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\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}+\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(\frac{2\sqrt{3}+\sqrt{18}+2\sqrt{3}-\sqrt{18}}{4-6}\right)-\frac{1}{\sqrt{2}}.\)
\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}-\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}.\left(2\sqrt{3}\right)-\frac{1}{\sqrt{2}}\)
\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}-\frac{2\sqrt{6}-6}{\sqrt{2}+1}-\frac{1}{\sqrt{2}}\)
Nhân cả tử và mẫu với \(\sqrt{3}\)
\(P=\frac{\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}}{\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}}=\frac{\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}}\)
\(=\frac{\sqrt{2}+1+\sqrt{2}-1}{\sqrt{2}+1-\sqrt{2}+1}=\frac{2\sqrt{2}}{2}=\sqrt{2}\)
M= \(\sqrt{2}+1-\) \(\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}+1-\sqrt{2}+1=2\)
N=\(\sqrt{1+2\sqrt{\left(\sqrt{2}+1\right)^2}}=\sqrt{1+2\left(\sqrt{2}+1\right)}=\) \(\sqrt{1+2\sqrt{2}+2}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
P= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}+\frac{2\sqrt{x}.\sqrt{x}}{\sqrt{x}}\) (dk \(x>0\))
=\(\sqrt{x}+1+2\sqrt{x}=3\sqrt{x}+1\)
Q= \(\sqrt{\left(\sqrt{x}+1\right)^2}+\sqrt{\left(\sqrt{x}-1\right)^2}\) (dk \(x\ge0\) )
=\(\left|\sqrt{x}+1\right|+\left|\sqrt{x}-1\right|\)
th1 \(\sqrt{x}\ge1\Leftrightarrow x\ge1\) Q=\(\sqrt{x}+1+\sqrt{x}-1=2\sqrt{x}\)
th2 \(0\le x< 1\) Q=\(\sqrt{x}+1+1-\sqrt{x}=2\)
a) \(M=\sqrt{2}+1-\sqrt{1,5.2-2.\sqrt{2}}\)
\(=\sqrt{2}+1-\sqrt{2.\left(1,5-\sqrt{2}\right)}\)\(=\sqrt{2}+1-\sqrt{2}.\sqrt{1,5-\sqrt{2}}\)
\(=\sqrt{2}.\left(1+1,5-\sqrt{2}\right)+1=\sqrt{2}.\left(2,5-\sqrt{2}\right)+1\)
\(=\sqrt{2}.2,5-2+1=\sqrt{2}.2,5-1\)
P/s: Theo em thì em nghĩ là đúng '-' Khoảng 90% :)
Câu 1,2,3 Ez quá rồi :3
Câu 4:
Tổng quát:
\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v
Câu 5 ko khác câu 4 lắm :v
Câu 5:
Tổng quát:
\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v
\(\frac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2}-\sqrt{2}-\sqrt{3}}\)
\(\Leftrightarrow\frac{2}{2\sqrt{2}+\sqrt{2}+\sqrt{6}}+\frac{1}{\sqrt{2}-\sqrt{2}-\sqrt{3}}\)
\(\Leftrightarrow\frac{2\left(3\sqrt{2}-\sqrt{6}\right)}{\left(3\sqrt{2}+\sqrt{6}\right)\left(3\sqrt{2}-\sqrt{6}\right)}+\frac{1}{\sqrt{2}-\sqrt{2}-\sqrt{3}}\)
\(\Leftrightarrow\frac{6\sqrt{2}-2\sqrt{6}}{18-6}+\frac{1}{-\sqrt{3}}\)
\(\Leftrightarrow\frac{3\sqrt{2}-\sqrt{6}}{6}-\frac{\sqrt{3}}{3}\)
\(\Leftrightarrow\frac{3\sqrt{2}-\sqrt{6}-2\sqrt{3}}{6}\)
Ps : anh gửi em nhé, có chỗ nào không hiểu thì hỏi anh nhé. Nhớ k :33
# Aeri #
\(\frac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2}-\sqrt{2}-\sqrt{3}}\)
\(\frac{\sqrt{2}}{2+\sqrt{4+2\sqrt{3}}}+\frac{1}{-\sqrt{3}}\)
\(=\frac{\sqrt{2}}{2+\sqrt{\sqrt{3}^2+2\sqrt{3}+1}}-\frac{1}{\sqrt{3}}\)
\(=\frac{\sqrt{2}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}-\frac{1}{\sqrt{3}}\)
\(=\frac{\sqrt{2}}{2+\sqrt{3}+1}-\frac{1}{\sqrt{3}}\)
\(=\frac{\sqrt{2}}{3+\sqrt{3}}-\frac{1}{\sqrt{3}}=\frac{\sqrt{2}}{\sqrt{3}\left(\sqrt{3}+1\right)}-\frac{1}{\sqrt{3}}\)
\(=\frac{\sqrt{2}-\sqrt{3}-1}{3+\sqrt{3}}\)