#)Giải :

Áp dụng BĐT Cauchy :

\(\left(ab+c\right)\left(bc+a\right)\le\left(\frac{ab+c+bc+a}{2}\right)^2=\frac{\left(b+1\right)^2\left(c+a\right)^2}{4}\)

Tương tự với các cặp còn lại, ta được :

\(\left(bc+a\right)\left(ca+b\right)\le\frac{\left(c+1\right)^2\left(a+b\right)^2}{4}\)

\(\left(ab+c\right)\left(ca+b\right)\le\frac{\left(a+1\right)^2\left(b+c\right)^2}{4}\)

Nhân theo vế :

\(\left[\left(ab+c\right)\left(ca+b\right)\left(bc+a\right)\right]^2\le\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\frac{\left[\left(a+1\right)\left(b+1\right)\left(c+1\right)\right]^2}{64}\)

Mà : \(\left(a+1\right)\left(b+1\right)\left(c+1\right)\le\left(\frac{a+1+b+1+c+1}{3}\right)^3=8\)

Do đó \(\left[\left(ab+c\right)\left(ac+b\right)\left(bc+a\right)\right]^2\le\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2.\frac{8^2}{64}\)

Từ đó suy ra \(\left(ab+c\right)\left(ca+b\right)\left(bc+a\right)\le\left(a+b\right)\left(b+c\right)\left(c+a\right)\Rightarrowđpcm\)

ai giúp mk với

\(pt\Leftrightarrow\sqrt{2x^2+8x+6}-4+\sqrt{x^2-1}-2x+2=0\)

\(\Leftrightarrow\frac{2\left(x-1\right)\left(x+5\right)}{\sqrt{2x^2+8x+6}+4}+\sqrt{x^2-1}-2\left(x-1\right)=0\)

Giải nốt nhá

\(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)

\(\Leftrightarrow\sqrt{2\left(x^2+4x+3\right)}+\sqrt{x^2-1}=2x+2\)

\(\Leftrightarrow\sqrt{2\left(x+1\right)\left(x-3\right)}+\sqrt{x^2-1}=2x+2\)

\(\Leftrightarrow\sqrt{2\left(x+1\right)\left(x+3\right)}+\sqrt{x^2-1^2}=2x+2\)

\(\Leftrightarrow\sqrt{2\left(x+1\right)\left(x+3\right)}+\sqrt{\left(x+1\right)\left(x-1\right)}=2x+2\)

\(\Leftrightarrow2x^2+8x+6+\left(2x+2\right)\sqrt{2\left(x+3\right)\left(x-1\right)}+\left(x+1\right)\left(x-1\right)=4\left(x+1\right)^2\)

\(\Leftrightarrow\left(2x+2\right)\sqrt{2\left(x+3\right)\left(x-1\right)}=4\left(x+1\right)^2-2x^2-8x-6-\left(x+1\right)\left(x-1\right)\)

\(\Leftrightarrow8\left(x+1\right)^3.\left(x+3\right)\left(x-1\right)=\left(x+1\right)^2.\left(x-1\right)^2\)

\(\Leftrightarrow8x^4-8x^3+24x^3-24x^2+16x^3-16x^2+48x^2-48x+8x^2-8x+24x-24\)\(=x^4-2x^3+x^2+2x^3-4x^2+2x+x-2x+1\)

\(\Leftrightarrow8x^4+32x^3+16x^3-32x=x^4-2x^3+x^2+2x^3-4x^2+2x+x^2-2x+1\)

\(\Leftrightarrow8x^4+32x^3+16x^2-32x-24=x^4-2x^2+1\)

\(\Leftrightarrow8x^4+32x^2+16x^2-32x-24-x^4+2x^2-1=0\)

\(\Leftrightarrow7x^4+32x^3+18x^2-32x-25=0\)

\(\Leftrightarrow\left(7x^3+39x^2+57x+25\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(7x^2+25x+7x+25\right)\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[x\left(7x+25\right)+\left(7x+25\right)\right]\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(7x+25\right)\left(x+1\right)\left(x-1\right)=0\)

Nhưng \(7x+25\ne0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=1\end{cases}}\)

Vậy: nghiệm phương trình là x = 1; x = -1

\(PT\Leftrightarrow\sqrt{8x+1}-3+\sqrt{46x-10}-6=-x^3+5x^2+4x+1-3-6\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{8}{\sqrt{8x+1}+3}-5+x^2-4x-3-\frac{10}{\sqrt{46-10x}+6}\right)=0\)

Xét \(\left(\frac{8}{\sqrt{8x+1}+3}-5+x^2-4x-3-\frac{10}{\sqrt{46-10x}+6}\right)\)(*) (đk\(\frac{23}{5}\ge x\ge-\frac{1}{8}\))

(*)\(=\frac{8-5\left(\sqrt{8x+1}+3\right)}{\sqrt{8x+1}+3}+\left(x^2-4x-3\right)-\frac{10}{\sqrt{46-10x}+6}\)

\(=\frac{-7-5\left(\sqrt{8x+1}\right)}{\sqrt{8x+1}+3}+\left(x^2-4x-3\right)-\frac{10}{\sqrt{46-10x}+6}< 0\)

\(\Rightarrow x-1=0\Leftrightarrow x=1\)

Vậy..................

Đề thi thuyển sinh lớp 10 môn Toán Chuyên, TP HCM năm 2012-2013

ĐK \(\frac{-1}{8}\le x\le\frac{23}{5}\)(*) Ta có:

\(\sqrt{8x+1}+\sqrt{46-10x}=-x^3+5x^2+4x+1\)

\(\Leftrightarrow\sqrt{8x+1}-3+\sqrt{46-10x}-6+x^3-x^2-4x^2+4x-8x+8=0\)

\(\Leftrightarrow\frac{8x-1}{\sqrt{8x+1}+3}+\frac{10-10x}{\sqrt{46-10x}+6}+x^2\left(x-1\right)-4x\left(x-1\right)-8\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{8}{\sqrt{8x+1}+3}+\frac{10}{\sqrt{46-10x}+6}+x^2-4x-8\right)=0\)(**)

(*) \(\Rightarrow-1< x< 5\Rightarrow\left(x+1\right)\left(x+5\right)< 0\Rightarrow x^2-4x-5< 0\)

Và \(\frac{8}{\sqrt{8x+1}+3}< \frac{9}{3}=3\Rightarrow\frac{8}{\sqrt{8x+1}+3}-3< 0\) Do vậy:

\(\frac{8}{\sqrt{8x+1}+3}-\frac{10}{\sqrt{46-10x}+6}+x^2-4x-8< 0\)Do đó:

(**)\(\Leftrightarrow x=1\)

Vậy S={1}