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

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

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

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

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

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

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

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: \(x=2\).

Có nhận xét sau: 

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

\(=\frac{\sqrt{a+1}-\sqrt{a}}{\sqrt{a}\sqrt{a+1}}=\frac{\sqrt{a+1}}{\sqrt{a}\sqrt{a+1}}-\frac{\sqrt{a}}{\sqrt{a}\sqrt{a+1}}=\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{a+1}}\). Từ đây suy ra:

\(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-....-\frac{1}{\sqrt{2025}}=1-\frac{1}{45}=\frac{44}{45}\).

Vậy: \(A=\frac{44}{45}\).

suy ra 2x+1=125

suy ra 2x=124

suy ra x=62