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TL:
\(\sqrt{8-3\sqrt{7}}-\sqrt{8+3\sqrt{7}}\)
\(=\frac{8-3\sqrt{7}-8-3\sqrt{7}}{\sqrt{8-3\sqrt{7}}+\sqrt{8+3\sqrt{7}}}\)
\(=\frac{-6\sqrt{7}}{\sqrt{8-3\sqrt{7}}+\sqrt{8+3\sqrt{7}}}\)
Cho \(A=\sqrt{8-3\sqrt{7}}-\sqrt{8+3\sqrt{7}}\)
CACH 1 : \(\Rightarrow A\sqrt{2}=\sqrt{16-6\sqrt{7}}-\sqrt{16+6\sqrt{7}}\)
\(\Rightarrow A\sqrt{2}=\sqrt{9-2.3.\sqrt{7}+7}-\sqrt{9+2.3.\sqrt{7}+7}\)
\(\Rightarrow A\sqrt{2}=\sqrt{\left(3-\sqrt{7}\right)^2}-\sqrt{\left(3+\sqrt{7}\right)^2}\)
\(\Rightarrow A\sqrt{2}=|3-\sqrt{7}|-|3+\sqrt{7}|\)
\(\Rightarrow A\sqrt{2}=3-\sqrt{7}-3-\sqrt{7}=-2\sqrt{7}=-\sqrt{28}\)
\(\Rightarrow A=-\sqrt{14}\)
CACH 2 : \(A^2=8-3\sqrt{7}+8+3\sqrt{7}-2.\sqrt{8^2-\left(3\sqrt{7}\right)^2}\)
\(\Rightarrow A^2=16-2\sqrt{64-63}=16-2=14\)
\(\Rightarrow A=\sqrt{14}\) hoặc \(A=-\sqrt{14}\)
Mà \(8+3\sqrt{7}>8-3\sqrt{7}\) \(\Rightarrow\sqrt{8+3\sqrt{7}}>\sqrt{8-3\sqrt{7}}\)
Vây A âm \(\Rightarrow A=-\sqrt{14}\)
a, \(\left(2\sqrt{2}-3\sqrt{2}+\sqrt{10}\right):\sqrt{2}-\sqrt{5}=\left(-\sqrt{2}+\sqrt{10}\right):\sqrt{2}-\sqrt{5}=-1\)
b.\(\sqrt{16+2\sqrt{16.5}+5}+\sqrt{16-2\sqrt{16.5}+5}=\sqrt{\left(4+\sqrt{5}\right)^2}+\sqrt{\left(4-\sqrt{5}\right)^2}=8\)
d,dat \(A=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\Rightarrow A^2=4+\sqrt{7}+2\sqrt{16-7}+4-\sqrt{7}\)\(A^2=8+6=14\Rightarrow A=\sqrt{14}\)
C,\(\sqrt{17-4\sqrt{\left(2+\sqrt{5}\right)^2}}=\sqrt{17-4\left(2+\sqrt{5}\right)}=\sqrt{17-8-4\sqrt{5}}=\sqrt{9-4\sqrt{5}}=\sqrt{5}-2\)
a, A= \(\frac{\sqrt{48-12\sqrt{7}}}{2}-\frac{\sqrt{48+12\sqrt{7}}}{2}\)
= \(\frac{\sqrt{\left(\sqrt{42}-\sqrt{6}\right)^2}}{2}-\frac{\sqrt{\left(\sqrt{42}+\sqrt{6}\right)^2}}{2}\)
= \(\frac{-2\sqrt{6}}{2}\)
= \(-\sqrt{6}\)
a) Kết quả rút gọn xấu (+dài) nữa. (có thể đề sai)
b)
\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left[\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right].\left(\sqrt{7}-\sqrt{5}\right)\)
\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=-\left(7-5\right)=-2\)
c) \(\frac{\sqrt{5-2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}=\frac{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)
\(=\frac{\sqrt{3}-\sqrt{2}+\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}=\frac{\left(\sqrt{5}-\sqrt{2}\right)^2}{3}\)
a) \(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right).\frac{1}{\sqrt{6}}=\left[\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right].\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{6}}{2}-2\sqrt{6}\right).\frac{1}{\sqrt{6}}=\frac{1}{2}-2=-\frac{3}{2}\)
Ta có:
\(A=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
\(\sqrt{2}A=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(\sqrt{2}A=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}=\sqrt{7}-1-\sqrt{7}-1=-2\)
\(\Rightarrow A=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)
a) \(\sqrt{36}.\sqrt{121}+\sqrt[3]{-64}-\sqrt[3]{125}\)
\(=6.11+\left(-4\right)-5=66-9=57\)
b) \(\sqrt{75}+\sqrt{\left(\sqrt{3}-2\right)^2}-30\sqrt{\frac{3}{25}}\)
\(=\sqrt{25.3}+\left|\sqrt{3}-2\right|-30.\frac{\sqrt{3}}{\sqrt{25}}\)
\(=5\sqrt{3}+2-\sqrt{3}-30.\frac{\sqrt{3}}{5}\)
\(=5\sqrt{3}+2-\sqrt{3}-6\sqrt{3}=2-2\sqrt{3}\)
c) \(\sqrt{11-4\sqrt{7}}-\frac{12}{1+\sqrt{7}}=\sqrt{7-4\sqrt{7}+4}-\frac{12}{1+\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}-2\right)^2}-\frac{12}{1+\sqrt{7}}=\left|\sqrt{7}-2\right|-\frac{12}{1+\sqrt{7}}\)
\(=\left(\sqrt{7}-2\right)-\frac{12}{\sqrt{7}+1}=\frac{\left(\sqrt{7}-2\right)\left(\sqrt{7}+1\right)}{\sqrt{7}+1}-\frac{12}{\sqrt{7}+1}\)
\(=\frac{5-\sqrt{7}}{\sqrt{7}+1}-\frac{12}{\sqrt{7}+1}=\frac{-7-\sqrt{7}}{\sqrt{7}+1}\)
\(=\frac{-\sqrt{7}\left(\sqrt{7}+1\right)}{\sqrt{7}+1}=-\sqrt{7}\)
Lời giải:
\(A=\sqrt{8-3\sqrt{7}}+\sqrt{4-\sqrt{7}}\)
$A\sqrt{2}=\sqrt{16-6\sqrt{7}}+\sqrt{8-2\sqrt{7}}$
$=\sqrt{(3-\sqrt{7})^2}+\sqrt{(\sqrt{7}-1)^2}$
$=|3-\sqrt{7}|+|\sqrt{7}-1|$
$=3-\sqrt{7}+\sqrt{7}-1=2$