Toán Lớp 9: giải giúp e vs ạ 2 tiếng nx e trả bài r
thực hiện phép tính
√(3-2 √2) ²+ √ (3+2 √2) ²
√(2-√3) ²+√(1-√3) ²
√( √5- √2) ²+ √( √5+√2) ²
√ (5- 2√6) ²- √ (5+2√6) ²
√(3+√2) ²- √(1- √2) ²
√( √2+1) ²- √(√2-5) ²
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Toán Lớp 9: giải giúp e vs ạ 2 tiếng nx e trả bài r
thực hiện phép tính
√(3-2 √2) ²+ √ (3+2 √2) ²
√(2-√3) ²+√(1-√3) ²
√( √5- √2) ²+ √( √5+√2) ²
√ (5- 2√6) ²- √ (5+2√6) ²
√(3+√2) ²- √(1- √2) ²
√( √2+1) ²- √(√2-5) ²
Comments ( 1 )
a)\\
\sqrt {{{\left( {3 – 2\sqrt 2 } \right)}^2}} + \sqrt {{{\left( {3 + 2\sqrt 2 } \right)}^2}} \\
= 3 – 2\sqrt 2 + 3 + 2\sqrt 2 \\
= 6\\
b)\\
\sqrt {{{\left( {2 – \sqrt 3 } \right)}^2}} + \sqrt {{{\left( {1 – \sqrt 3 } \right)}^2}} \\
= 2 – \sqrt 3 + \sqrt 3 – 1\\
= 1\\
c)\\
\sqrt {{{\left( {\sqrt 5 – \sqrt 2 } \right)}^2}} + \sqrt {{{\left( {\sqrt 5 + \sqrt 2 } \right)}^2}} \\
= \sqrt 5 – \sqrt 2 + \sqrt 5 + \sqrt 2 \\
= 2\sqrt 5 \\
d)\sqrt {{{\left( {5 – 2\sqrt 6 } \right)}^2}} – \sqrt {{{\left( {5 + 2\sqrt 6 } \right)}^2}} \\
= 5 – 2\sqrt 6 – \left( {5 + 2\sqrt 6 } \right)\\
= – 4\sqrt 6 \\
e)\sqrt {{{\left( {3 + \sqrt 2 } \right)}^2}} – \sqrt {{{\left( {1 – \sqrt 2 } \right)}^2}} \\
= 3 + \sqrt 2 – \left( {\sqrt 2 – 1} \right)\\
= 4\\
f)\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} – \sqrt {{{\left( {\sqrt 2 – 5} \right)}^2}} \\
= \sqrt 2 + 1 – \left( {5 – \sqrt 2 } \right)\\
= \sqrt 2 + 1 – 5 + \sqrt 2 \\
= 2\sqrt 2 – 4
\end{array}$