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Ta có:\(\left(\sqrt{7-\sqrt{5}}+\sqrt{7+\sqrt{5}}\right)^2=7-\sqrt{5}+7+\sqrt{5}+2\sqrt{\left(7-\sqrt{5}\right)\left(7+\sqrt{5}\right)}=14+2\sqrt{44}=14+4\sqrt{11}\)
=>\(\sqrt{7-\sqrt{5}}+\sqrt{7+\sqrt{5}}=\sqrt{14+4\sqrt{11}}=\sqrt{2}.\sqrt{7+2\sqrt{11}}\)
=>B=\(\dfrac{\sqrt{2}.\sqrt{7+2\sqrt{11}}}{\sqrt{7+2\sqrt{11}}}\cdot\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
=\(\sqrt{2}\cdot\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)(mình làm tắt tách 4=2+2=\(\sqrt{4}+\sqrt{4}\))
=\(\sqrt{2}\)\(\cdot\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\sqrt{2}\cdot\left(1+\sqrt{2}\right)=2+\sqrt{2}\)
\(B=\dfrac{\sqrt{7-\sqrt{5}}+\sqrt{7+\sqrt{5}}}{\sqrt{7+2\sqrt{11}}}.\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(B=\dfrac{\sqrt{14-2\sqrt{5}}+\sqrt{14+2\sqrt{5}}}{\sqrt{2}.\sqrt{7+2\sqrt{11}}}.\dfrac{\sqrt{2}+\sqrt{3}+2+\sqrt{6}+\sqrt{8}+2}{\sqrt{2}+\sqrt{3}+2}\)
\(B=\dfrac{\sqrt{\left(\left(\sqrt{7+2\sqrt{11}}\right)-\left(\sqrt{7-2\sqrt{11}}\right)\right)^2}+\sqrt{\left(\left(\sqrt{7+2\sqrt{11}}\right)+\left(7-2\sqrt{11}\right)\right)^2}}{\sqrt{2}.\sqrt{7+2\sqrt{11}}}.\dfrac{\sqrt{2}+\sqrt{3}+2+\sqrt{2}\left(\sqrt{3}+2+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+2}\)
\(B=\dfrac{\sqrt{7+2\sqrt{11}}-\sqrt{7-2\sqrt{11}}+\sqrt{7+2\sqrt{11}}+\sqrt{7-2\sqrt{11}}}{\sqrt{2}.\sqrt{7+2\sqrt{11}}}.\dfrac{\left(\sqrt{2}+\sqrt{3}+2\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+2}\)
\(B=\dfrac{2.\sqrt{7+2\sqrt{11}}}{\sqrt{2}.\sqrt{7+2\sqrt{11}}}.\left(1+\sqrt{2}\right)\)
\(B=\sqrt{2}.\left(1+\sqrt{2}\right)=\sqrt{2}+2\)
\(=\sqrt{5.\left(\sqrt{3}+1\right)}.\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}\)
\(=\sqrt{5}.\left(\sqrt{3}+1\right).\sqrt{48-10.\left(2+\sqrt{3}\right)}\)
\(=\left(\sqrt{15}+\sqrt{5}\right).\sqrt{28-10\sqrt{3}}\)
\(=\left(\sqrt{15}+\sqrt{5}\right).\sqrt{\left(5-\sqrt{3}\right)^2}\)
\(=\left(\sqrt{15}+\sqrt{5}\right).\left(5-\sqrt{3}\right)\)
Vậy...
~ Chắc chắn đúng cậu nhé ~ Tiếc gì 1 tk cho tớ nào?
a: Ta có: \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}-11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}-11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}-11\right)\)
\(=127-22\sqrt{6}\)
b: Ta có: \(\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
\(=\left(1-\sqrt{5}\right)\left(-1-\sqrt{5}\right)\)
=-1+5
=4
1)
a) \(\sqrt{2x-4}\) có nghĩa khi:
\(2x-4\ge0\)
\(\Leftrightarrow2x\ge4\)
\(\Leftrightarrow x\ge\dfrac{4}{2}\)
\(\Leftrightarrow x\ge2\)
b) \(\sqrt{\dfrac{-7}{4-x}}\) có nghĩa khi
\(\dfrac{-7}{4-x}\ge0\) mà \(-7< 0\)
\(\Rightarrow4-x\le0\)
\(\Leftrightarrow x\ge4\)
\(\left(\sqrt{200}+5\sqrt{150}-7\sqrt{600}\right):\sqrt{50}=2+5\sqrt{3}-7\sqrt{12}\)
\(2+5\sqrt{3}-14\sqrt{3}=2-9\sqrt{3}\)
\(A=\sqrt{4+\sqrt{15}}-\sqrt{7-3\sqrt{5}}\)
\(\Rightarrow A\sqrt{2}=\sqrt{8+2\sqrt{15}}-\sqrt{14-2\sqrt{45}}\)
\(A\sqrt{2}=\sqrt{\left(\sqrt{3}\right)^2+\left(\sqrt{5}\right)^2+2.\sqrt{3}.\sqrt{5}}-\sqrt{\left(\sqrt{5}\right)^2+\left(\sqrt{9}\right)^2-2.\sqrt{5}.\sqrt{9}}\)
\(A\sqrt{2}=\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{9}-\sqrt{5}\right)^2}\)
\(A\sqrt{2}=\sqrt{3}+\sqrt{5}-\sqrt{9}+\sqrt{5}=2\sqrt{5}+\sqrt{3}-\sqrt{9}\Rightarrow A=\frac{2\sqrt{5}+\sqrt{3}-\sqrt{9}}{\sqrt{2}}\)\(B=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(\Rightarrow B\sqrt{2}=\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)
\(B\sqrt{2}=\sqrt{1^2+\left(\sqrt{3}\right)^2+2.1.\sqrt{3}}+\sqrt{\left(\sqrt{3}\right)^2+1^2-2.1.\sqrt{3}}\)\(B\sqrt{2}=\sqrt{\left(1+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}=1+\sqrt{3}+\sqrt{3}-1=2\sqrt{3}\)
Phép 1:
Ta có: \(3\cdot\sqrt{7-4\sqrt{3}}\)
\(=3\cdot\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\)
\(=3\cdot\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=3\cdot\left|2-\sqrt{3}\right|\)
\(=3\cdot\left(2-\sqrt{3}\right)\)(Vì \(2>\sqrt{3}\))
\(=6-3\sqrt{3}\)
Phép 2:
Ta có: \(\sqrt{11+4\sqrt{7}}\)
\(=\sqrt{7+2\cdot\sqrt{7}\cdot2+4}\)
\(=\sqrt{\left(\sqrt{7}+2\right)^2}\)
\(=\left|\sqrt{7}+2\right|\)
\(=\sqrt{7}+2\)(Vì \(\sqrt{7}+2>0\))
Phép 3:
Ta có: \(2\cdot\sqrt{11-4\sqrt{7}}\)
\(=2\cdot\sqrt{7-2\cdot\sqrt{7}\cdot2+4}\)
\(=2\cdot\sqrt{\left(\sqrt{7}-2\right)^2}\)
\(=2\cdot\left|\sqrt{7}-2\right|\)
\(=2\cdot\left(\sqrt{7}-2\right)\)(Vì \(\sqrt{7}>2\))
\(=2\sqrt{7}-4\)
Phép 4:
Ta có: \(\sqrt{19-4\sqrt{15}}\)
\(=\sqrt{15-2\cdot\sqrt{15}\cdot2+4}\)
\(=\sqrt{\left(\sqrt{15}-2\right)^2}\)
\(=\left|\sqrt{15}-2\right|\)
\(=\sqrt{15}-2\)(Vì \(\sqrt{15}>2\))