\(a,PTHH:2Cu+O_2\xrightarrow{t^o}2CuO\\ n_{Cu}=\dfrac{11,52}{64}=0,18(mol)\\ b,n_{CuO}=n_{Cu}=0,18(mol)\\ \Rightarrow m_{CuO}=0,18.80=14,4(g)\\ c,n_{O_2}=0,5.n_{Cu}=0,09(mol)\\ \Rightarrow V_{O_2}=0,09.22,4=2,016(l)\\ \Rightarrow V_{kk}=2,016.5=10,08(l)\)

\(a,=\dfrac{1}{3}\cdot9\sqrt{3}-6\cdot\dfrac{2}{\sqrt{3}}+\dfrac{\sqrt{6}\left(\sqrt{3}+1\right)}{\sqrt{6}}=3\sqrt{3}-4\sqrt{3}+\sqrt{3}+1=1\\ b,=\sqrt{\left(\sqrt{6}+\sqrt{2}\right)^2}+\dfrac{14\left(3-\sqrt{2}\right)}{7}-\dfrac{\sqrt{3}\left(\sqrt{6}-\sqrt{2}\right)}{\sqrt{3}}\\ =\sqrt{6}+\sqrt{2}+6-2\sqrt{2}-\sqrt{6}+\sqrt{2}=6\)

\(=\left(2x^3-x^2+4x^2-2x-6x+3\right):\left(2x-1\right)\\ =\left[x^2\left(2x-1\right)+2x\left(2x-1\right)-3\left(2x-1\right)\right]:\left(2x-1\right)\\ =x^2+2x-3\)

\(\sqrt{1-a^2}+\sqrt{1-b^2}\le2\sqrt{1-\left(\dfrac{a+b}{2}\right)^2}\\ \Leftrightarrow\sqrt{\left(1-a^2\right)\left(1-b^2\right)}\le1-ab\\ \Leftrightarrow\left(1-a^2\right)\left(1-b^2\right)\le1-ab\\ \Leftrightarrow\left(a-b\right)^2\ge0\left(\text{luôn đúng}\right)\)

Vậy ta được đpcm, dấu \("="\Leftrightarrow a=b\)

\(1,\\ a,=6+7-8=5\\ b,=2\sqrt{2}+5\sqrt{2}+\sqrt{2}=8\sqrt{2}\\ 2,\\ \dfrac{3+\sqrt{3}}{1+\sqrt{3}}=\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{1+\sqrt{3}}=\sqrt{3}\)

\(7,\\ a,ĐK:a>0;a\ne1\\ b,K=\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}-1+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\\ K=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\dfrac{a-1}{\sqrt{a}}\\ c,a=3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\\ \Leftrightarrow K=\dfrac{3+2\sqrt{2}-1}{\sqrt{2}+1}=\dfrac{2+2\sqrt{2}}{\sqrt{2}+1}=\dfrac{2\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=2\\ d,K< 0\Leftrightarrow a-1< 0\left(\sqrt{a}>0\right)\Leftrightarrow0< a< 1\)

\(a,\Delta=m^2+20\ge20>0,\forall m\)

Vậy PT luôn có 2 nghiệm phân biệt với mọi m

\(b,\Delta=m^2+4\left(m+1\right)=m^2+4m+4=\left(m+2\right)^2\ge0,\forall m\)

Vậy PT luôn có 2 nghiệm với mọi m

\(c,\Delta=\left(m+2\right)^2-4\left(2m-5\right)=m^2-4m+24=\left(m-2\right)^2+20\ge20>0\)

Vậy PT luôn có 2 nghiệm phân biệt với mọi m

\(d,\Delta'=\left(2m-1\right)^2-\left(m-20\right)=4m^2-5m+21=\left(4m^2-2\cdot2\cdot\dfrac{5}{4}+\dfrac{25}{16}\right)+\dfrac{311}{16}=\left(2m-\dfrac{5}{4}\right)^2+\dfrac{311}{16}\ge\dfrac{311}{16}>0,\forall m\)

Vậy PT luôn có 2 nghiệm phân biệt với mọi m

\(a,\Leftrightarrow\Delta=\left(2m+1\right)^2-4\left(m^2-3\right)=0\\ \Leftrightarrow4m^2+4m+1-4m^2+12=0\\ \Leftrightarrow4m+13=0\Leftrightarrow m=-\dfrac{13}{4}\\ b,\Leftrightarrow\Delta=9m^2-4\left(m-2\right)< 0\Leftrightarrow9m^2-4m+8< 0\)

Ta thấy \(9m^2-4m+8=\left(9m^2-2\cdot3\cdot\dfrac{2}{3}m+\dfrac{4}{9}\right)+\dfrac{68}{9}=\left(3m-\dfrac{2}{3}\right)^2+\dfrac{68}{9}>0\)

Vậy không có gt của m thỏa đề

\(c,\Leftrightarrow\Delta'=\left(m-1\right)^2-m^2\ge0\\ \Leftrightarrow-2m+1\ge0\Leftrightarrow m\le\dfrac{1}{2}\)

\(a,\text{Sửa: }x^2+5x-3=0\\ \Delta=25+12=37\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5-\sqrt{37}}{2}\\x=\dfrac{-5+\sqrt{37}}{2}\end{matrix}\right.\)

\(b,\Delta=\left(2\sqrt{3}-1\right)^2-4\left(4\sqrt{3}-6\right)=37-20\sqrt{3}=\left(5-2\sqrt{3}\right)^2>0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\sqrt{3}-1-5+2\sqrt{3}}{2}=-3+2\sqrt{3}\\x=\dfrac{2\sqrt{3}-1+5-2\sqrt{3}}{2}=2\end{matrix}\right.\)

\(c,\Delta'=\left(2\sqrt{3}\right)^2+4=4+12=16\\ \Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{3}-4\\x=2\sqrt{3}+4\end{matrix}\right.\)