\(x^2-4y^2\)

\(=x^2-\left(2y\right)^2\)

\(=\left(x+2y\right)\left(x-2y\right)\)

\(=\left(35+2.3,25\right)\left(35-2.3,25\right)\)

\(=\left(35+6,5\right)\left(35-6,5\right)\)

\(=41,5+28,5\)

\(=70\)

Hok tốt

\(A=\left(x-1\right)^2+\left(x-2\right)^2+5.\)

\(A=\left(x^2-2.x.1+1^2\right)+\left(x^2-2.x.2+2^2\right)+5.\)

\(A=\left(x^2-2x+1\right)+\left(x^2-4x+4\right)+5.\)

( suy nghĩ tiếp nha)

Hok tốt

Thanhs!!!!! Uyên Trần

Ta có: \(5^{2000}+5^{1998}=5^{1998}\left(5^2+1\right)=5^{1998}.26\)

Vì \(26⋮13\Rightarrow5^{1998}.26⋮13\)

              hay \(5^{2000}+5^{1998}⋮13\)

TK NHOA!!

Ta có :

\(5^{2000}+5^{1998}\)

\(=5^{1998}\times5^2+5^{1998}\)

\(=5^{1998}\times\left(5^2+1\right)\)

\(=5^{1998}\times26\)

\(=5^{1998}\times13\times2\)

Vậy \(5^{2000}+5^{1998}⋮13\)

_Chúc bạn học tốt_

\(R=x^2-4xy+5y^2+10x-22y+28\)

\(R=\left(x^2-4xy+4y^2\right)+y^2+10x-22y+28\)

\(R=\left[\left(x-2y\right)^2+2\left(x-2y\right).5+25\right]+\left(y^2-2y+1\right)+2\)

\(R=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\)

Mà  \(\left(x-2y+5\right)^2\ge0\forall x;y\)

      \(\left(y-1\right)^2\ge0\forall y\)

\(\Rightarrow R\ge2\)

Dấu "=" xảy ra khi : 

\(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

Vậy ...

\(Q=x^2-2x+2y^2+4y+8\)

\(Q=\left(x^2-2x+1\right)+2\left(y^2+2y+1\right)+5\)

\(Q=\left(x-1\right)^2+2\left(y+1\right)^2+5\)

Mà  \(\left(x-1\right)^2\ge0\forall x\)

      \(\left(y+1\right)^2\ge0\forall y\Rightarrow2\left(y+1\right)^2\ge0\forall y\)

\(\Rightarrow Q\ge5\)

Dấu "=" xảy ra khi : 

\(\hept{\begin{cases}x-1=0\\y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-1\end{cases}}\)

Vậy ...

\(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=1-3ab\)

\(a^2+b^2=\left(a+b\right)^2-2ab=1-2ab\)

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

\(=4\left(1-3ab\right)-6\left(1-2ab\right)\)

\(=4-12ab-6+12ab=-2\)