Giải thích các bước giải: Ta có: \(\begin{array}{l}a,\\5 – 2\sqrt 5 = {\sqrt 5 ^2} – 2\sqrt 5 = \sqrt 5 \left( {\sqrt 5 – 2} \right)\\b,\\5\sqrt {10} – 2\sqrt 5 = 5.\sqrt 2 .\sqrt 5 – 2\sqrt 5 = \sqrt 5 .\left( {5\sqrt 2 – 2} \right)\\c,\\\sqrt {20} – \sqrt {30} = \sqrt {10} .\sqrt 2 – \sqrt {10} .\sqrt 3 = \sqrt {10} .\left( {\sqrt 2 – \sqrt 3 } \right)\\d,\\3\sqrt 3 – 9 = 3\sqrt 3 – {3^2} = 3\sqrt 3 .\left( {1 – \sqrt 3 } \right)\\e,\\2\sqrt 5 – 8 = 2\sqrt 5 – 2\sqrt 2 = 2\left( {\sqrt 5 – \sqrt 2 } \right)\\f,\\2\sqrt {10} – 4\sqrt 5 = 2\sqrt 2 .\sqrt 5 – 4\sqrt 5 = 2\sqrt 5 \left( {\sqrt 2 – 2} \right)\\h,\\a + 4\sqrt a = {\sqrt a ^2} + 4\sqrt a = \sqrt a \left( {\sqrt a + 4} \right)\\i,\\a\sqrt a + 3\sqrt a = \sqrt a \left( {a + 3} \right)\\k,\\2\sqrt a + 3\sqrt {ab} = \sqrt a \left( {2 + 3\sqrt b } \right)\\l,\\a – 4\sqrt a + 4 = {\sqrt a ^2} – 2.\sqrt a .2 + {2^2} = {\left( {\sqrt a – 2} \right)^2}\\m,\\a + 6\sqrt a + 9 = {\left( {\sqrt a + 3} \right)^2}\\n,\\a\sqrt a + 1 = {\sqrt a ^3} + {1^3} = \left( {\sqrt a + 1} \right)\left( {a – \sqrt a + 1} \right)\\o,\\a – 9 = {\sqrt a ^2} – {3^2} = \left( {\sqrt a – 3} \right)\left( {\sqrt a + 3} \right)\\a\sqrt a + 8 = {\sqrt a ^3} + {2^3} = \left( {\sqrt a + 2} \right)\left( {a – 2\sqrt a + 4} \right)\\a – 4\sqrt a – 7 = \left( {a – 4\sqrt a + 4} \right) – 11 = {\left( {\sqrt a – 2} \right)^2} – 11 = \left( {\sqrt a – 2 – \sqrt {11} } \right)\left( {\sqrt a – 2 + \sqrt {11} } \right)\end{array}\) Reply
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
5 – 2\sqrt 5 = {\sqrt 5 ^2} – 2\sqrt 5 = \sqrt 5 \left( {\sqrt 5 – 2} \right)\\
b,\\
5\sqrt {10} – 2\sqrt 5 = 5.\sqrt 2 .\sqrt 5 – 2\sqrt 5 = \sqrt 5 .\left( {5\sqrt 2 – 2} \right)\\
c,\\
\sqrt {20} – \sqrt {30} = \sqrt {10} .\sqrt 2 – \sqrt {10} .\sqrt 3 = \sqrt {10} .\left( {\sqrt 2 – \sqrt 3 } \right)\\
d,\\
3\sqrt 3 – 9 = 3\sqrt 3 – {3^2} = 3\sqrt 3 .\left( {1 – \sqrt 3 } \right)\\
e,\\
2\sqrt 5 – 8 = 2\sqrt 5 – 2\sqrt 2 = 2\left( {\sqrt 5 – \sqrt 2 } \right)\\
f,\\
2\sqrt {10} – 4\sqrt 5 = 2\sqrt 2 .\sqrt 5 – 4\sqrt 5 = 2\sqrt 5 \left( {\sqrt 2 – 2} \right)\\
h,\\
a + 4\sqrt a = {\sqrt a ^2} + 4\sqrt a = \sqrt a \left( {\sqrt a + 4} \right)\\
i,\\
a\sqrt a + 3\sqrt a = \sqrt a \left( {a + 3} \right)\\
k,\\
2\sqrt a + 3\sqrt {ab} = \sqrt a \left( {2 + 3\sqrt b } \right)\\
l,\\
a – 4\sqrt a + 4 = {\sqrt a ^2} – 2.\sqrt a .2 + {2^2} = {\left( {\sqrt a – 2} \right)^2}\\
m,\\
a + 6\sqrt a + 9 = {\left( {\sqrt a + 3} \right)^2}\\
n,\\
a\sqrt a + 1 = {\sqrt a ^3} + {1^3} = \left( {\sqrt a + 1} \right)\left( {a – \sqrt a + 1} \right)\\
o,\\
a – 9 = {\sqrt a ^2} – {3^2} = \left( {\sqrt a – 3} \right)\left( {\sqrt a + 3} \right)\\
a\sqrt a + 8 = {\sqrt a ^3} + {2^3} = \left( {\sqrt a + 2} \right)\left( {a – 2\sqrt a + 4} \right)\\
a – 4\sqrt a – 7 = \left( {a – 4\sqrt a + 4} \right) – 11 = {\left( {\sqrt a – 2} \right)^2} – 11 = \left( {\sqrt a – 2 – \sqrt {11} } \right)\left( {\sqrt a – 2 + \sqrt {11} } \right)
\end{array}\)