Toán Lớp 9: tính A=√(4-√7)-√(4+√7) và B=11*∛(5-√17)*∛(5+√17)
ai đúng mình vote 5 sao
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Toán Lớp 9: tính A=√(4-√7)-√(4+√7) và B=11*∛(5-√17)*∛(5+√17)
ai đúng mình vote 5 sao
Comments ( 1 )
A = – \sqrt 2 \\
B = 22
\end{array}\)
A = \sqrt {4 – \sqrt 7 } – \sqrt {4 + \sqrt 7 } \\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 2 .\sqrt {4 – \sqrt 7 } – \sqrt 2 .\sqrt {4 + \sqrt 7 } } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {2.\left( {4 – \sqrt 7 } \right)} – \sqrt {2.\left( {4 + \sqrt 7 } \right)} } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {8 – 2\sqrt 7 } – \sqrt {8 + 2\sqrt 7 } } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {7 – 2\sqrt 7 + 1} – \sqrt {7 + 2\sqrt 7 + 1} } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {{{\sqrt 7 }^2} – 2.\sqrt 7 .1 + {1^2}} – \sqrt {{{\sqrt 7 }^2} + 2.\sqrt 7 .1 + {1^2}} } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {{{\left( {\sqrt 7 – 1} \right)}^2}} – \sqrt {{{\left( {\sqrt 7 + 1} \right)}^2}} } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\left| {\sqrt 7 – 1} \right| – \left| {\sqrt 7 + 1} \right|} \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left[ {\left( {\sqrt 7 – 1} \right) – \left( {\sqrt 7 + 1} \right)} \right]\\
= \dfrac{1}{{\sqrt 2 }}.\left( { – 2} \right)\\
= – \sqrt 2 \\
B = 11.\sqrt[3]{{5 – \sqrt {17} }}.\sqrt[3]{{5 + \sqrt {17} }}\\
= 11.\sqrt[3]{{\left( {5 – \sqrt {17} } \right).\left( {5 + \sqrt {17} } \right)}}\\
= 11.\sqrt[3]{{{5^2} – {{\sqrt {17} }^2}}}\\
= 11.\sqrt[3]{{25 – 17}}\\
= 11.\sqrt[3]{8}\\
= 11.\sqrt[3]{{{2^3}}}\\
= 11.2\\
= 22
\end{array}\)