\(\sqrt{55-6\sqrt{6}}=\sqrt{3\sqrt{6}-2\cdot3\sqrt{6}+1}=\sqrt{\left(3\sqrt{6}-1\right)^2}=3\sqrt{6}-1=3\sqrt{6}+\left(-1\right)\)

\(=>a=-1;b=3\)

\(=>a-b=-1-3=-4\)

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=?

phantuananh cảm ơn nka

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

\(\sqrt{14-6\sqrt{5}}=\sqrt{14-2\sqrt{9.5}}=\sqrt{\left(3-\sqrt{5}\right)^2}\)

=\(3-\sqrt{5}\)

=> a=3 và b=-1

=> a+b=3-1=2

ta có : \(\sqrt{55-6\sqrt{6}}=\sqrt{55-2\sqrt{54}}\)

= \(\sqrt{54-2\sqrt{54.1}+1}=\sqrt{\left(\sqrt{54}-1\right)^2}\)

=\(\left|3\sqrt{6}-1\right|=3\sqrt{6}-1\)

=>a=-1 và b=3

=> a-b=-1-3=-4

\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{6-2\times\sqrt{6}\times3+9}+\sqrt{\left(2\sqrt{6}\right)^2-2\times2\sqrt{6}\times3+9}\)

\(=\sqrt{\left(\sqrt{6}-3\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2=\sqrt{6}-3+2\sqrt{6}-3=3\sqrt{6}-3}\)

Vậy \(a=-3;b=3\) => \(a+b=3-3=0\)

Cứ thu gọn VT đi xong sẽ thấy

\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{6-2\times\sqrt{6}\times3+9}\)\(+\sqrt{\left(2\sqrt{6}\right)^2-2\times2\sqrt{6}\times3+9}\)

\(=\sqrt{\left(\sqrt{6}-3\right)^2}\)\(+\sqrt{\left(2\sqrt{6}-3\right)^2=\sqrt{6}-3+2\sqrt{6}-3=3\sqrt{6}-3}\)

\(V\text{ậya=-3;b=3a=-3;b=3}\)

\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\)

\(=\sqrt{3^2-2\cdot3\cdot\sqrt{6}+\left(\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}\right)^2-2\cdot2\sqrt{6}\cdot3+3^2}\)

\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)

Suy ra: a= 0 và b = 1 => a+b = 1.

\(\sqrt{14-6\sqrt{5}}=\left(3-\sqrt{5}\right)^{ }\)

suy ra a=3 ; b=-1

suy ra a+b=3+(-1)=2

ai cha tui nhầm mất òi xin lỗi nha