\(cminh:\)\(\dfrac{a}{a^3+a+1}\le\dfrac{4-a}{9}\left(1\right)\)

\(\left(1\right)\Leftrightarrow\)\(9a-\left(4-a\right)\left(a^3+a+1\right)\le0\Leftrightarrow\left(a-1\right)^2\left(a^2-2a-4\right)\le0\left(đúng\right)\Rightarrow\left(1\right)đúng\)

\(tương\) \(tự\Rightarrow\Sigma\dfrac{a}{a^3+a+1}\le\dfrac{4-a+4-b+4-c}{9}=\dfrac{-\left(a+b+c\right)+12}{9}\le\dfrac{-3\sqrt[3]{abc}+12}{9}=\dfrac{-3+12}{9}=1\left(đpcm\right)\)

\(\left\{{}\begin{matrix}x^2+2y=xy+2x\left(1\right)\\\sqrt{x+y}=xy-2\left(2\right)\end{matrix}\right.\)\(\left(đk:x\ge-y\right)\)

\(\left(1\right)\Leftrightarrow\left(x-2\right)\left(x-y\right)=0\Leftrightarrow\left[{}\begin{matrix}x=y\Rightarrow\sqrt{2y}=y^2-2\left(3\right)\\x=2\Rightarrow\left(2\right)\Leftrightarrow\sqrt{2+y}=2y-2\left(4\right)\end{matrix}\right.\)

\(giải\left(3\right)và\left(4\right)\Rightarrow x;y\)

\(f\left(x\right)=\dfrac{\sqrt{3x-3}}{\sqrt{3}}+\sqrt{6-3x}\le\sqrt{\left(\dfrac{1}{3}+1\right)\left(3x-3+6-3x\right)}=2\Rightarrow max=2\Leftrightarrow x=\dfrac{5}{4}\)

\(f\left(x\right)=\sqrt{x-1}-1+\sqrt{6-3x}+1=\dfrac{x-2}{\sqrt{x-1}+1}+\sqrt{3\left(2-x\right)}+1=\left(2-x\right)\left[\sqrt{3}-\dfrac{1}{\sqrt{x-1}+1}\right]+1\ge1\Rightarrow min=1\Leftrightarrow x=2\)

\(2,,\sqrt{2x^2+12x+5}+\sqrt{2x^2-3x+5}=8\sqrt{x}\)\(\left(1\right)\)

\(đặt:\sqrt{2x^2+12x+5}=a\ge0;\sqrt{2x^2-3x+5}=b\ge0\Rightarrow a^2-b^2=15x\Rightarrow\sqrt{x}=\sqrt{\dfrac{a^2-b^2}{15}}\)

\(\Rightarrow pt\left(1\right)\Leftrightarrow a+b=8\sqrt{\dfrac{a^2-b^2}{15}}\Leftrightarrow a+b-\dfrac{8}{\sqrt{15}}\sqrt{\left(a+b\right)\left(a-b\right)}=0\Leftrightarrow\sqrt{a+b}\left[\sqrt{a+b}-\dfrac{8}{\sqrt{15}}\sqrt{a-b}\right]=0\Leftrightarrow\left[{}\begin{matrix}a=-b\left(loại\right)do:\sqrt{2x^2+12x+5}=-\sqrt{2x^2-3x+5}< 0\\\sqrt{a+b}-8\sqrt{\dfrac{a-b}{15}}=0\left(2\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow\sqrt{a+b}=8\sqrt{\dfrac{a-b}{15}}\Leftrightarrow a+b=\dfrac{64}{15}\left(a-b\right)\Rightarrow a=\dfrac{79}{49}b\Rightarrow\sqrt{2x^2+12x+5}=\dfrac{79}{49}\sqrt{2x^2-3x+5}\Rightarrow x=.....\)

\(1.\) \(xem\) \(lại\) \(do\) \(pt\Leftrightarrow8x^2=-\left(\sqrt{x+1}+\sqrt{x^2+x+1}\right)^2< 0\)

\(4\sqrt{x^2-x+10-4\sqrt{2-2x}}=x-7+8\sqrt{3-x}=x-7+4\sqrt{\left(3-x\right).4}\le x-7+\dfrac{4\left(3-x+4\right)}{2}=x-7+2\left(3-x+4\right)=7-x\)

\(\Rightarrow4\sqrt{x^2-x+10-4\sqrt{2-2x}}\le7-x\)\(\left(đkxđ:x\le1\right)\)

\(\Leftrightarrow16\left(x^2-x+10\right)-64\sqrt{2-2x}\le\left(7-x\right)^2\)

\(\Leftrightarrow15x^2-2x+111\le64\sqrt{2-2x}\Leftrightarrow\left(x+1\right)^2\left(225x^2-510x+4129\right)\le0\left(1\right)\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\left(tm\right)\)  

\(\Sigma\dfrac{x^3}{y+z}=\Sigma\dfrac{x^4}{xy+xz}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(xy+yz+xz\right)}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2.\sqrt{\left(x^2+y^2+z^2\right)\left(x^2+y^2+z^2\right)}}=\dfrac{x^2+y^2+z^2}{2}\ge\dfrac{\left(x+y+z\right)^2}{2+2+2}\ge\dfrac{3^2}{6}=\dfrac{3}{2}\Rightarrow min=\dfrac{3}{2}\Leftrightarrow x=y=z=1\)

\(a;b;c>0\) \((a;b;c\ge0\) \(là\) \(sai)\)

\(\Sigma\dfrac{4}{\left(a+b\right)^3}\ge\Sigma\dfrac{c}{a+b}\Leftrightarrow\Sigma\dfrac{4}{\left(3-c\right)^3}\ge\Sigma\dfrac{c}{3-c}\left(1\right)\)

\(\dfrac{4}{\left(3-c\right)^3}\ge\dfrac{c}{3-c}\Leftrightarrow4\left(3-c\right)\ge c\left(3-c\right)^3\Leftrightarrow4\left(c-3\right)-c\left(3-c\right)^3\ge0\Leftrightarrow-\left(3-c\right)\left(c-4\right)\left(c-1\right)^2\ge0\left(2\right)\)

\(do:a,b,c>0;a+b+c=3\Rightarrow0< a,b,c< 3\Rightarrow\left(2\right)\) \(đúng\)

\(tương\) \(tự\Rightarrow\dfrac{4}{\left(3-a\right)^3}\ge\dfrac{a}{3-a};\dfrac{4}{\left(3-b\right)^3}\ge\dfrac{b}{3-b}\)

\(\Rightarrow\)\(\left(1\right)đúng\Rightarrowđpcm\)

\(\sqrt{2x-3}+\sqrt{5-2x}=3x^2-12x+14\)\(\left(đk:\dfrac{3}{2}\le x\le\dfrac{5}{2}\right)\)

\(VT\le\sqrt{2\left(2x-3+5-2x\right)}=\sqrt{2.2}=2\)\(\left(bunhiacopxki\right)\)

\(VP=3\left(x-2\right)^2+2\ge2\)

\(\Rightarrow dấu"="\Leftrightarrow\left\{{}\begin{matrix}x=2\\2x-3=5-2x\end{matrix}\right.\)\(\Leftrightarrow x=2\left(thỏa\right)\)

\(\left(đk:-\dfrac{1}{3}\le x\le6\right)\sqrt{6-x}-\sqrt{3x+1}=3x^2-14x-8\)

2.xem lại đề

\(3;đặt:\sqrt{2x+1}=y\Rightarrow y^2=2x+1\left(1\right)\)

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

\(lấy\left(2\right)-\left(1\right)\Rightarrow x^2-y^2=2y+1-2x-1\Leftrightarrow\left(x-y\right)\left(x+y+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=y\left(3\right)\\x=-\left(y+2\right)\left(4\right)\end{matrix}\right.\)

thay(3) và (4) lần lượt vào(1) ta giải ra được x

ý 1 nháp: \(đặt:\sqrt[3]{x-9}=ay+b\Rightarrow a^3y^3+3a^2y^2b+3ayb^2-x=-b^3-9\left(1\right)\)

\(bài:\Rightarrow x^3-9x^2+27x-21=ay+b\Leftrightarrow x^3-9x^2+27x-ay=b+21\left(2\right)\)

\(lấy\left(2\right)-\left(1\right)\Rightarrow x^3-9x^2+27x-ay-a^3y^3-3a^2y^2b-3ayb^2+x=b+21-\left(-b^3-9\right)=b^3+b+30\Leftrightarrow x^3-a^3y^3-\left(9x^2+3a^2y^2b\right)+28x-\left(a+3ab^2\right)y=b^3+b+30\)

\(chọn:\left\{{}\begin{matrix}a^3=1\\3a^2b=-9\\a+3ab^2=28\\b^3+b+30=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=-3\end{matrix}\right.\) \(\Rightarrowđặt:\sqrt[3]{x-9}=y-3\)(như cách làm dưới)

ý 2: bấm máy tính nhận thấy \(x=3\) \(\Rightarrow4x+4>0\Rightarrow cosi\) để đánh giá 

\(2.\left(đk:x\ge-1\right)\) \(x^3-3x^2-8x+40=8\sqrt[4]{4x+4}=\sqrt{16.16.16.\left(4x+4\right)}\le\dfrac{16+16+16+4x+4}{4}=\dfrac{4x+52}{4}\Rightarrow4\left(x^3-3x^2-8x+40\right)-\left(4x+52\right)\le0\Leftrightarrow4\left(x-3\right)^2\left(x+3\right)\le0\Leftrightarrow\left(x-3\right)^2\left(x+3\right)\le0\left(1\right)\)

\(xét\) \(đk:x\ge-1\Rightarrow x+3\ge2>0\Rightarrow\left(x-3\right)^2\left(x+3\right)\le0\Leftrightarrow\left(x-3\right)^2\le0\Leftrightarrow x=3\)

\(thử:x=3\) \(vào\) \(pt\) \(thấy\) \(thỏa\)