áp dụng bdt (a²+b²+c²)(x²+y²+z²)\(\ge\left(ax+by+cz\right)^2\) dấu '=" khi \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)

(\(\left(\sqrt{2b+c}\right)^2+\left(\sqrt{2c+a}\right)^2+\left(\sqrt{2a+b}\right)^2\)). P\(\ge\left(a+b+c\right)^2\)

<=> P\(\ge\frac{\left(a+b+c\right)^2}{3\left(a+b+c\right)}=\frac{a+b+c}{3}=\frac{2018}{3}\)=> P min= \(\frac{2018}{3}\)

P min khi \(\frac{a}{2b+c}=\frac{b}{2c+a}=\frac{c}{2b+a}\)<=> a=b=c= \(\frac{2018}{3}\)

Mẫu = y -6 căn y + 9 + 2

= (căn y - 3)² + 2 >=2

=
> 1/(( căn y -3 )² + 2) <= 1/2

=> Max_B = 1/2 khi căn y - 3 = 0 (=) y =9

học tốt

A= \(a^{2017}\left(a^2-8a+11\right)+b^{2017}\left(b^2-8b+11\right)=\)\(a^{2017}\left(a^2-8a+16-5\right)+b^{2017}\left(b^2-8b+16-5\right)=\)\(a^{2017}\left(\left(a-4\right)^2-\sqrt{5^2}\right)+b^{2017}\left(\left(b-4\right)^2-\sqrt{5^2}\right)\)=\(a^{2017}\left(a-4-\sqrt{5}\right)\left(a-4+\sqrt{5}\right)+b^{2017}\left(b-4-\sqrt{5}\right)\left(b-4+\sqrt{5}\right)\)= 0+0= 0

Kẻ đường cao BH

diện tích tam giác ABC là \(\frac{1}{2}BH.AC=\frac{1}{2}ABsin\widehat{A}.\left(6-AB\right)\le\frac{9}{2}sin\widehat{A}\) vì AB(6-AB)= 6AB-AB² = 9- (AB-3)² \(\le9\)

vậy diện tích ABC lớn nhất khi AB-3=0 hay AB=AC =3

\(\sqrt{a-2}.b^2=b-\sqrt{a-2}\left(a\ge2\right)\)

\(\Leftrightarrow\sqrt{a-2}.b^2-b+\sqrt{a-2}=0\)

Để pt có nghiệm \(\Leftrightarrow\Delta A\ge0\)

\(\Leftrightarrow1-4\left(a-2\right)=0\Leftrightarrow9-4a\ge0\Leftrightarrow a\le2,25\)

Khi đó a đạt GTLN là 2,25

Với a = 2,25 ta có \(\frac{1}{2}b^2=b-\frac{1}{2}\Leftrightarrow b^2-2b+1=0\Leftrightarrow b=1\)

Vậy cặp (a;b) cần tìm là : ( 2,25 ; 1 )

Chúc bạn học tốt !!!

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2- căn 3

Trả lời:

\(\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}+\frac{3+6\sqrt{3}}{\sqrt{3}}-\frac{13}{\sqrt{3}+4}\)

\(=-\frac{\sqrt{3}.\left(\sqrt{2}-1\right)}{\sqrt{2}-1}+\frac{\sqrt{3}.\left(\sqrt{3}+6\right)}{\sqrt{3}}-\frac{13.\left(\sqrt{3}-4\right)}{3-16}\)

\(=-\sqrt{3}+\sqrt{3}+6-\frac{13.\left(\sqrt{3}-4\right)}{-13}\)

\(=-\sqrt{3}+\sqrt{3}+6+\sqrt{3}-4\)

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