\(x=\sqrt[3]{13-7\sqrt{6}}+\sqrt[3]{13+7\sqrt{6}}\Rightarrow x^3=26-15x\)

\(x^3+15x-25=1\Rightarrow\left(x^3+15x-25\right)^{2013}=1\)

Vậy P(x)=1 với .....

=> \(A^2=13+\sqrt{7+\sqrt{13+\sqrt{7+\sqrt{13+\sqrt{7+....}}}}}\)

=>\(\left(A^2-13\right)^2=7+\sqrt{13+\sqrt{7+\sqrt{13+\sqrt{7...}}}}\)

=>\(\left(A^2-13\right)^2=7+A\)

Đến đây tách ra giải PT bậc 4 nha!

Bạn giải giúp mình với bấm không ra nghiệm nơi

*\(A=2\sqrt{80\sqrt{7}}-2\sqrt{45\sqrt{7}}-5\sqrt{20\sqrt{7}}\)

\(A=16\sqrt{5\sqrt{7}}-6\sqrt{5\sqrt{7}}-10\sqrt{5\sqrt{7}}\)

\(A=\left(16-6-10\right)\sqrt{5\sqrt{7}}=0\)

* \(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)

\(B^3=5+2\sqrt{13}+5-2\sqrt{13}+3\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\right).\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\)

\(B^3=10-9B\)

\(\Rightarrow B^3+9B-10=0\)

\(\Rightarrow B^3-B^2+B^2-B+10B-10=0\)

\(\Rightarrow B^2\left(B-1\right)+B\left(B-1\right)+10\left(B-1\right)=0\)

\(\Rightarrow\left(B-1\right)\left(B^2+B+10\right)=0\)

\(\Rightarrow B=1\)

\(=\frac{\left(\sqrt{13}+\sqrt{11}\right)^2+\left(\sqrt{13}-\sqrt{11}\right)^2}{\left(\sqrt{13}-\sqrt{11}\right)\left(\sqrt{13}+\sqrt{11}\right)}\)

\(=\frac{13+2\sqrt{143}+11+13-2\sqrt{143}+11}{13-11}\)

\(=\frac{48}{2}=24\)

ĐKXĐ: x>=0

Để A là số nguyên thì \(\sqrt{x}+13⋮\sqrt{x}+5\)

=>\(\sqrt{x}+5+8⋮\sqrt{x}+5\)

=>\(\sqrt{x}+5\inƯ\left(8\right)\)

mà \(\sqrt{x}+5>=5\)

nên \(\sqrt{x}+5=8\)

=>x=9

ĐK: \(x\ge0\) 

Để \(\dfrac{\sqrt{x}+13}{\sqrt{x}+5}\) có giá trị nguyên 

Mà:  \(\dfrac{\sqrt{x}+13}{\sqrt{x}+5}=\dfrac{\sqrt{x}+5+8}{\sqrt{x}+5}\)

\(=\dfrac{\sqrt{x}+5}{\sqrt{x}+5}+\dfrac{8}{\sqrt{x}+5}=1+\dfrac{8}{\sqrt{x}+5}\)

Vậy:  \(8\) ⋮ \(\sqrt{x}+5\)

\(\Rightarrow\sqrt{x}+5\inƯ\left(8\right)=\left\{1;-1;2;-2;4;-4;8;-8\right\}\)

Mà: \(\sqrt{x}+5\ge5\)

\(\Rightarrow\sqrt{x}+5\in\left\{8\right\}\)

\(\Rightarrow x=9\left(tm\right)\)

áp dụng bất đẳng thức bunhia-copxki ta có

A² <= (1+1)*(x-5+13-x)=16 

=> \(-\sqrt{16}< =A< =\sqrt{16}\)

Vậy giá trị lớn nhất của A là \(\sqrt{16}\)

Có : \(A=\sqrt{x-5}+\sqrt{13-x}>0\)

\(\Leftrightarrow P^2=\sqrt{x-5^2}+2\cdot\sqrt{x-5;13-x}+\sqrt{13-x^2}\)

\(\Leftrightarrow P^2=x-5+13-x+2\cdot\sqrt{-x^2+18x-81+16}\)

\(\Leftrightarrow P^2=8+2\cdot\sqrt{16-x-9^2}\)

Nhận xét : Để \(Pmax\Rightarrow P^2max;8+2\cdot\sqrt{16-x-9^2}max\)

\(\Rightarrow2\cdot\sqrt{16-x-9^2}max\Rightarrow16-x-9^2max\)

Nhận xét : \(x-9^2>=\Rightarrow-x-9< =16\)

Để \(\Rightarrow16-x-9^2max\)thì \(16-x-9^2=16\Rightarrow x=9\)

Khi \(x=9\Rightarrow P^2=8+2\cdot\sqrt{16}=16\)

\(\Rightarrow P=4\)

Vậy ta kết luật: \(Amax=4\Leftrightarrow x=9\)

P/s: Chị thay P thành A nha coi chừng sai đề nha

       Em ko chắc đâu ạ

còn cách này

\(A^2=x-5+2\sqrt{\left(x-5\right)\left(13-x\right)}+13-x=8+2\sqrt{\left(x-5\right)\left(13-x\right)}\)

Áp dụng bđt Cô-si: \(2\sqrt{\left(x-5\right)\left(13-x\right)}\le x-5+13-x=8\)

=>\(A^2=8+2\sqrt{\left(x-5\right)\left(13-x\right)}\le8+8=16\)

=>\(-4\le A\le4\)=>\(maxA=4\Leftrightarrow x=9\)

\(A=\frac{\sqrt{x}-13}{\sqrt{x}+3}=\frac{-\frac{13}{3}\sqrt{x}-13+\frac{16}{3}\sqrt{x}}{\sqrt{x}+3}=\frac{-\frac{13}{3}\left(\sqrt{x}+3\right)+\frac{16}{3}\sqrt{x}}{\sqrt{x}+3}\)

\(=-\frac{13}{3}+\frac{\frac{16}{3}\sqrt{x}}{\sqrt{x}+3}=-\frac{13}{3}+\frac{16\sqrt{x}}{3\sqrt{x}+9}\ge-\frac{13}{3}\)có GTNN là \(-\frac{13}{3}\)

sữa lại câu cuối cho Nhã Doanh

\(\sqrt{22-2\sqrt{21}-\sqrt{22+2\sqrt{21}}}=\sqrt{22-2\sqrt{21}-\sqrt{\left(\sqrt{21}+1\right)^2}}\)

\(=\sqrt{22-2\sqrt{21}-\sqrt{21}-1}=\sqrt{21-3\sqrt{21}}\)

\(a.\sqrt{8+2\sqrt{7}}-\sqrt{7}=\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{7}=\sqrt{7}+1-\sqrt{7}=1\)

\(b.\sqrt{7+4\sqrt{3}}-2\sqrt{3}=\sqrt{\left(2+\sqrt{3}\right)^2}-2\sqrt{3}=2+\sqrt{3}-2\sqrt{3}=2-\sqrt{3}\)

\(c.\sqrt{14-2\sqrt{13}}+\sqrt{14+2\sqrt{13}}=\sqrt{\left(\sqrt{13}-1\right)^2}+\sqrt{\left(\sqrt{13}+1\right)^2}=\sqrt{13}-1+\sqrt{13}+1=2\sqrt{13}\)\(d.\sqrt{22-2\sqrt{21}-\sqrt{22+2\sqrt{21}}}=\sqrt{\left(\sqrt{21}-1\right)^2-\sqrt{\left(\sqrt{21}+1\right)^2}}=\sqrt{21}-1-\sqrt{\sqrt{21}+1}\)