\(a,\dfrac{3}{8}+\dfrac{5}{6}=\dfrac{9+20}{24}=\dfrac{29}{24}\\ b,2,9\cdot\sqrt{100}-13,4=2,9\cdot10-13,4=29-13,4=15,6\)

\(PT\Leftrightarrow\left[\left(x+1\right)\left(x-3\right)\right]\left[\left(x+2\right)\left(x-4\right)\right]=36\\ \Leftrightarrow\left(x^2-2x-3\right)\left(x^2-2x-8\right)=36\\ \Leftrightarrow\left(x^2-2x-3\right)^2-5\left(x^2-2x-3\right)-36=0\\ \Leftrightarrow\left(x^2-2x-3\right)^2-9\left(x^2-2x-3\right)+4\left(x^2-2x-3\right)-36=0\\ \Leftrightarrow\left(x^2-2x-3\right)\left(x^2-2x-12\right)+4\left(x^2-2x-12\right)=0\\ \Leftrightarrow\left(x^2-2x-12\right)\left(x^2-2x+1\right)=0\\ \Leftrightarrow\left(x^2-2x-12\right)\left(x-1\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x-1\right)^2-13=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=1+\sqrt{13}\\x=1-\sqrt{13}\end{matrix}\right.\)

\(A=\left[\left(x^2-2xy+y^2\right)+2\left(x-y\right)+1\right]+\left(y^2-8y+16\right)-17\\ A=\left(x-y+1\right)^2+\left(y-4\right)^2-17\ge-17\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-y+1=0\\y-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y-1=3\\y=4\end{matrix}\right.\)

\(a,P=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\\ P=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)

\(b,P=\dfrac{1}{2}\Leftrightarrow4-10\sqrt{x}=\sqrt{x}+3\Leftrightarrow\sqrt{x}=\dfrac{7}{11}\Leftrightarrow x=\dfrac{49}{121}\left(tm\right)\)

\(c,P-\dfrac{2}{3}=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}-\dfrac{2}{3}=\dfrac{6-15\sqrt{x}-2\sqrt{x}-6}{3\left(\sqrt{x}+3\right)}=\dfrac{-17\sqrt{x}}{3\left(\sqrt{x}+3\right)}\)

Ta có \(3\left(\sqrt{x}+3\right)>0;-17\sqrt{x}\le0,\forall x\)

\(\Rightarrow P-\dfrac{2}{3}\le0\Leftrightarrow P\le\dfrac{2}{3}\left(đpcm\right)\)

\(a,P=\dfrac{-x+2\sqrt{x}-1+x-2\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}:\dfrac{2\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ P=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}+1}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

\(b,x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\\ \Rightarrow P=\dfrac{\sqrt{5}-1}{\sqrt{5}-1+1}=\dfrac{\sqrt{5}-1}{\sqrt{5}}=\dfrac{5-\sqrt{5}}{5}\\ c,\dfrac{P}{\sqrt{x}}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\cdot\dfrac{1}{\sqrt{x}}=\dfrac{1}{\sqrt{x}-1}\le\dfrac{1}{0-1}=-1\)

Vậy \(\left(\dfrac{P}{\sqrt{x}}\right)_{max}=-1\Leftrightarrow x=0\)

$PTHH:2NaOH+H_2SO_4\to Na_2SO_4+2H_2O$

$n_{H_2SO_4}=\dfrac{98.20\%}{98}=0,2(mol)$

Theo PT: $n_{NaOH}=0,2.2=0,4(mol)$

$\Rightarrow C\%_{NaOH}=\dfrac{0,4.40}{200}.100\%=8\%$

Đáp án B

$1)PTHH:C_2H_8O_2+5O_2\xrightarrow{t^o}4CO_2\uparrow+4H_2O$

$n_{C_4H_8O_2}=\dfrac{4,4}{88}=0,05(mol)$

Theo PT: $n_{O_2}=5.0,05=0,25(mol)$

$\Rightarrow V_{O_2}=0,25.22,4=5,6(l)$

$2)PTHH:4Al+3O_2\xrightarrow{t^o}2Al_2O_3$

$n_{O_2}=\dfrac{33,6}{22,4}=1,5(mol)$

Theo PT: $n_{Al}=\dfrac{4}{3}n_{O_2}=2(mol)$

$\Rightarrow m_{Al}=2.27=54(g)$

$n_{Fe}=\dfrac{2,24}{56}=0,04(mol)$

$a,PTHH:Fe+2HCl\to FeCl_2+H_2$

$b,$ Theo PT: $n_{H_2}=n_{Fe}=0,04(mol)$

$\Rightarrow V_{H_2}=0,04.22,4=0,896(l)$

$a,PTHH:2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2$

$b,$ Áp dụng định luật BTKL:

$m_{KMnO_4}=m_{K_2MnO_4}+m_{MnO_2}+m_{O_2}$

$c,m_{O_2}=31,6-19,7-8,7=3,2(g)$

":))" cái này là ok hả :)???