Đặt \(\left(\frac{1}{x};\frac{1}{y}\right)=\left(a;b\right)\Rightarrow ab+a+b=3\)

\(\Rightarrow ab+2\sqrt{ab}\le3\Rightarrow\left(\sqrt{ab}+3\right)\left(\sqrt{ab}-1\right)\le0\)

\(\Rightarrow\sqrt{ab}\le1\Rightarrow ab\le1\)

\(P=\frac{a}{\sqrt{3+a^2}}+\frac{b}{\sqrt{3+b^2}}=\frac{a}{\sqrt{ab+a+b+a^2}}+\frac{b}{\sqrt{ab+a+b+b^2}}\)

\(=\frac{a}{\sqrt{\left(a+b\right)\left(a+1\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+1\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+1}+\frac{b}{a+b}+\frac{b}{b+1}\right)\)

\(P\le\frac{1}{2}\left(1+\frac{a}{a+1}+\frac{b}{b+1}\right)=\frac{1}{2}\left(1+\frac{ab+a+ab+b}{ab+a+b+1}\right)=\frac{1}{2}\left(1+\frac{ab+3}{4}\right)\)

\(P\le\frac{1}{2}\left(1+\frac{1+3}{4}\right)=1\)

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

Chúc bạn học tốt !!!

\(\left(x^2-3x+9\right)\left(x^2+5x+9\right)=9x^2\)

\(\Leftrightarrow x^4+5x^3+9x^2-3x^3-15x^2-27x+9x^2+45x+81=9x^2\)

\(\Leftrightarrow x^4+2x^3+3x^2+18x+81=9x^2\)

\(\Leftrightarrow x^4+2x^3+3x^2+18x+81-9x^2=0\)

\(\Leftrightarrow x^4+2x^2-6x^2+18x+81=0\)

\(\Leftrightarrow\left(x^3-x^2-3x+27\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x^2-4x+9\right)\left(x+3\right)\left(x+3\right)=0\)

Vì \(x^2-4x+9\ne0\) nên: 

\(\Rightarrow x+3=0\)

     \(x=-3\)

Vậy: nghiệm phương trình là: {-3}

Đặt \(K=a\sqrt{b^3+1}+b\sqrt{c^3+1}+c\sqrt{a^3+1}\)

\(\Rightarrow2K=2a\sqrt{b^3+1}+2b\sqrt{c^3+1}+2c\sqrt{a^3+1}=\)\(2a\sqrt{\left(b+1\right)\left(b^2-b+1\right)}+2b\sqrt{\left(c+1\right)\left(c^2-c+1\right)}\)\(+2c\sqrt{\left(a+1\right)\left(a^2-a+1\right)}\)\(\le a\left[\left(b+1\right)+\left(b^2-b+1\right)\right]+b\left[\left(c+1\right)+\left(c^2-c+1\right)\right]\)\(+c\left[\left(a+1\right)+\left(a^2-a+1\right)\right]\)(Theo BĐT AM - GM)

\(=a\left(b^2+2\right)+b\left(c^2+2\right)+c\left(a^2+2\right)\)\(=ab^2+bc^2+ca^2+6\)

Đặt \(M=ab^2+bc^2+ca^2\)

Không mất tính tổng quát, giả sử \(a\ge c\ge b\)thì ta có \(b\left(a-c\right)\left(c-b\right)\ge0\Leftrightarrow abc+b^2c\ge ab^2+bc^2\)

\(\Leftrightarrow ab^2+bc^2+ca^2\le abc+b^2c+ca^2\)

hay \(M\le abc+b^2c+ca^2\le2abc+b^2c+ca^2=c\left(a+b\right)^2\)\(=4c.\frac{a+b}{2}.\frac{a+b}{2}\le\frac{4}{27}\left(c+\frac{a+b}{2}+\frac{a+b}{2}\right)^3\)\(=\frac{4\left(a+b+c\right)^3}{27}=4\)

\(\Rightarrow2K\le10\Rightarrow K\le10\)

Vậy \(a\sqrt{b^3+1}+b\sqrt{c^3+1}+c\sqrt{a^3+1}\le5\)

Đẳng thức xảy ra khi \(\left(a,b,c\right)=\left(2,0,1\right)\)

Kiệt cop sai đáp án rồi kìa :))
Đoạn cuối không giả sử \(a\ge c\ge b\) được đâu nhá

Mà phải giả sử b là số nằm giữa a và c

Khi đó:

\(\left(b-a\right)\left(b-c\right)\le0\Leftrightarrow b^2+ac\le ab+bc\)

\(\Leftrightarrow ab^2+a^2c\le a^2b+abc\Leftrightarrow ab^2+bc^2+ca^2\le a^2b+abc+bc^2=b\left(a^2+ac+c^2\right)\)

\(\le b\left(a^2+2ac+c^2\right)=b\left(a+c\right)^2=b\left(3-b\right)^2\)

Ta chứng minh \(b\left(3-b\right)^2\le4\Leftrightarrow\left(b-1\right)^2\left(b-4\right)\le0\) *đúng *

Vậy ............................

\(\hept{\begin{cases}-0,5x+1,2y=2,7\\x-4,5y=-7,5\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-x+2,4y=5,4\\x-4,5y=-7,5\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-4,5y=-7,5\\-21y=-2,1\\x=4,5y-7,5\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=1\\x=-3\end{cases}}\)

\(\hept{\begin{cases}-0,5x+1,2y=2,7\\x-4,5y=7,5\end{cases}}\) <=> \(\hept{\begin{cases}-0,5x+1,2y=2,7\\0,5x-2,25y=3,75\end{cases}}\) <=> \(\hept{\begin{cases}-0,5x+1,2y=2,7\\-1,05y=6,45\end{cases}}\) <=> \(\hept{\begin{cases}x=\frac{1,2y-2,7}{0,5}=\frac{1,2.-\frac{43}{7}-2,7}{0,5}=-\frac{141}{7}\\y=-\frac{43}{7}\end{cases}}\)

Vậy nghiệm của hpt là: \(\left(-\frac{141}{7};-\frac{43}{7}\right)\)

Chú ý ghi lời giải, không khi đáp số !