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1.Nếu $\sqrt{55-6\sqrt{6}}=a+b\sqrt{6}$√55−6√6=a+b√6 với $a,b\in Z$a,b∈Z thì a-b=?
2. Nếu $\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=a+b\sqrt{6}$√15−6√6+√33−12√6=a+b√6 với $a,b\in Z$a,b∈Z thì a+b=?
\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\)
\(=\sqrt{3^2-2\cdot3\cdot\sqrt{6}+\left(\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}\right)^2-2\cdot2\sqrt{6}\cdot3+3^2}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
Suy ra: a= 0 và b = 1 => a+b = 1.
1/ Ta có √(14 - 6√5) = √(9 - 6√5 +5) = 3 - √5
Từ đó a + b = 2
2/ Đề sai sửa lại là
√(15 - 6√6) = √(9 - 6√6 + 6) = (3 - √6)
Vậy a = 3; b = -1
=> a + b = 2
a) \(\sqrt{3-2\sqrt{2}}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\sqrt{\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{2}-1\right|+\left|2-\sqrt{2}\right|\)
\(=\sqrt{2}-1+2-\sqrt{2}\)
\(=1\)
b) \(\sqrt{33-12\sqrt{6}}-\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(2\sqrt{6}\right)^2-2\cdot2\sqrt{6}\cdot3+3^2}-\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(2\sqrt{6}-3\right)^2}-\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=\left|2\sqrt{6}-3\right|-\left|5-2\sqrt{6}\right|\)
\(=2\sqrt{6}-3-5+2\sqrt{6}\)
\(=4\sqrt{6}-8\)
c) \(\sqrt{7-2\sqrt{6}}+\sqrt{15-6\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}+\sqrt{3^2-2\cdot3\cdot\sqrt{6}+\left(\sqrt{6}\right)^2}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}+\sqrt{\left(3-\sqrt{6}\right)^2}\)
\(=\left|\sqrt{6}-1\right|+\left|3-\sqrt{6}\right|\)
\(=\sqrt{6}-1+3-\sqrt{6}\)
\(=2\)
\(a,\sqrt{3-2\sqrt{2}}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\sqrt{\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1}+\left|2-\sqrt{2}\right|\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}+2-\sqrt{2}\)
\(=\left|\sqrt{2}-1\right|+2-\sqrt{2}\)
\(=\sqrt{2}-1+2-\sqrt{2}\)
\(=1\)
\(---\)
\(b,\sqrt{33-12\sqrt{6}}-\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(2\sqrt{6}\right)^2-2\cdot2\sqrt{6}\cdot3+3^2}-\left|5-2\sqrt{6}\right|\)
\(=\sqrt{\left(2\sqrt{6}-3\right)^2}-5+2\sqrt{6}\)
\(=\left|2\sqrt{6}-3\right|-5+2\sqrt{6}\)
\(=2\sqrt{6}-3-5+2\sqrt{6}\)
\(=4\sqrt{6}-8\)
\(---\)
\(c,\sqrt{7-2\sqrt{6}}+\sqrt{15-6\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}+\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot3+3^2}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}+\sqrt{\left(\sqrt{6}-3\right)^2}\)
\(=\left|\sqrt{6}-1\right|+\left|\sqrt{6}-3\right|\)
\(=\sqrt{6}-1+3-\sqrt{6}\)
\(=2\)
#\(Toru\)
Ta có $\sqrt{55-6\sqrt{6}}$ = $\sqrt{55-2.3.\sqrt{6}}$ = $\sqrt{55-2\sqrt{54}}$ = $\sqrt{\left(54^2\right)-2.54+1}$ = $\sqrt{\left(\sqrt{54}-1\right)^2}$ = $\sqrt{54-1}$ = $3\sqrt{6}$ -1
$\Rightarrow $ a=-1 và b=3
$\Rightarrow $ a-b=-1-3=-4
a) \(\sqrt{33-12\sqrt{6}}+\sqrt{15+6\sqrt{6}}=\sqrt{24-2.2\sqrt{6}.3+9}+\sqrt{6+2.\sqrt{6}.3+9}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(\sqrt{6}+3\right)^2}=\left|2\sqrt{6}-3\right|+\left|\sqrt{6}+3\right|=2\sqrt{6}-3+\sqrt{6}+3=3\sqrt{6}\)
b) \(\dfrac{\sqrt{99}}{\sqrt{11}}+\dfrac{\sqrt{28}}{\sqrt{7}}-\sqrt{\sqrt{81}}=\sqrt{\dfrac{99}{11}}+\sqrt{\dfrac{28}{7}}-\sqrt{9}=\sqrt{9}+\sqrt{4}-\sqrt{9}=\sqrt{4}=2\)
a) \(\sqrt{33-12\sqrt{6}}\) + \(\sqrt{15+6\sqrt{6}}\)
= \(\sqrt{9-2.3.2\sqrt{6}+24}\)+\(\sqrt{9+2.3\sqrt{6}+6}\)
= \(\sqrt{\left(3-2\sqrt{6}\right)^2}\)+\(\sqrt{\left(3+\sqrt{6}\right)^2}\)
=\(\left|3-2\sqrt{6}\right|+\left|3+\sqrt{6}\right|\)
=\(2\sqrt{6}-3+3+\sqrt{6}\)
=\(\sqrt{6}\)
b)\(\dfrac{\sqrt{99}}{\sqrt{11}}\)+\(\dfrac{\sqrt{28}}{\sqrt{7}}\)\(-\sqrt{\sqrt{81}}\)
= \(\sqrt{\dfrac{99}{11}}+\sqrt{\dfrac{28}{7}}-3\)
=\(\sqrt{9}+\sqrt{4}-3\)
= 3+2-3
= 2
\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{6-2\times\sqrt{6}\times3+9}+\sqrt{\left(2\sqrt{6}\right)^2-2\times2\sqrt{6}\times3+9}\)
\(=\sqrt{\left(\sqrt{6}-3\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2=\sqrt{6}-3+2\sqrt{6}-3=3\sqrt{6}-3}\)
Vậy \(a=-3;b=3\) => \(a+b=3-3=0\)
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