\(\Rightarrow B^2=5+\sqrt{13+B}\Rightarrow\left(B^2-5\right)^2=13+B\)

\(\Leftrightarrow B^4-10B^2-B+12=0\)

\(\Leftrightarrow\left(B-3\right)\left(B^3+3B^2-B-4\right)=0\)

\(\Leftrightarrow B=3\text{ hoặc }B^3+3B^2-B-4=0\text{ (1)}\)

Lấy máy tính thấy (1) có 2 nghiệm âm và một nghiệm B = 1,11....

Mà \(B>\sqrt{5}>2>0\) nên loại hết các nghiệm của (1) :))

Vậy B = 3.

\(\Rightarrow B^2=5+\sqrt{13+\sqrt{5+\sqrt{13+...}}}\left(B>\sqrt{4}=2\right)\)

\(B^4=25+13+\sqrt{5+\sqrt{13+...}}+2.5.\sqrt{13+\sqrt{5+\sqrt{13+...}}}\)

\(B^4=38+B+10\left(B^2-5\right)\)

\(B^4=10B^2-50+B+38=10B^2+B-12\)

\(\Rightarrow B^4-10B^2-B+12=0\)

\(\Leftrightarrow\left(B-3\right)\left(B^3+3B^2-B-4\right)=0\)

\(\Leftrightarrow\left(B-3\right)\left[B^2\left(B+3\right)-\left(B+3\right)-1\right]=0\)

\(\Leftrightarrow\left(B-3\right)\left[\left(B+3\right)\left(B-1\right)\left(B+1\right)-1\right]=0\left(1\right)\)

Vì B > 2 =>\(\left[\left(B+3\right)\left(B-1\right)\left(B+1\right)-1\right]>0\)

Do đó, (1) => B - 3 = 0 => B = 3 (TMĐK)

\(A^2=2+\sqrt{2+\sqrt{2+...}}\)

\(A^2=2+A\)

\(A^2-A-2=0\Rightarrow A^2-2A+A-2=0\)

\(\left(A-2\right)\left(A+1\right)=0\)

=> A = 2 hoặc A = -1 ( loại A > 0 )

Vậy A = 2

A = \(\sqrt[4]{3^2+2.3.\left(2\sqrt{2}\right)+\left(2\sqrt{2}\right)^2}-\sqrt{2}=\sqrt[4]{\left(3+2\sqrt{2}\right)^2}-\sqrt{2}\)

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

A = \(\sqrt{2}+1-\sqrt{2}=1\)

\(\sqrt[4]{\left(3+2\sqrt{2}\right)^2}-\sqrt{2}=\sqrt{3+2\sqrt{2}}-\sqrt{2}=\sqrt{\left(1+\sqrt{2}\right)^2}-\sqrt{2}=\sqrt{2}+1-\sqrt{2}=1\)

nếu sửa lại a+b+c=0 thì

a+b+c=0

=>0²=(a+b+c)²

<=>0=a²+b²+c²+2ab+2bc+2ac

<=>0=1+2.(ab+bc+ac)

<=>ab+bc+ac=-1/2

=>(ab+bc+ac)²=1/4

<=>a²b²+b²c²+a²c²+2ab²c+2a²bc+2abc²=1/4

<=>a²b²+b²c²+a²c²+2abc(a+b+c)=1/4

<=>a²b²+b²c²+a²c²+2abc.0=1/4

<=>a²b²+b²c²+a²c²=1/4

ta có:

(a²+b²+c²)²=a⁴+b⁴+c⁴+2a²b²+2b²c²+2a²c²

<=>1²=a⁴+b⁴+c⁴+2.(a²b²+b²c²+a²c²)

<=>1=a⁴+b⁴+c⁴+2.1/4

<=>a⁴+b⁴+c⁴=1/2

\(\text{ĐKXĐ :}x\ge-1\)

\(\sqrt{x\left(x+1\right)}+1=0\)

\(\Leftrightarrow\sqrt{x\left(x+1\right)}=-1\)

\(\text{Mà }\sqrt{x\left(x+1\right)}\ge0\text{ Nên PT vô nghiệm}\)

Áp dụng: \(sin^2a+cos^2a=1\)

\(bt=\frac{sin^2a+cos^2a-2sina.cosa}{\left(sina-cosa\right)\left(sina+cosa\right)}=\frac{\left(sina-cosa\right)^2}{\left(sina-cosa\right)\left(sina+cosa\right)}=\frac{sina-cosa}{sina+cosa}\)

Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\). Dấu "=" xảy ra khi \(ab\ge0\)

Ta có: \(A=\left|\sqrt{x-1}-2\right|+\left|4-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}-2+4-\sqrt{x-1}\right|=2\)

Dấu "=" xảy ra khi \(\left(\sqrt{x-1}-2\right)\left(4-\sqrt{x-1}\right)\ge0\)\(\Leftrightarrow\sqrt{x-1}\in\left[2;4\right]\Leftrightarrow x\in\left[5;17\right]\)

Vậy GTNN của A là 2.