=>a^2-ab-2ab+2b^2=0

=>(a-b)(a-2b)=0

=>a=b(loại) hoặc a=2b

Khi a=2b thì G=(4b+b)/(2b+2b)=5/4

1.

- Với \(a+b\ge4\Rightarrow A\le0\)

- Với \(a+b< 4\Rightarrow4-a-b>0\)

\(\Rightarrow A=\dfrac{a}{2}.\dfrac{a}{2}.b.\left(4-a-b\right)\)

\(\Rightarrow A\le\dfrac{1}{64}\left(\dfrac{a}{2}+\dfrac{a}{2}+b+4-a-b\right)^4=4\)

\(A_{max}=4\) khi \(\left(a;b\right)=\left(2;1\right)\)

2.

\(P=a+\dfrac{1}{2}.a.2b\left(1+2c\right)\le a+\dfrac{a}{8}\left(2b+1+2c\right)^2\)

\(P\le a+\dfrac{a}{8}\left(7-2a\right)^2=\dfrac{1}{8}\left(4a^3-28a^2+57a-36\right)+\dfrac{9}{2}\)

\(P\le\dfrac{1}{8}\left(a-4\right)\left(2a-3\right)^2+\dfrac{9}{2}\le\dfrac{9}{2}\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\dfrac{3}{2};1;\dfrac{1}{2}\right)\)

Câu 3 bạn xem lại đề, mình có thể chắc chắn với bạn là đề sai

Ví dụ bạn cho \(x=98,y=100\) thì vế trái chỉ lớn hơn 8 một chút

Đề đúng phải là: \(\left(x+y\right)\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+\dfrac{16xy}{\left(x-y\right)^2}\ge12\)

Nếu câu 3 đề là \(\left(x+y\right)\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+\dfrac{16xy}{\left(x-y\right)^2}\ge12\)

Ta có:

\(VT=2+\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{16xy}{\left(x-y\right)^2}=\dfrac{x^2+y^2}{xy}+\dfrac{16xy}{\left(x-y\right)^2}+2\)

\(VT=\dfrac{x^2+y^2-2xy+2xy}{xy}+\dfrac{16xy}{\left(x-y\right)^2}+2\)

\(VT=\dfrac{\left(x-y\right)^2}{xy}+\dfrac{16xy}{\left(x-y\right)^2}+4\ge2\sqrt{\dfrac{16xy\left(x-y\right)^2}{xy\left(x-y\right)^2}}+4=12\)

Ta có:

\(2A+54\ge2\left(3ab+bc+ca\right)+3\left(a^2+b^2+c^2\right)\)

\(=\left(a+b+c\right)^2+2\left(a+b\right)^2+2c^2\ge0\)

\(\Rightarrow2A\ge-54\Rightarrow A\ge-27\)

Dấu = khi a=3;b=-3;c=0

Lời giải:

$1=a+b+3ab\leq (a+b)+3.\frac{(a+b)^2}{4}$

$\Rightarrow a+b\geq \frac{2}{3}$

$\Rightarrow a^2+b^2\geq \frac{(a+b)^2}{2}=\frac{2}{9}$

\(p=\sqrt{1-a^2}+\sqrt{1-b^2}+\frac{1-(a+b)}{a+b}=\sqrt{1-a^2}+\sqrt{1-b^2}+\frac{1}{a+b}-1\)

\(\leq \sqrt{(1-a^2+1-b^2)(1+1)}+\frac{1}{\frac{2}{3}}-1=\sqrt{2(2-a^2-b^2)}+\frac{1}{2}\)

Mà \(2-a^2-b^2\leq 2-\frac{2}{9}=\frac{16}{9}\)

Do đó:

\(P\leq \sqrt{\frac{32}{9}}+\frac{1}{2}=\frac{3+8\sqrt{2}}{6}\) và đây chính là giá trị max.

Ta có: \(A=\left(a+b\right)\left(a^2-ab+b^2\right)+\dfrac{6}{a^2+b^2}+3ab\)

               \(=2\left(a^2+b^2\right)+\dfrac{6}{a^2+b^2}+ab\)

               \(=\left[\dfrac{3}{2}\left(a^2+b^2\right)+\dfrac{6}{a^2+b^2}\right]+\dfrac{a^2+b^2}{2}+ab\)

               \(\ge2\sqrt{\dfrac{3}{2}\left(a^2+b^2\right).\dfrac{6}{a^2+b^2}}+\dfrac{\left(a+b\right)^2}{2}=2.3+\dfrac{2^2}{2}=8\)

Dấu "=" xảy ra ⇔ a=b=1

Bạn tham khảo nhé!!!!

a3+b3=3ab−1

⇔a³+b³−3ab+1=0⇔a3+b3−3ab+1=0

⇔(a+b)³−3ab(a+b)−3ab+1=0

⇔(a+b)³+1−3ab(a+b+1)=0

⇔(a+b+1)[(a+b)²−(a+b)+1]−3ab(a+b+1)=0

⇔(a+b+1)(a²+b²+1−ab−a−b)=0

Vì a,b>0a,b>0 nên a+b+1≠0

Do đó:

a²+b²+1−a−b−ab=0

⇔\(\frac{\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2}{2}\)=0

⇔a=b=1

Do đó: a²⁰¹⁸+b²⁰¹⁹=1+1=2

Ta có đpcm.

đề lm j cho a³+b³=3ab-1 đâu bạn

đặt \(t=a+b\) từ GT => \(3=t^2-ab\ge\frac{3}{4}t^2\)\(\Leftrightarrow\)\(-2\le t\le2\)

\(P=-4t^3-3t^2+18t+9=\hept{\begin{cases}\frac{-1}{4}\left(2t+3\right)^2\left(4t-9\right)-\frac{45}{4}\ge\frac{-45}{4}\left(dungvoit\le2\right)\\-\left(t-1\right)^2\left(4t+11\right)+20\le20\left(dungvoit\ge-2\right)\end{cases}}\)

\(P_{min}=\frac{-45}{4}\) tại 

\(\hept{\begin{cases}a^2+b^2+ab=3\\a+b=\frac{-3}{2}\end{cases}}\Leftrightarrow\left(a;b\right)=\left\{\left(\frac{-3-\sqrt{21}}{4};\frac{-3+\sqrt{21}}{4}\right);\left(\frac{-3+\sqrt{21}}{4};\frac{-3-\sqrt{21}}{4}\right)\right\}\)

\(P_{max}=20\) tại \(\hept{\begin{cases}a^2+b^2+ab=3\\a+b=1\end{cases}}\Leftrightarrow\left(a;b\right)=\left\{\left(2;-1\right);\left(-1;2\right)\right\}\)