ta có :

\(P=ab\le\left(\frac{a+b}{2}\right)^2=\frac{121}{4}\)

vậy GTLN của P là \(121\text{ khi }\hept{\begin{cases}a+b=11\\a=b\end{cases}\Leftrightarrow a=b=\frac{11}{2}}\)

mình nhầm đề bài ạ!a\(\le\)3 ạ

a)=1-4a
b) = 2x - 4y
c) = 2x - 2 (nếu x>5)
=2x(nếu x<5)
 

-1.       2x.        2x

-\(x+3+\sqrt{x^2-6x+9}\)

\(=x+3+\left|x\right|-6x+9\)

\(x< 0\)

\(--->x+3-x-6x+9\)

\(=\left(x-x\right)-6x+3+9\)

\(=-6x+\left(3+9\right)=-6x+12\)

\(x>0\)

\(--->3+x+x-6x+9\)

\(=\left(x+x-6x\right)+\left(3+9\right)\)

\(=\left(2x-6x\right)+12\)

\(=4x+12\)

a) A=6
b) B=1
 

\(\sqrt{13+30\sqrt{2}+\sqrt{9+4\sqrt{2}}}\)

\(=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}\right)^2}+4\sqrt{2}+1^2}}\)

\(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1^2}}\)

\(=\sqrt{13+30\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}+1^2}}\)

\(=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)

\(=\sqrt{5^2+2.5.3\sqrt{2}+\left(3+\sqrt{2}\right)^2}\)

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

\(=\sqrt{\left(5+6\right)}=\sqrt{11}\)

\(=5+6=11\)

a)=1
b)=\(3\sqrt{2}+5\)

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

\(=\sqrt{15+2.3.\sqrt{6}}\)\(-\sqrt{10+2.2\sqrt{6}}\)

\(=\sqrt{9+2.3\sqrt{6}+6}\)\(-\sqrt{6+2.\sqrt{6}.2+4}\)

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

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

\(=1+0=1\)

a)  \((\sqrt{3}-\sqrt{2}).\sqrt{(\sqrt{3}+\sqrt{2})^2}\)

\(\left(\sqrt{3}-\sqrt{2}\right).\left(\sqrt{3}+\sqrt{2}\right)\)

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

b)  \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)

=\(\sqrt{(2+2\sqrt{5})^2}+\sqrt{(\sqrt{5}-2)^2}\)

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