\(\sqrt{16-2\sqrt{55}}=\sqrt{11}-\sqrt{5}\)

=>a=11; b=5

=>a-b=6

\(\sqrt{16-2\sqrt{55}}=\sqrt{\left(\sqrt{11}-\sqrt{5}\right)^2}=\sqrt{11}-\sqrt{5}\)

suy ra a=11;b=5

suy ra a+b=11+5=16

 a+b=16

Nếu \(\sqrt{16-2\sqrt{55}}=\) \(\sqrt{a}-\sqrt{b}\) với \(a,b\)\(EZ\)thì a + b = 16.

a) Pt \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=5\Leftrightarrow\left|x-2\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

Vậy...

b)Đk: \(x\ge-1\)

Pt \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}\)

\(\Leftrightarrow4\sqrt{x+1}=16\)\(\Leftrightarrow x+1=16\)\(\Leftrightarrow x=15\) (tm)

Vậy...

\(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\) (a>0)

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

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

b) \(A=a-\sqrt{a}=a-2.\dfrac{1}{2}\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Dấu "=" xảy ra khi \(\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\left(tmđk\right)\) 

Vậy \(A_{min}=-\dfrac{1}{4}\)

a) \(\sqrt{x^2-4x+4}=5\Rightarrow\sqrt{\left(x-2\right)^2}=5\Rightarrow\left|x-2\right|=5\)

\(\Rightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

b) \(\sqrt{16x+16}-3\sqrt{x+1}+\sqrt{4x+4}=16-\sqrt{x+1}\)

\(\Rightarrow\sqrt{16\left(x+1\right)}-3\sqrt{x+1}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)

\(\Rightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Rightarrow4\sqrt{x+1}=16\Rightarrow\sqrt{x+1}=4\Rightarrow x=15\)

a) \(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)

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

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

b) Ta có: \(a-\sqrt{a}=\left(\sqrt{a}\right)^2-2.\sqrt{a}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)

\(=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

\(\Rightarrow A_{min}=-\dfrac{1}{4}\) khi \(a=\dfrac{1}{4}\)

✱ giải pt:

a.\(\sqrt{x^2-4x+4}\)\(=5\)

⇔\(\sqrt{\left(x-2\right)^2}=5\)

⇒\(\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

vậy....

b.\(\sqrt{16x+16}-3\sqrt{x+1}+\sqrt{4x+4}=16-\sqrt{x+1}\)

⇔ \(4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

⇔ \(4\sqrt{x+1}=16\)

⇔ \(\sqrt{x+1}=16\)

⇒ \(x+1=256\)

⇔ \(x=255\)

vậy.....

\(\sqrt{16-2\sqrt{55}}=\sqrt{\left(\sqrt{11}-\sqrt{5}\right)^2}\)

=\(\sqrt{11}-\sqrt{5}\)

=> a=11 và b=5

=> a-b=6

a: Khi x=16 thì \(A=\dfrac{4+1}{4-1}=\dfrac{5}{3}\)

b: \(P=\dfrac{x+4\sqrt{x}+4-3\sqrt{x}+6-12}{x-4}=\dfrac{x+\sqrt{x}-2}{x-4}=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)

c: \(P=A\cdot B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}=1+\dfrac{3}{\sqrt{x}-2}\)

Để P lớn nhất thì căn x-2=1

=>căn x=3

=>x=9

a) \(\sqrt{7}.\sqrt{55}.\sqrt{35}.\sqrt{11}=\sqrt{7.55.35.11}=\sqrt{7.5.11.5.7.11}=\sqrt{\left(5.7.11\right)^2}\)

\(=5.7.11=385\)

b) \(\frac{\sqrt{144}}{23}:\frac{\sqrt{16}}{23}=\frac{\sqrt{144}}{23}.\frac{23}{\sqrt{16}}=\frac{\sqrt{144}}{\sqrt{16}}=\sqrt{\frac{144}{16}}=\sqrt{9}=3\)

c) \(\frac{\sqrt{5}}{\sqrt{125}}=\sqrt{\frac{5}{125}}=\sqrt{\frac{1}{25}}=\frac{1}{5}\)

d) \(\frac{\sqrt{135}}{\sqrt{15}}=\sqrt{\frac{135}{15}}=\sqrt{9}=3\)

a)\(\sqrt{7}.\sqrt{55}.\sqrt{35}.\sqrt{11}=\left(\sqrt{7}.\sqrt{355}\right).\left(\sqrt{35}.\sqrt{11}\right)=\sqrt{385}.\sqrt{385}=385\)

b) \(\frac{\sqrt{144}}{23}:\frac{\sqrt{16}}{23}=\frac{12}{23}.\frac{23}{4}=3\)

c) \(\frac{\sqrt{5}}{\sqrt{125}}=\sqrt{\frac{5}{125}}=\sqrt{\frac{1}{25}}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}\)

d) \(\frac{\sqrt{135}}{\sqrt{15}}=\sqrt{\frac{135}{15}}=\sqrt{9}=3\)

           Bài làm :

\(\text{a)}\sqrt{7}.\sqrt{55}.\sqrt{35}.\sqrt{11}=\left(\sqrt{7}.\sqrt{355}\right).\left(\sqrt{35}.\sqrt{11}\right)=\sqrt{385}.\sqrt{385}=385\)

 \(\text{b)}\frac{\sqrt{144}}{23}\div\frac{\sqrt{16}}{23}=\frac{12}{23}.\frac{23}{4}=\frac{12}{4}=3\)

 \(\text{c)}\frac{\sqrt{5}}{\sqrt{125}}=\sqrt{\frac{5}{125}}=\sqrt{\frac{1}{25}}=\frac{1}{5}\)

 \(\text{d)}\frac{\sqrt{135}}{\sqrt{15}}=\sqrt{\frac{135}{15}}=\sqrt{9}=3\)

\(\sqrt{16-2\sqrt{55}}=\sqrt{11}-\sqrt{5}\)

=> A-B= 11-5 =6

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