Môn Toán Lớp 9 giúp mình bài 4,5,7 với mình cần gấp
Question
Allison
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Trả lời ( )
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}4,\\
a,\\
\sqrt 5 – \sqrt {{{\left( {1 + \sqrt 5 } \right)}^2}} = \sqrt 5 – \left| {1 + \sqrt 5 } \right| = \sqrt 5 – \left( {1 + \sqrt 5 } \right) = – 1\\
b,\\
\sqrt {{{\left( {2 – \sqrt 3 } \right)}^2}} + \sqrt 3 = \left| {2 – \sqrt 3 } \right| + \sqrt 3 = 2 – \sqrt 3 + \sqrt 3 = 2\\
c,\\
\sqrt {{{\left( {\sqrt 7 – 3} \right)}^2}} + \sqrt 7 = \left| {\sqrt 7 – 3} \right| + \sqrt 7 = \left( {3 – \sqrt 7 } \right) + \sqrt 7 = 3\\
d,\\
\sqrt {{{\left( {\sqrt 8 – 7} \right)}^2}} + \sqrt 8 = \left| {\sqrt 8 – 7} \right| + \sqrt 8 = \left( {7 – \sqrt 8 } \right) + \sqrt 8 = 7\\
e,\\
\sqrt {{{\left( {\sqrt 2 – \sqrt 7 } \right)}^2}} + \sqrt 2 = \left| {\sqrt 2 – \sqrt 7 } \right| + \sqrt 2 = \left( {\sqrt 7 – \sqrt 2 } \right) + \sqrt 2 = \sqrt 7 \\
f,\\
\sqrt {{{\left( {\sqrt {21} – \sqrt {33} } \right)}^2}} + \sqrt {33} + \sqrt {21} = \left| {\sqrt {21} – \sqrt {33} } \right| + \sqrt {33} + \sqrt {21} \\
= \sqrt {33} – \sqrt {21} + \sqrt {33} + \sqrt {21} = 2\sqrt {33} \\
5,\\
a,\\
11 + 6\sqrt 2 = 9 + 2.3.\sqrt 2 + 2 = {3^2} + 2.3.\sqrt 2 + {\sqrt 2 ^2} = {\left( {3 + \sqrt 2 } \right)^2}\\
b,\\
8 – 2\sqrt 7 = 7 – 2\sqrt 7 + 1 = {\sqrt 7 ^2} – 2.\sqrt 7 .1 + {1^2} = {\left( {\sqrt 7 – 1} \right)^2}\\
c,\\
28 – 10\sqrt 3 = 25 – 2.5.\sqrt 3 + 3 = {5^2} – 2.5.\sqrt 3 + {\sqrt 3 ^2} = {\left( {5 – \sqrt 3 } \right)^2}\\
d,\\
\sqrt {4 + 2\sqrt 3 } – \sqrt {4 – 2\sqrt 3 } = \sqrt {3 + 2.\sqrt 3 + 1} – \sqrt {3 – 2\sqrt 3 + 1} \\
= \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} – \sqrt {{{\left( {\sqrt 3 – 1} \right)}^2}} = \left( {\sqrt 3 + 1} \right) – \left( {\sqrt 3 – 1} \right) = 2\\
e,\\
\sqrt {8 + 2\sqrt {15} } + \sqrt {8 – 2\sqrt {15} } = \sqrt {5 + 2.\sqrt 5 .\sqrt 3 + 3} + \sqrt {5 – 2.\sqrt 5 .\sqrt 3 + 3} \\
= \sqrt {{{\left( {\sqrt 5 + \sqrt 3 } \right)}^2}} + \sqrt {{{\left( {\sqrt 5 – \sqrt 3 } \right)}^2}} = \left( {\sqrt 5 + \sqrt 3 } \right) + \left( {\sqrt 5 – \sqrt 3 } \right) = 2\sqrt 5 \\
f,\\
\sqrt {5 – 2\sqrt 6 } – \sqrt {4 – 2\sqrt 3 } = \sqrt {3 – 2.\sqrt 3 .\sqrt 2 + 2} – \sqrt {3 – 2.\sqrt 3 .1 + 1} \\
= \sqrt {{{\left( {\sqrt 3 – \sqrt 2 } \right)}^2}} – \sqrt {{{\left( {\sqrt 3 – 1} \right)}^2}} = \left( {\sqrt 3 – \sqrt 2 } \right) – \left( {\sqrt 3 – 1} \right) = 1 – \sqrt 2
\end{array}\)