\(1.đặt:\sqrt[3]{x-9}=y-3\)

\(\Rightarrow y^3-9y^2+27y-27=x-9\Leftrightarrow y^3-9y^2+27y-x=18\left(1\right)\)

\(x^3-9x^2+27x-21=y-3\Leftrightarrow x^3-9x^2+27x-y=18\left(2\right)\)

\(\left(2\right)-\left(1\right)\Rightarrow\left(x-y\right)\left(x^2+xy-9x+y^2-9y+28\right)=0\Leftrightarrow\left[{}\begin{matrix}x=y\left(3\right)\\x^2+xy-9x+y^2-9y+28=0\left(4\right)\end{matrix}\right.\)

\(\left(3\right)thay\left(1\right)\Rightarrow x=1\)

\(\left(4\right)\Leftrightarrow x^2+x\left(y-9\right)+y^2-9y+28=0\Rightarrow\Delta=\left(y-9\right)^2-4\left(y^2-9y+28\right)=-\left(3y^2-18y+31\right)< 0\Rightarrow x;y=\varnothing\)

\(vậy:x=1\)

\(\dfrac{4}{x}+\dfrac{5}{y}\ge23?\)

\(a=8x+\dfrac{2}{x}+18y+\dfrac{2}{y}+\dfrac{4}{x}+\dfrac{5}{y}\ge2\sqrt{8.2}+2\sqrt{18.2}+23=43\)

\(dấu"="\Leftrightarrow x\)\(=\dfrac{1}{2};y=\dfrac{1}{3}\)

\(1.\Leftrightarrow\sqrt{x+3}=x^2-2\Leftrightarrow\left\{{}\begin{matrix}x^2-2\ge0\\x+3=\left(x^2-2\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\le-\sqrt{2}\\x\ge\sqrt{2}\end{matrix}\right.\\x^4-4x^2-x+1=0\left(1\right)\end{matrix}\right.\)

\(giải\left(1\right)\Rightarrow x\)

\(2.pt\Leftrightarrow\sqrt{x+1}+\sqrt{4-x}+\sqrt{\left(4-x\right)\left(x+1\right)}=5\)

\(đặt:\sqrt{x+1}+\sqrt{4-x}=t\)

\(t\le\sqrt{2\left(x+1+4-x\right)}=\sqrt{2.5}=\sqrt{10}\)

\(t\ge\sqrt{x+1+4-x}=\sqrt{5}\)

\(\Rightarrow\sqrt{5}\le t\le\sqrt{10}\)(1)

\(\Rightarrow pt\Leftrightarrow t+\dfrac{t^2-5}{2}=5\left(2\right)\)

\(giải\left(2\right)\Rightarrow t=...\)\(\)(đối chiếu điều kiện (1)) rồi tính ra x

\(3.bn\) \(xem\) \(lại\) \(đề\)

\(5.pt\Leftrightarrow x^2-2x-11+2x^2-2x-5-2\sqrt{2x^2-2x-5}+1=0\)

\(\Leftrightarrow x^2-2x-11+\left(\sqrt{2x^2-2x-5}-1\right)^2=0\left(1\right)\)

\(đặt:\sqrt{2x^2-2x-5}=t\ge0\Leftrightarrow t^2=2x^2-2x-5\Leftrightarrow x^2-2x=\dfrac{t^2}{2}+\dfrac{5}{2}\)

\(pt\Leftrightarrow t^2+\dfrac{5}{2}-11+\left(t-1\right)^2=0\left(2\right)\)

giải (2) \(\Rightarrow t\Rightarrow x=....\)

\(4..x^2+22x+5=16\sqrt{2x+51}\)

\(đặt:\sqrt{2x+51}=t\ge0\Leftrightarrow2x+51=t^2\Leftrightarrow x=\dfrac{t^2-51}{2}\Rightarrow pt\Leftrightarrow\dfrac{\left(t^2-51\right)^2}{4}+11\left(t^2-51\right)+5-16t=0\Leftrightarrow\dfrac{1}{4}\left(t^2-4t-29\right)\left(t^2+4t-13\right)=0\Leftrightarrow\left[{}\begin{matrix}t=....\\t=....\end{matrix}\right.\) đối chiếu \(t\ge0\Rightarrow t\Rightarrow x\)

\(2x+3y=\dfrac{1}{\sqrt{2}}.2.\sqrt{2}x+\dfrac{1}{\sqrt{3}}.3\sqrt{3}y=7\le\sqrt{[\left(\dfrac{2}{\sqrt{2}}\right)^2+\left(\dfrac{3}{\sqrt{3}}\right)^2]\left(2x^2+3y^2\right)}=\sqrt{5\left(2x^2+3y^2\right)}\Leftrightarrow2x^2+3y^2\ge\dfrac{49}{5}\)

\(dấu"="\Leftrightarrow x=y=\dfrac{7}{5}\)

\(a+b+c=a+\dfrac{1}{\sqrt{2}}.\sqrt{2}.b+c=1\le\sqrt{\left(1+\dfrac{1}{2}+1\right)\left(a^2+2b^2+c^2\right)}=\sqrt{\dfrac{5}{2}\left(a^2+2b^2+c^2\right)}\left(bunhiacopxki\right)\)

\(\Rightarrow\sqrt{\dfrac{5}{2}\left(a^2+2b^2+c^2\right)}\ge1\Leftrightarrow\dfrac{5}{2}\left(a^2+2b^2+c^2\right)\ge1\Leftrightarrow a^2+2b^2+c^2\ge\dfrac{2}{5}\)

\(dâu"="\Leftrightarrow a=c=0,4;b=0,2\)

\(\Sigma\dfrac{a^3}{b^2+c}=\Sigma\dfrac{a^4}{ab^2+ac}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab^2+bc^2+ca^2+ac+ab+bc}\)

\(\ge\dfrac{1}{\sqrt{\left(a^2+b^2+c^2\right)\left(a^2c^2+b^2c^2+a^2b^2\right)}+a^2+b^2+c^2}\ge\dfrac{1}{\sqrt{\dfrac{\left(a^2+b^2+c^2\right)^2}{3}}+1}=\dfrac{1}{\dfrac{1}{\sqrt{3}}+1}=\dfrac{\sqrt{3}}{1+\sqrt{3}}\)

\(dau"="\Leftrightarrow a=b=c=\dfrac{1}{\sqrt{3}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(\sqrt{3}+2\right)^2}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(\sqrt{3}+2\right)}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}=\sqrt{4+\sqrt{25}}=\sqrt{5+4}=3\)

\(b;\sqrt{x+4\sqrt{x-4}}=5\Rightarrow x+4\sqrt{x-4}=25\Rightarrow4\sqrt{x-4}=25-x\Rightarrow16\left(x-4\right)=x^2-50x+625\Leftrightarrow\left[{}\begin{matrix}x=53\\x=13\end{matrix}\right.\)

thử lại vào pt thấy \(x=13\left(thỏa\right);x=53\left(loại\right)\)

\(gọi:AB=x;AC=y\)

\(có:x^2+y^2-2xy.cos135^o=BC^2=25\left(1\right)\)

\(\dfrac{BC}{sin135^o}=5\sqrt{2}=\dfrac{x}{sinC}=\dfrac{y}{sinB}\Leftrightarrow\dfrac{x}{\dfrac{AH}{y}}+\dfrac{y}{\dfrac{AH}{x}}=5\sqrt{2}\Leftrightarrow2xy=5\sqrt{2}\left(2\right)\)

\(\)\(\left(1\right)\left(2\right)\Rightarrow\left\{{}\begin{matrix}x^2+y^2-2xycos135^o=25\Leftrightarrow\left(x+y\right)^2-2xy-2xycos135^2=25\\2xy=5\sqrt{2}\end{matrix}\right.\)

giải bằng cách đặt \(\left\{{}\begin{matrix}S=x+y\\P=xy\end{matrix}\right.\)\(\left(S^2\ge4P\right)\)

\(\left(đk:x\ge0\right)\sqrt{x}+9\le31\Leftrightarrow\sqrt{x}\le22\Leftrightarrow x\le22\Rightarrow0\le x\le22\)

\(b;\left(đk:x\ge\dfrac{1}{2}\right)\sqrt{2x-1}>6\Leftrightarrow2x-1>36\Leftrightarrow x>\dfrac{37}{2}\)

\(c;đk:x\ge-3\Rightarrow\sqrt{x+3}\ge5\Leftrightarrow x+3\ge25\Leftrightarrow x\ge22\)

\(d;đk:x\ge\dfrac{1}{2};\sqrt{2x-1}< -3\left(vôli\right)\)