\(5,A=\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)

\(A=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)

\(A=\left|2x-1\right|+\left|2x-3\right|\)

\(A=\left|2x-1\right|+\left|3-2x\right|\ge\left|2x-1+3-2x\right|\)

\(A\ge2\)

\(< =>MIN:A=2\)dấu = xảy khi \(\frac{1}{2}\le x\le\frac{3}{2}\)

thế cũng nói tui báo cáo á

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

\(\Leftrightarrow x^2y+xy^2+4xy+x+y+4=5\)

\(\Leftrightarrow\left(xy+1\right)\left(x+y+4\right)=5\)

Đến đây bạn xét bảng giá trị, thu được kết quả là: 

\(\left(x,y\right)\in\left\{\left(0,1\right),\left(1,0\right),\left(-6,1\right),\left(1,-6\right)\right\}\).

\(x\left(4-x\right)\ge0\)

\(\Leftrightarrow\hept{\begin{cases}x\ge0\\4-x\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x\le0\\4-x\le0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge0\\x\le4\end{cases}}\)hoặc \(\hept{\begin{cases}x\le0\\x\ge4\end{cases}}\)

\(\Leftrightarrow0\le x\le4\).

\(x=\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4+\sqrt{15}}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8+2\sqrt{15}}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)\)

\(=2\)

Với \(x=2\):

\(\frac{\sqrt{\frac{1}{x}+4+4x}}{\sqrt{x}\left(2x^2-x-1\right)}=\frac{\sqrt{\frac{1}{2}+4+8}}{\sqrt{2}\left(8-2-1\right)}=\frac{1}{2}\)

\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{2}\sqrt{3}+\sqrt{2}\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(1+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

Thôi không cần nữa

\(A=\left(\sqrt{3}+1\right)\left(\sqrt{2}-\sqrt{2+\sqrt{3}}\right)\)

\(\sqrt{2}A=\left(\sqrt{3}+1\right)\left(2-\sqrt{4+\sqrt{3}}\right)\)

\(\sqrt{2}A=\left(\sqrt{3}+1\right)\left(2-\sqrt{\left(\sqrt{3}+1\right)^2}\right)\)

\(\sqrt{2}A=\left(\sqrt{3}+1\right)\left(2-\sqrt{3}-1\right)\)

\(\sqrt{2}A=\left(\sqrt{3}+1\right)\left(1-\sqrt{3}\right)\)

\(\sqrt{2}A=1-3\)

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

Mình làm đc rồi thôi thì cảm ơn bạn nha

sửa đề : \(\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)

\(=\sqrt{5^2+2.5\sqrt{2}+2}-\sqrt{4^2+2.4\sqrt{2}+2}\)

\(=\sqrt{\left(5+\sqrt{2}\right)^2}-\sqrt{\left(4+\sqrt{2}\right)^2}=\left|5+\sqrt{2}\right|-\left|4+\sqrt{2}\right|\)

\(=5+\sqrt{2}-4-\sqrt{2}=1\)

=1 nha

t.i.c.k mình nha

bạn nào 10sp gúp mình đi