a: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}+4}{x\sqrt{x}-3x+2\sqrt{x}}-\dfrac{3\sqrt{x}+3}{-x+\sqrt{x}+2}\right):\left(\dfrac{x-\sqrt{x}-6}{x-3\sqrt{x}}-\dfrac{x-2\sqrt{x}}{x-4\sqrt{x}+4}\right)+\sqrt{x}\)

\(=\left(\dfrac{\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{3}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)+\sqrt{x}\)

\(=\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{x-4-x}{\sqrt{x}\left(\sqrt{x}-2\right)}+\sqrt{x}\)

\(=\dfrac{4\left(\sqrt{x}-1\right)}{-4}+\sqrt{x}\)

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

=-1

\(\frac{5}{a+b\sqrt{2}}-\frac{4}{a-b\sqrt{2}}+18\sqrt{2}=3\)

<=> \(\frac{5\left(a-b\sqrt{2}\right)}{a^2-2b^2}-\frac{4\left(a+b\sqrt{2}\right)}{a^2-2b^2}+18\sqrt{2}=3\) trục căn thức

<=> \(\frac{5a}{a^2-2b^2}-\frac{5b\sqrt{2}}{a^2-2b^2}-\frac{4a}{a^2-2b^2}-\frac{4b\sqrt{2}}{a^2-2b^2}+18\sqrt{2}=3\)

Vì a; b nguyên => \(\hept{\begin{cases}\frac{5a}{a^2-2b^2}-\frac{4a}{a^2-2b^2}=3\\-\frac{5b\sqrt{2}}{a^2-2b^2}-\frac{4b\sqrt{2}}{a^2-2b^2}+18\sqrt{2}=0\end{cases}}\)

<=> \(\hept{\begin{cases}\frac{a}{a^2-2b^2}=3\\\frac{9b}{a^2-2b^2}=18\end{cases}}\)<=> \(\hept{\begin{cases}\frac{a}{a^2-2b^2}=3\\\frac{b}{a^2-2b^2}=2\end{cases}}\)

Với b = 0 => loại 

Với b khác 0: 

=> \(\frac{a}{b}=\frac{3}{2}\Leftrightarrow a=\frac{3}{2}b\)

=> \(\frac{b}{\frac{9}{4}b^2-2b^2}=2\)=> b = 2 => a = 3  thử lại  thỏa mãn 

Vậy a = 3 và b = 2.

\(\frac{5}{a+b\sqrt{2}}-\frac{4}{a-b\sqrt{2}}+18\sqrt{2}=3\)

\(\Leftrightarrow5a-5b\sqrt{2}-4a-4b\sqrt{2}+18\sqrt{2}\left(a^2-2b^2\right)=3\left(a^2-2b^2\right)\)

\(\Leftrightarrow5a-5b\sqrt{2}-4a-4b\sqrt{2}+18a^2\sqrt{2}-36b^2\sqrt{2}=3a^2-6b^2\)

\(\Leftrightarrow\left(18a^2-36b^2-9b\right)\sqrt{2}=3a^2-6b^2-a\)

-Nếu \(18a^2-36b^2-9b\ne0\Rightarrow\sqrt{2}=\frac{3a^2-6b^2-a}{18a^2-36b^2-9b}\)

Vì a,b nguyên nên \(\frac{3a^2-6b^2-a}{18a^2-36b^2-9b}\inℚ\Rightarrow\sqrt{2}\inℚ\)=> Vô lý vì \(\sqrt{2}\)là số vô tỷ

-Vậy ta có: \(18a^2-36b^2-9b=0\Rightarrow\hept{\begin{cases}18a^2-36b^2-9b=0\\3a^2-6b^2-a=0\end{cases}\Rightarrow\hept{\begin{cases}3a^2-6b^2=\frac{3}{2}b\\3a^2-6b^2=2\end{cases}}\Leftrightarrow a=\frac{3}{2}b}\)

Thay a=\(\frac{3}{2}b\)vào \(3a^2-6b^2-a=0\)

ta có \(3\cdot\frac{9}{4}b^2-6b^2-\frac{3}{2}b=0\Leftrightarrow27b^2-6b=0\Leftrightarrow3b\left(b-2\right)=0\)

Ta có b=0 (loại), b=2 (tm) => a=3

Vậy b=2; a=3

Áp dụng BĐT Minicopski, ta có:

\(P=\sqrt{a^2+\dfrac{1}{a^2}}+\sqrt{b^2+\dfrac{1}{b^2}}\ge\sqrt{\left(a+b\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2}\\ \Rightarrow P\ge\sqrt{4^2+\left(\dfrac{4}{a+b}\right)^2}=\sqrt{16+\left(\dfrac{4}{4}\right)^2}=\sqrt{17}\)

Đẳng thức xảy ra \(\Leftrightarrow a=b=2\)

Áp dụng BĐT Cô si

⇒ P≥ \(\sqrt{2\sqrt{a^2.\dfrac{1}{a^2}}}+\sqrt{2\sqrt{b^2.\dfrac{1}{b^2}}}\)

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

\(=2\sqrt{2}\)

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\((a^2+\frac{1}{b^2})(1+4^2)\geq (a+\frac{4}{b})^2\Rightarrow \sqrt{a^2+\frac{1}{b^2}}\geq \frac{1}{\sqrt{17}}(a+\frac{4}{b})\)

Hoàn toàn tương tự với những cái còn lại và cộng theo vế suy ra:

$S\geq \frac{1}{\sqrt{17}}(a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c})$

$\geq \frac{1}{\sqrt{17}}(a+b+c+\frac{36}{a+b+c})$ theo BĐT Cauchy-Schwarz.

Áp dụng BĐT AM-GM:

\(a+b+c+\frac{9}{4(a+b+c)}\geq 3\)

\(\frac{135}{4(a+b+c)}\geq \frac{135}{4.\frac{3}{2}}=\frac{45}{2}\)

\(\Rightarrow a+b+c+\frac{36}{a+b+c}\geq \frac{51}{2}\)

\(\Rightarrow S\geq \frac{3\sqrt{17}}{2}\)

Vậy $S_{\min}=\frac{3\sqrt{17}}{2}$

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\((a^2+\frac{1}{b^2})(1+4^2)\geq (a+\frac{4}{b})^2\Rightarrow \sqrt{a^2+\frac{1}{b^2}}\geq \frac{1}{\sqrt{17}}(a+\frac{4}{b})\)

Hoàn toàn tương tự với những cái còn lại và cộng theo vế suy ra:

$S\geq \frac{1}{\sqrt{17}}(a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c})$

$\geq \frac{1}{\sqrt{17}}(a+b+c+\frac{36}{a+b+c})$ theo BĐT Cauchy-Schwarz.

Áp dụng BĐT AM-GM:

\(a+b+c+\frac{9}{4(a+b+c)}\geq 3\)

\(\frac{135}{4(a+b+c)}\geq \frac{135}{4.\frac{3}{2}}=\frac{45}{2}\)

\(\Rightarrow a+b+c+\frac{36}{a+b+c}\geq \frac{51}{2}\)

\(\Rightarrow S\geq \frac{3\sqrt{17}}{2}\)

Vậy $S_{\min}=\frac{3\sqrt{17}}{2}$

\(6a+3b+2c=abc\Leftrightarrow\dfrac{2}{ab}+\dfrac{3}{ac}+\dfrac{6}{bc}=1\)

Đặt \(\left(\dfrac{1}{a};\dfrac{2}{b};\dfrac{3}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\)

\(Q=\dfrac{1}{\sqrt{\dfrac{1}{x^2}+1}}+\dfrac{2}{\sqrt{\dfrac{4}{y^2}+4}}+\dfrac{3}{\sqrt{\dfrac{9}{z^2}+9}}=\dfrac{x}{\sqrt{x^2+1}}+\dfrac{y}{\sqrt{y^2+1}}+\dfrac{z}{\sqrt{z^2+1}}\)

\(Q=\dfrac{x}{\sqrt{x^2+xy+yz+zx}}+\dfrac{y}{\sqrt{y^2+xy+yz+zx}}+\dfrac{z}{\sqrt{z^2+xy+yz+zx}}\)

\(Q=\dfrac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}+\dfrac{y}{\sqrt{\left(x+y\right)\left(y+z\right)}}+\dfrac{z}{\sqrt{\left(x+z\right)\left(y+z\right)}}\)

\(Q\le\dfrac{1}{2}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}+\dfrac{y}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{x+z}+\dfrac{z}{y+z}\right)=\dfrac{3}{2}\)

\(Q_{max}=\dfrac{3}{2}\) khi \(x=y=z=\dfrac{1}{\sqrt{3}}\) hay \(\left(a;b;c\right)=\left(\sqrt{3};2\sqrt{3};3\sqrt{3}\right)\)

⚽⚽

Đề bài sai

Đề đúng: \(\dfrac{1}{\sqrt{a}+2\sqrt{b}+3}+\dfrac{1}{\sqrt{b}+2\sqrt{c}+3}+\dfrac{1}{\sqrt{c}+2\sqrt{a}+3}\le\dfrac{1}{2}\)

Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x^2;y^2;z^2\right)\Rightarrow xyz=1\)

Đặt vế trái BĐT cần chứng minh là P, ta có:

\(P=\dfrac{1}{x^2+2y^2+3}+\dfrac{1}{y^2+2z^2+3}+\dfrac{1}{z^2+2x^2+3}\)

\(P=\dfrac{1}{\left(x^2+y^2\right)+\left(y^2+1\right)+2}+\dfrac{1}{\left(y^2+z^2\right)+\left(z^2+1\right)+2}+\dfrac{1}{\left(z^2+x^2\right)+\left(x^2+1\right)+2}\)

\(P\le\dfrac{1}{2xy+2y+2}+\dfrac{1}{2yz+2z+2}+\dfrac{1}{2zx+2x+2}\)

\(P\le\dfrac{1}{2}\left(\dfrac{xz}{xz\left(xy+y+1\right)}+\dfrac{x}{x\left(yz+z+1\right)}+\dfrac{1}{zx+x+1}\right)\)

\(P\le\dfrac{1}{2}\left(\dfrac{xz}{x.xyz+xyz+xz}+\dfrac{x}{xyz+xz+1}+\dfrac{1}{xz+x+1}\right)\)

\(P\le\dfrac{1}{2}\left(\dfrac{xz}{x+1+xz}+\dfrac{x}{1+xz+1}+\dfrac{1}{xz+x+1}\right)=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)