\(P=\left(1+2a\right)\left(1+2bc\right)\le\left(1+2a\right)\left(1+b^2+c^2\right)=\left(1+2a\right)\left(2-a^2\right)\)

\(=\frac{3}{2}\left(\frac{2}{3}+\frac{4}{3}a\right)\left(2-a^2\right)\le\frac{3}{8}\left(\frac{8}{3}+\frac{4}{3}a-a^2\right)^2=\frac{3}{8}\left[\frac{28}{9}-\left(a-\frac{2}{3}\right)^2\right]^2\)

\(\le\frac{3}{8}.\left(\frac{28}{9}\right)^2=\frac{98}{27}\)

Dấu \(=\)khi \(\hept{\begin{cases}b=c\\\frac{2}{3}+\frac{4}{3}a=2-a^2,a-\frac{2}{3}=0\\a^2+b^2+c^2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}a=\frac{2}{3}\\b=c=\frac{\sqrt{\frac{5}{2}}}{3}\end{cases}}\).

Vậy \(maxP=\frac{98}{27}\).

Ta co : \(P=2a+2bc+2abc+1\)

Ap dung bdt Co-si : \(P\le a^2+b^2+c^2+2abc+2=2abc+3\)

Tiep tuc ap dung Co-si : \(1=a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}< =>\sqrt[3]{a^2b^2c^2}\le\frac{1}{3}\)

\(< =>a^2b^2c^2\le\frac{1}{27}< =>abc\le\frac{1}{\sqrt{27}}\)

Khi do : \(2abc+3\le2.\frac{1}{\sqrt{27}}+3=\frac{2}{\sqrt{27}}+3\)

Suy ra \(P\le a^2+b^2+c^2+2abc+2\le\frac{2}{\sqrt{27}}+3\)

Dau "=" xay ra khi va chi khi \(a=b=c=\frac{1}{\sqrt{3}}\)

Vay Max P = \(\frac{2}{\sqrt{27}}+3\)khi a = b = c = \(\frac{1}{\sqrt{3}}\) 

p/s : khong biet dau = co dung k nua , minh lam bay do

\(\sqrt{\frac{2a}{9}}.\sqrt{\frac{25a}{32}}=\sqrt{\frac{2a}{9}.\frac{25a}{32}}=\sqrt{\frac{2.\left(5a\right)^2}{2.\left(3.4\right)^2}}=\frac{5a}{12}\)

\(\sqrt{\frac{2a}{9}}\). \(\sqrt{\frac{25a}{32}}\)

\(=\)\(\sqrt{\frac{2a.25a}{9.32}}\)

\(=\)\(\sqrt{\frac{2.\left(5a\right)^2}{2.\left(3.4\right)^2}}\)

\(=\)\(\frac{5a}{12}\)

\(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}+\frac{12}{\sqrt{6}-3}-\sqrt{6}\)

\(\frac{15\left(\sqrt{6}-1\right)}{6-1}+\frac{4\left(\sqrt{6}+2\right)}{6-2^2}+\frac{12\left(\sqrt{6}+3\right)}{6-3^2}-\sqrt{6}\)

\(\frac{15\sqrt{6}-15}{5}+\frac{4\sqrt{6}+8}{2}+\frac{12\sqrt{6}+36}{-3}-\sqrt{6}\)

\(\frac{5\left(3\sqrt{6}-3\right)}{5}+\frac{2\left(2\sqrt{6}+4\right)}{2}-\frac{3\left(4\sqrt{6}+12\right)}{3}-\sqrt{6}\)

\(3\sqrt{6}-3+2\sqrt{6}+4-4\sqrt{6}-12-\sqrt{6}\)

\(=-11\)

\(\frac{a\sqrt{a}-8+2a-4\sqrt{a}}{a-4}\)

\(=\frac{\sqrt{a}\left(a-4\right)+2\left(a-4\right)}{a-4}\)

\(=\frac{\left(a-4\right)\left(\sqrt{a}+2\right)}{a-4}\)

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

Trả lời :

2*căn bậc hai(a)-2

~ HT ~

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

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

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

\(TH1:a\ge4\)

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

\(TH2:a< 4\)

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

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

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

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

+) Nếu \(\sqrt{x-1}-1\ge0\Leftrightarrow x\ge2\) thì \(C=2\)

+) Nếu \(\sqrt{x-1}-1< 0\Leftrightarrow x< 2\Rightarrow1\le x< 2\) thì \(C=2\sqrt{x-1}\)

1) \(\frac{\sqrt{12}-\sqrt{15}}{\sqrt{5}-2}-\frac{1}{\sqrt{3}+2}=\frac{\sqrt{3}\left(2-\sqrt{5}\right)}{\sqrt{5}-2}-\frac{2-\sqrt{3}}{\left(\sqrt{3}+2\right)\left(2-\sqrt{3}\right)}\)

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

2) \(\frac{10}{\sqrt{6}-1}+\frac{\sqrt{12}-\sqrt{18}}{\sqrt{3}-\sqrt{2}}-\frac{2\sqrt{3}}{\sqrt{2}}=\frac{10\left(\sqrt{6}+1\right)}{\left(\sqrt{6}-1\right)\left(\sqrt{6}+1\right)}+\frac{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{3}-\sqrt{2}}-\sqrt{6}\)

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

CM bđt theo phương pháp tương đương:

Ta có: \(\sqrt{14}-\sqrt{13}< 2\sqrt{3}-\sqrt{11}\)

<=> \(\sqrt{14}+\sqrt{11}< \sqrt{12}+\sqrt{13}\)

<=> \(14+11+2\sqrt{14.11}< 12+13+2\sqrt{12.13}\)

<=> \(\sqrt{14.11}< \sqrt{12.13}\)

<=> \(14.11< 12.13\)

Ta có: 14.11 = 12.11 + 2.11 = 12.13 - 2.12 + 2.11 = 12.13 - 2(12 - 11) = 12.13 - 2 < 12.13

=> 14.11 < 12.13 (luôn đúng)

=> \(\sqrt{14}-\sqrt{13}< 2\sqrt{3}-\sqrt{11}\)(luôn đúng)