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

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

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

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.

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

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

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

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

Nhầm a - b = 4 

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) \(\sqrt{7-2\sqrt{6}}+\sqrt{15-6\sqrt{6}}\)

\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}+\sqrt{3^2-2\cdot3\cdot\sqrt{6}+\left(\sqrt{6}\right)^2}\)

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

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

\(=\sqrt{6}-1+3-\sqrt{6}\)

\(=2\)

\(a,\sqrt{3-2\sqrt{2}}+\sqrt{\left(2-\sqrt{2}\right)^2}\)

\(=\sqrt{\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1}+\left|2-\sqrt{2}\right|\)

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

\(=\left|\sqrt{2}-1\right|+2-\sqrt{2}\)

\(=\sqrt{2}-1+2-\sqrt{2}\)

\(=1\)

\(---\)

\(b,\sqrt{33-12\sqrt{6}}-\sqrt{\left(5-2\sqrt{6}\right)^2}\)

\(=\sqrt{\left(2\sqrt{6}\right)^2-2\cdot2\sqrt{6}\cdot3+3^2}-\left|5-2\sqrt{6}\right|\)

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

\(=\left|2\sqrt{6}-3\right|-5+2\sqrt{6}\)

\(=2\sqrt{6}-3-5+2\sqrt{6}\)

\(=4\sqrt{6}-8\)

\(---\)

\(c,\sqrt{7-2\sqrt{6}}+\sqrt{15-6\sqrt{6}}\)

\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}+\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot3+3^2}\)

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

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

\(=\sqrt{6}-1+3-\sqrt{6}\)

\(=2\)

#\(Toru\)

C1: Bình phương 2 vế ta có: \(55-6\sqrt{6}=\left(a+b\sqrt{6}\right)^2\)

<=> \(55-6\sqrt{6}=a^2 +6b^2+2ab\sqrt{6}\)

=>  a² + 6b² = 55 và 2ab = - 6

=> a² + 6b² = 55   (1)   và ab = -3  => a = -3/b (2)

thế (2) vào (1) ta được : \(\left(-\frac{3}{b}\right)^2+6b^2=55\) => \(9+6b^4=55b^2\)

=> 6b⁴ - 55b² + 9 = 0 => 6b⁴ - 54b² - b² + 9 =0 <=> 6b².(b² - 9) - (b² - 9) = 0 <=> (6b² - 1).(b² - 9 ) = 0 

<=> b² = 1/6 (Loại; vì b nguyên )  hoặc b² = 9 

+) b² = 9 => a² = 1 => a = 1 hoặc - 1 ; b = 3 hoặc - 3

Do \(a+b\sqrt{6}\) > 0  và a; b trái dấu nên a =  -1; b = 3 => a+ b = 2

Vậy a +  b  = 2

C2: \(\sqrt{55-6\sqrt{6}}=\sqrt{\left(3\sqrt{6}\right)^2-2.3\sqrt{6}.1+1}=\sqrt{\left(3\sqrt{6}-1\right)^2}\)

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

=> a = -1; b = 3 => a + b = 2