

a)\(\left(2+1\right)\left(2^2+1\right)....\left(2^{256}+1\right)-1\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^{256}+1\right)-1\)

\(=\left(2^2-1\right)\left(2^2+1\right)...\left(2^{256}+1\right)-1\)

Tiếp tục như thế, ta được:

\(=\left(2^{256}-1\right)\left(2^{256}+1\right)-1=2^{512}-1-1=2^{512}-2\)

b) \(24\left(5^2+1\right)\left(5^4+1\right)...\left(5^{32}+1\right)-5^{64}\)

\(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)...\left(5^{32}+1\right)-5^{64}\)

\(=\left(5^4-1\right)\left(5^4+1\right)...\left(5^{32}+1\right)-5^{64}\)

Tiếp tục như thế, ta được:

\(=\left(5^{32}-1\right)\left(5^{32}+1\right)-5^{64}=5^{64}-1-5^{64}=-1\)

\(\left(2+1\right).\left(2^2+1\right)....\left(2^{256}+1\right)-1\)

\(\left(2-1\right).\left(2+1\right).\left(2^2+1\right).....\left(2^{256}+1\right)-1\)

\(=\left(2^2-1\right).\left(2^2+1\right)....\left(2^{256}+1\right)-1\)

\(=\left(2^{256}-1\right).\left(2^{256}+1\right)+1=2^{512}+1\)

Xét các tam giác vuông BKC và BHC có:

BC chung

^KBC=^HBC

=>\(\Delta\)BKC=\(\Delta\)BHC ( ch-gn )

=> BK=HC;KC=BH ( 1 ) 

Mà AB=AC=>AK=AH

Xét tam giác cân AKH có ^AKH=180⁰-^KAH-^KHA=\(\frac{180^0-\widehat{A}}{2}\)

Mà tam giác \(ABC\) cân tại A nên \(\widehat{B}=\frac{180^0-\widehat{A}}{2}\)

=> KH//BC ( 2 )
Từ ( 1 );( 2 ) suy ra đpcm