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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
\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{6-2\times\sqrt{6}\times3+9}+\sqrt{\left(2\sqrt{6}\right)^2-2\times2\sqrt{6}\times3+9}\)
\(=\sqrt{\left(\sqrt{6}-3\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2=\sqrt{6}-3+2\sqrt{6}-3=3\sqrt{6}-3}\)
Vậy \(a=-3;b=3\) => \(a+b=3-3=0\)
\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\)
\(=\sqrt{3^2-2\cdot3\cdot\sqrt{6}+\left(\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}\right)^2-2\cdot2\sqrt{6}\cdot3+3^2}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
Suy ra: a= 0 và b = 1 => a+b = 1.
\(A=\sqrt{5-2\sqrt{5}+1}-\sqrt{5+2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\sqrt{5}-1-\sqrt{5}-1=-2\)
Vậy \(A\in Z\)
Làm tương tự với B.
a) \(\left(3+1\sqrt{6}-\sqrt{33}\right)\left(\sqrt{22}+\sqrt{6}+4\right)\)
\(=\sqrt{3}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right).\sqrt{2}\left(\sqrt{11}+\sqrt{3}+2\sqrt{2}\right)\)
\(=\sqrt{6}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right)\left(\sqrt{3}+2\sqrt{2}+\sqrt{11}\right)\)
\(=\sqrt{6}\left[\left(\sqrt{3}+2\sqrt{2}\right)^2-11\right]=\sqrt{6}\left(11+4\sqrt{6}-11\right)=\sqrt{6}.4\sqrt{6}=6.4=24\)
b) \(\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)=\left(\frac{5+2\sqrt{6}+10-4\sqrt{6}}{5^2-\left(2\sqrt{6}\right)^2}\right)\left(15+2\sqrt{6}\right)\)
\(=\left(15-2\sqrt{6}\right)\left(15+2\sqrt{6}\right)=15^2-24=201\)
C) \(\left(\frac{4}{3}.\sqrt{3}+\sqrt{2}+\sqrt{3\frac{1}{3}}\right)\left(\sqrt{1,2}+\sqrt{2}-4\sqrt{\frac{1}{5}}\right)\)
\(=\left(\frac{4}{\sqrt{3}}+\frac{\sqrt{6}}{\sqrt{3}}+\frac{\sqrt{10}}{\sqrt{3}}\right)\left(\frac{\sqrt{6}}{\sqrt{5}}+\frac{\sqrt{10}}{\sqrt{5}}-\frac{4}{\sqrt{5}}\right)\)
\(=\frac{1}{\sqrt{15}}\left(\sqrt{6}+\sqrt{10}+4\right)\left(\sqrt{6}+\sqrt{10}-4\right)=\frac{1}{\sqrt{15}}\left[\left(\sqrt{6}+\sqrt{10}\right)^2-16\right]\)
\(=\frac{1}{\sqrt{15}}\left(16+4\sqrt{15}-16\right)=\frac{4\sqrt{15}}{\sqrt{15}}=4\)
d) \(\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1990+2\sqrt{1989}}=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1989+2\sqrt{1989}+1}\)
\(=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{\left(\sqrt{1989}+1\right)^2}=\left(\sqrt{1989}-1\right)\left(\sqrt{1989}+1\right)=1989-1=1988\)
e) \(\frac{a-\sqrt{ab}+b}{a\sqrt{a}+b\sqrt{b}}-\frac{1}{a-b}=\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}-1}{a-b}\)
\(\sqrt{14-6\sqrt{5}}=\left(3-\sqrt{5}\right)^{ }\)
suy ra a=3 ; b=-1
suy ra a+b=3+(-1)=2
cần mk giải chi tiết ko
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=?