\(PT\Leftrightarrow\dfrac{5\left(a-b\sqrt{2}\right)}{a^2-2b^2}-\dfrac{4\left(a+b\sqrt{2}\right)}{a^2-2b^2}+18\sqrt{2}-3=0\\ \Leftrightarrow\left(\dfrac{5a}{a^2-2b^2}-\dfrac{4a}{a^2-2b^2}-3\right)+\left(18\sqrt{2}-\dfrac{5b\sqrt{2}}{a^2-2b^2}-\dfrac{4b\sqrt{2}}{a^2-2b^2}\right)=0\\ \Leftrightarrow\left(\dfrac{5a}{a^2-2b^2}-\dfrac{4a}{a^2-2b^2}-3\right)+\sqrt{2}\left(18-\dfrac{5b}{a^2-2b^2}-\dfrac{4b}{a^2-2b^2}\right)=0\)

Vì a,b nguyên mà vế trái có \(\sqrt{2}\) vô tỉ nên 2 biểu thức còn lại phải bằng 0

 \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5a}{a^2-2b^2}-\dfrac{4a}{a^2-2b^2}=3\\\dfrac{5b}{a^2-2b^2}+\dfrac{4b}{a^2-2b^2}=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{a^2-2b^2}=3\\\dfrac{b}{a^2-2b^2}=2\end{matrix}\right.\left(a,b\ne0\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^2-2b^2=\dfrac{a}{3}\\b=2\left(a^2-2b^2\right)=2\cdot\dfrac{a}{3}=\dfrac{2}{3}a\end{matrix}\right.\)

\(\Leftrightarrow a^2-\dfrac{8}{9}a^2=\dfrac{a}{3}\Leftrightarrow\dfrac{1}{9}a^2-\dfrac{1}{3}a=0\Leftrightarrow\dfrac{1}{3}a\left(\dfrac{1}{3}a-1\right)=0\\ \Leftrightarrow a=3\left(a\ne0\right)\)

\(\Leftrightarrow b=\dfrac{2}{3}\cdot3=2\left(tm\right)\)

Vậy \(\left(a;b\right)=\left(3;2\right)\)

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:

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

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

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

1) Áp dụng bất đẳng thức AM - GM và bất đẳng thức Schwarz:

\(P=\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\ge\dfrac{1}{a}+\dfrac{1}{\dfrac{a+b}{2}}\ge\dfrac{4}{a+\dfrac{a+b}{2}}=\dfrac{8}{3a+b}\ge8\).

Đẳng thức xảy ra khi a = b = \(\dfrac{1}{4}\).

2.

\(4=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\sqrt{2}\)

Đồng thời \(\left(a+b\right)^2\ge a^2+b^2\Rightarrow a+b\ge2\)

\(M\le\dfrac{\left(a+b\right)^2}{4\left(a+b+2\right)}=\dfrac{x^2}{4\left(x+2\right)}\) (với \(x=a+b\Rightarrow2\le x\le2\sqrt{2}\) )

\(M\le\dfrac{x^2}{4\left(x+2\right)}-\sqrt{2}+1+\sqrt{2}-1\)

\(M\le\dfrac{\left(2\sqrt{2}-x\right)\left(x+4-2\sqrt{2}\right)}{4\left(x+2\right)}+\sqrt{2}-1\le\sqrt{2}-1\)

Dấu "=" xảy ra khi \(x=2\sqrt{2}\) hay \(a=b=\sqrt{2}\)

3. Chia 2 vế giả thiết cho \(x^2y^2\)

\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\ge\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)

\(\Rightarrow0\le\dfrac{1}{x}+\dfrac{1}{y}\le4\)

\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\right)=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le16\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)

Khúc đầu là: \(\dfrac{1}{a^4+b^2+2b^2}\) hay \(\dfrac{1}{a^4+b^2+2ab^2}\) ??

\(2a^2b\) không phải \(2ab^2\)

Bài 1: 

b) ĐKXĐ: \(x\ne3\)

Ta có: \(\dfrac{3-x}{20}=\dfrac{-5}{x-3}\)

\(\Leftrightarrow\dfrac{x-3}{-20}=\dfrac{-5}{x-3}\)

\(\Leftrightarrow\left(x-3\right)^2=100\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\left(nhận\right)\\x=-7\left(nhận\right)\end{matrix}\right.\)

Vậy: \(x\in\left\{13;-7\right\}\)