Chép đề đúng chưa bạn? 2 phân số đầu có ngoặc không vậy?

Nguyễn Công Tỉnh đúng r bạn, mình sửa lại r

\(D=x^2+2x\left(y+2\right)+2y^2+6y+10\)

\(=x^2+2x\left(y+2\right)+\left(y^2+4y+4\right)+\left(y^2+2y+1\right)+5\)

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

\(=\left(x+y+2\right)^2+\left(y+1\right)^2+5\ge5\forall x\)

\(\Rightarrow\)Min D = 5 tại \(\hept{\begin{cases}x+y+2=0\\y+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\y=-1\end{cases}}}\)

=.= hk tốt!!

\(E=x^2+4xy+5y^2=x^2+4xy+4y^2+y^2=\left(x+2y\right)^2+y^2\ge0\forall x,y\)

=> Min E = 0 tại \(\hept{\begin{cases}x+2y=0\\y=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\y=0\end{cases}}}\)

\(a,\left(\sqrt{27}-2\sqrt{17}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)

\(=3\sqrt{21}-2\sqrt{119}+7+7\sqrt{8}\)

Đề sai chăng???

\(b,\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}\)

\(=\sqrt{2-2\sqrt{2}+1}+\sqrt{2+2\sqrt{2}+1}\)

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

\(=\sqrt{2}-1+\sqrt{2}+1\)

\(=2\sqrt{2}\)

\(c,9\sqrt{2}-4\sqrt{8}-\sqrt{50}+2\sqrt{32}\)

\(=9\sqrt{2}-8\sqrt{2}-5\sqrt{2}+8\sqrt{2}\)

\(=\sqrt{2}\left(9-8-5+8\right)\)

\(=4\sqrt{2}\)

\(d,\sqrt{3-2\sqrt{2}}-\sqrt{6+4\sqrt{2}}\)

\(=\sqrt{2-2\sqrt{2}+1}-\sqrt{4+2.2\sqrt{2}+2}\)

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

\(=\sqrt{2}-1-2-\sqrt{2}\)

\(=-3\)

\(a,\frac{\sqrt{108x^3}}{\sqrt{12x}}=\frac{\sqrt{36.3.x^3}}{\sqrt{3.4.x}}=\frac{6\sqrt{3}.\sqrt{x}^3}{2\sqrt{3}.\sqrt{x}}=3\sqrt{x}^2=3x\)

\(b,\frac{\sqrt{13x^4y^6}}{\sqrt{208x^6y^6}}=\frac{\sqrt{13}.\sqrt{x^4}.\sqrt{y^6}}{\sqrt{16.13}.\sqrt{x^6}.\sqrt{y^6}}=\frac{\sqrt{13}.x^2y^3}{4\sqrt{13}x^3y^3}=\frac{1}{4x}\)

\(c,\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}+\sqrt{y}\right)^2\)

\(=\frac{\sqrt{x}^3+\sqrt{y}^3}{\sqrt{x}+\sqrt{y}}-\left(x+2\sqrt{xy}+y\right)\)

\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x-2\sqrt{xy}-y\)

\(=x-\sqrt{xy}+y-x-2\sqrt{xy}-y=-3\sqrt{xy}\)

\(d,\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\frac{\sqrt{\left(\sqrt{x}-1\right)^2}}{\sqrt{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

Đk chỗ này là \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge\sqrt{1}\Rightarrow x\ge1\)nhé 

\(e,\frac{x-1}{\sqrt{y}-1}.\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}.\frac{y-2\sqrt{y}+1}{\left(x-1\right)^2}\)

\(=\frac{\left(x-1\right)\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}-1\right)\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)