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

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

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

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

\(\Rightarrow2y=x^3+3x\)

\(\Rightarrow2I=2x^4+x^3\left(x^3+3x\right)+6x^2+x\left(x^3+3x\right)-\left(x^3+3x\right)^2+2\)

\(=2x^4+x^6+3x^4+6x^2+x^4+3x^2-\left(x^6+6x^4+9x^2\right)+2\)

\(=2\)

ỨDFGHLJHGFD

Đk: \(x\ge2\).

\(\sqrt{x^2-4}+2\sqrt{x-2}=0\Leftrightarrow\sqrt{x-2}\left(\sqrt{x+2}+2\right)=0\).

Thấy ngay: \(\sqrt{x+2}+2\ge0+2=2>0\)nên từ đây suy ra: \(\sqrt{x-2}=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)(thỏa mãn).

Vậy: \(x=2\).