\(\dfrac{9}{x+y+z}\le\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\le1\Rightarrow x+y+z\ge9\)

\(3\sqrt[3]{\dfrac{1}{xyz}}\le1\Rightarrow xyz\ge27\)

\(Q=\Sigma\dfrac{1}{x\left(\sqrt{2}-1\right)+x+y+z}\le\Sigma\dfrac{1}{x\left(\sqrt{2}-1\right)+9}\le\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{2}\)

\(\)thật vậy tương đương

\(\sqrt{2}\left(\sqrt{2}-1\right)\left[x\left(\sqrt{2}-1\right)+9\right]\left[y\left(\sqrt{2}-1\right)+9\right]\left[z\left(\sqrt{2}-1\right)+9\right]\ge2\left[\left(x\left(\sqrt{2}-1\right)+9\right)\left(z\left(\sqrt{2}-1\right)+9\right)+\left(x\left(\sqrt{2}-1\right)+9\right)\left(y\left(\sqrt{2}-1\right)+9\right)+\left(y\left(\sqrt{2}-1\right)\left(z\left(\sqrt{ }\right)\right)+9\left(\right)\right)\left(\right)2-1\left(\right)+9\right]\)

\(\Leftrightarrow xyz\sqrt{2}\left(\sqrt{2}-1\right)^4+9\sqrt{2}\left(\sqrt{2}-1\right)^3\left(xy+yz+xz\right)+81\sqrt{2}\left(\sqrt{2}-1\right)^2\left(x+y+z\right)+729\sqrt{2}\left(\sqrt{2}-1\right)\ge2\left(\sqrt{2}-1\right)^2\left(xy+yz+xz\right)+36\left(\sqrt{2}-1\right)\left(x+y+z\right)+486\)

\(\)\(\Leftrightarrow xyz\sqrt{2}\left(\sqrt{2}-1\right)^4+9\sqrt{2}\left(\sqrt{2}-1\right)^3\left(xy+yz+xz\right)-2\left(\sqrt{2}-1\right)^2\left(xy+Yz+xz\right)+81\sqrt{2}\left(\sqrt{2}-1\right)^2\left(x+y+z\right)-36\left(\sqrt{2}-1\right)\left(x+y+z\right)+729\sqrt{2}\left(\sqrt{2}-1\right)-486\ge0\left(1\right)\)

\(\left(1\right)\) là đúng do:

 \(xyz\sqrt{2}\left(\sqrt{2}-1\right)^4\ge27\sqrt{2}\left(\sqrt{2}-1\right)^4\)

\(9\sqrt{2}\left(\sqrt{2}-1\right)^3\left(xy+yz+xz\right)-2\left(\sqrt{2}-1\right)^2\left(xy+yz+xz\right)=\left(xy+yz+xz\right)\left[9\sqrt{2}\left(\sqrt{2}-1\right)^3-2\left(\sqrt{2}-1\right)^2\right]\ge3\sqrt[3]{27^2}\left[9\sqrt{2}\left(\sqrt{2}-1\right)^3-2\left(\sqrt{2}-1\right)^2\right]\)

\([81\sqrt{2}\left(\sqrt{2}-1\right)^2-36\left(\sqrt{2}-1\right)]\left(x+y+z\right)\ge9\left[81\sqrt{2}\left(\sqrt{2}-1\right)^2-36\left(\sqrt{2}-1\right)\right]\)

=>vế trái của (1) 

\(\ge27\sqrt{2}\left(\sqrt{2}-1\right)^4+3\sqrt[3]{27^2}\left[9\sqrt{2}\left(\sqrt{2}-1\right)^3-2\left(\sqrt{2}-1\right)^2\right]+9\left[81\sqrt{2}\left(\sqrt{2}-1\right)^2-36\left(\sqrt{2}-1\right)\right]+729\sqrt{2}\left(\sqrt{2}-1\right)-486=0\)

do đó \(Q\le\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{2}\)

\(=\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}.2\sqrt{5}+\dfrac{2}{5}.5\sqrt{2}\right):\left(\dfrac{4}{15}.\dfrac{1}{2}\sqrt{\dfrac{1}{2}}\right)=\dfrac{15}{4}-\dfrac{45\sqrt{10}}{2}+30\)

\(e;=\)đề sai sửa \(12xy^2->12x^2y\Rightarrow e;=\left(2x-y+1\right)\left(4x^2-4xy-2x+y^2+y+1\right)\)

\(g;=\left(x-3\right)^3\)

\(i;=\left(x-7\right)\left(x+y\right)\)

\(k;=\left(3x-5\right)\left(y-2x\right)\)

\(l;=\left(x+y+1\right)\left(x^2-xy-x+y^2+2y+1\right)\)

\(h;=\left(x^2-1\right)\left(x^{21}+x^{19}+x^{17}+x^{15}+x^{13}+x^{11}+x^9+x^7+x^5+x^3+x-2\right)\)

\(i;Fx?\)

\(k;=\left(3x-5\right)\left(y-2x\right)\)

\(l;\) ko phân tích đc

\(đk:-1< x< 1\Rightarrow P=\left(\dfrac{2}{\sqrt{1+x}}+\dfrac{\sqrt{1-x^2}}{\sqrt{1+x}}\right):\left(\dfrac{2+\sqrt{1-x^2}}{\sqrt{1-x^2}}\right)=\left(\dfrac{2+\sqrt{1-x^2}}{\sqrt{1+x}}\right):\left(\dfrac{2+\sqrt{1-x^2}}{\sqrt{1-x^2}}\right)=\dfrac{\sqrt{1-x^2}}{\sqrt{1+x}}=\sqrt{1-x}\)

\(P< 1\Leftrightarrow\sqrt{1-x}< 1\Leftrightarrow1-x< 1\Leftrightarrow x>0\Rightarrow0< x< 1\)

pytago\(\Rightarrow BH=\sqrt{AB^2-AH^2}=\sqrt{30^2-24^2}=18cm\)

\(có:AB^2=BH.BC\Leftrightarrow BC=\dfrac{AB^2}{BH}=50cm\)

\(pytago\Rightarrow AC=\sqrt{BC^2-AB^2}=\sqrt{50^2-30^2}=40cm\)

\(P=\dfrac{\left(2-\sqrt{6}\right)\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}=\dfrac{6+2\sqrt{6}-3\sqrt{6}-6}{3^2-6}=\dfrac{-\sqrt{6}}{3}\)

\(M=\dfrac{\sqrt{3-2\sqrt{3}+1}-\sqrt{3+2\sqrt[]{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}=\dfrac{-2}{\sqrt{2}}=-\sqrt{2}\)\(\sqrt{ }\)

\(a;\Leftrightarrow-2x< -6\Leftrightarrow x>3\)

\(b;\Leftrightarrow-4x>-8\Leftrightarrow x< 2\)

\(c;\Leftrightarrow x-2x\le7\Leftrightarrow-x\le7\Leftrightarrow x\ge-7\)

\(\sqrt{x-1}=\sqrt{3-x}\left(đk:1\le x\le3\right)\)

\(pt\Leftrightarrow x-1=3-x\Leftrightarrow x=2\left(tm\right)\)

\(a;\left(đk:x>0\right)\Rightarrow\dfrac{x+\sqrt{x}}{\sqrt{x}}+\dfrac{x-1}{\sqrt{x}+1}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}=\sqrt{x}+1+\sqrt{x}-1=2\sqrt{x}\)

\(b;\left(đk:x\ge0;x\ne4\right)\Rightarrow\dfrac{x+2\sqrt{x}}{\sqrt{x}+2}+\dfrac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}+2}+\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-2}=\sqrt{x}+\sqrt{x}-2=2\sqrt{x}-2\)

\(a+b=\sqrt{28+6\sqrt{3}}-\sqrt{31-12\sqrt{3}}=\sqrt{\left(3\sqrt{3}\right)^2+2.3\sqrt{3}+1}-\sqrt{\left(3\sqrt{3}\right)^2-2.2.3\sqrt{3}+2^2}=\sqrt{\left(3\sqrt{3}+1\right)^2}-\sqrt{\left(3\sqrt{3}-2\right)^2}=3\sqrt{3}+1-3\sqrt{3}+2=3\)

\(\Rightarrow P=\left(a+b\right)^2-2ab+4\left(a+b\right)+2ab=\left(a+b\right)^2+4\left(a+b\right)=3^2+4.3=21\)

\(d;pt\Leftrightarrow\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3+2.4\sqrt{2x-3}+16}=5\Leftrightarrow\sqrt{\left(\sqrt{2x-3}+1\right)^2}+\sqrt{\left(\sqrt{2x-3}+4\right)^2}=5\Leftrightarrow2\sqrt{2x-3}+1+4=5\Leftrightarrow\sqrt{2x-3}=0\Leftrightarrow x=\dfrac{3}{2}\)

\(\)

hình bạn tự vẽ 

xét tam giác AMC có K,N lần lượt là trung điểm của MC;AC

=>NK là đường trung bình tam giác ACM\(\Rightarrow NK//AM;NK=\dfrac{1}{2}AM\)(1)

xét tam giác MQB có I;J lần lượt là trung điuểm MQ;QB

=>IJ là đường trung bình \(\Rightarrow IJ//MB;IJ=\dfrac{1}{2}MB\left(2\right)\)

có M là trung điểm AB=>AM=MB(3)

từ (1)(2) và(3) \(\Rightarrow\left\{{}\begin{matrix}NK=IJ\\NK//IJ\end{matrix}\right.\)=>đpcm