a,\(x^2+5xy+y^2=x^2+2.x.\frac{5}{2}y^2+\frac{25}{4}y^2-\frac{21}{4}y^2\)

\(=\left(x+\frac{5}{2}y\right)^2-\left(\frac{\sqrt{21}}{2}y\right)^2=\left(x+\frac{5}{2}y-\frac{\sqrt{21}}{2}y\right)\left(x+\frac{5}{2}y+\frac{\sqrt{21}}{2}y\right)\)

b,\(x^2-2x-11=y^2\)\(< =>\left(x-1\right)^2-y^2=12\)

\(< =>\left(x-y-1\right)\left(x-1+y\right)=12\)

tùy vô đk của x;y rồi xét các th

dat \(x^2=a\) ta co

\(4a^2-32a+1\)=\(4a^2-16a+6a\sqrt{7}+64-16x-24\sqrt{7}-6x\sqrt{7}+24\sqrt{7}-63\)=\(2a\left(2a-8+3\sqrt{7}\right)-8\left(2a-8+3\sqrt{7}\right)-3\sqrt{7}\left(2a-8+3\sqrt{7}\right)\)

=\(2a-8-3\sqrt{7}\left(2x-8+3\sqrt{7}\right)\)cuoi cung ban thay a=\(x^2\)vao la xong

có vẻ như nó rất xấu nhỉ

`4x^4 - 32x^2 + 1 = (2x^2)^2 - 2.2x^2 . 8 + 64 - 63 = (2x^2 - 8)^2 - 63`

\(=\left(2x^2-8\right)^2-63 =\left(2x^2-8\right)^2-\left(\sqrt{63}\right)^2=\left(2x^2-8-\sqrt{63}\right)\left(2x^2-8+\sqrt{63}\right)\)

\(\frac{200}{x}+\frac{100}{x-10}-\frac{300}{x}=\frac{1}{2}\left(ĐKXĐ:x\ne0;10\right)\)

\(\Leftrightarrow\frac{100}{x-10}-\frac{100}{x}=\frac{1}{2}\Leftrightarrow\frac{100.2x}{2x\left(x-10\right)}-\frac{100.2\left(x-10\right)}{2x\left(x-10\right)}=\frac{x\left(x-10\right)}{2x\left(x-10\right)}\)

\(\Leftrightarrow\frac{200x-100.2\left(x-10\right)}{2x\left(x-10\right)}=\frac{x\left(x-10\right)}{2x\left(x-10\right)}\Rightarrow200x-200x+2000=x\left(x-10\right)\)

\(x\left(x-10\right)=2000\). Xét nghiệm tính được  \(x=50\left(tm\right)\)

giả sử 3ⁿ+19=a² (\(a\inℕ\)). dễ thấy a chẵn nên \(a^2\equiv0\)(mod 4)

=> 3ⁿ \(\equiv\)1 (mod 4)

Mặt khắc vì 3\(\equiv\)-1 nên \(3^n\equiv\left(-1\right)^n\)(mod 4)

Vậy n là số chẵn hay n=2m (\(m\inℕ\)) Ta có 3^2m+19=a² nên \(\left(a-3^m\right)\left(a+3^m\right)=19\Rightarrow\hept{\begin{cases}a-3^m=1\\a+3^m=19\end{cases}\Rightarrow m=2\Rightarrow n=4}\)

Ta có: \(P=x^2-3x+\frac{1}{2x}+2=\left(x-2\right)^2+\left(\frac{x}{8}+\frac{1}{2x}\right)+\frac{7x}{8}-2\ge\frac{1}{4}\)

Đẳng thức xảy ra khi x = 2