`a.` \(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)

`b.`\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)

\(Fe_2\left(SO_4\right)_3+3Ba\left(OH\right)_2\rightarrow2Fe\left(OH\right)_3\downarrow+3BaSO_4\)

\(2Fe\left(OH\right)_3\rightarrow\left(t^o\right)Fe_2O_3+3H_2O\)

\(2H_2O\rightarrow\left(đp\right)2H_2+O_2\)

\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(FeSO_4+Ba\left(OH\right)_2\rightarrow Fe\left(OH\right)_2\downarrow+BaSO_4\downarrow\)

`c.`\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(CuSO_4+Ba\left(OH\right)_2\rightarrow Cu\left(OH\right)_2\downarrow+BaSO_4\downarrow\)

\(n_{CuO}=\dfrac{2,3}{80}=\dfrac{23}{800}\left(mol\right)\) ; \(n_{H_2SO_4}=\dfrac{100.19,6\%}{98}=0,2\left(mol\right)\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(\dfrac{23}{800}\)   <    0,2                                  ( mol )

\(\dfrac{23}{800}\)         \(\dfrac{23}{800}\)              \(\dfrac{23}{800}\)              ( mol )

\(m_{ddspứ}=2,3+100=102,3\left(g\right)\)

\(\left\{{}\begin{matrix}C\%_{CuSO_4}=\dfrac{\dfrac{23}{800}.160}{102,3}.100=4,49\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{\left(0,2-\dfrac{23}{800}\right).98}{102,3}.100=16,4\%\end{matrix}\right.\)

\(A=\dfrac{x+4}{x-4}-\dfrac{2}{\sqrt{x}-2}\) ; \(\left(x\ge0;x\ne4\right)\)

\(A=\dfrac{x+4-2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(A=\dfrac{x+4-2\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(A=\dfrac{x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(A=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(A=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)

`5.`\(\left(x+3\right)^2-\left(x-2\right)\left(x+5\right)=9\)

\(\Rightarrow x^2+6x+9-x^2-5x+2x+10-9=0\)

\(\Rightarrow3x+10=0\)

\(\Rightarrow x=-\dfrac{10}{3}\)

`6.`\(3x-\left(x+1\right)\left(x^2-x+1\right)+x\left(x^2-4\right)=5\)

\(\Rightarrow3x-x^3-1+x^3-4x-5=0\)

\(\Rightarrow-x-6=0\)

\(\Rightarrow x=-6\)

`7.`\(x^2-2x=24\)

\(\Rightarrow x^2-2x-24=0\)

\(\Rightarrow\left(x-6\right)\left(x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

`8.`\(x^3+6x^2+12x=19\)

\(\Rightarrow\left(x^3-x^2\right)+\left(7x^2-7x\right)+\left(19x-19\right)=0\)

\(\Rightarrow x^2\left(x-1\right)+7x\left(x-1\right)+19\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x^2+7x+19\right)=0\)

\(\Rightarrow x-1=0\)

`=>x=1`

Ta có: 

\(x^{2016}+y^{2016}+z^{2016}=x^{2017}+y^{2017}+z^{2017}=1\)

Ta có: \(0\le x^{2016};y^{2016};z^{2016}\le1\)

\(\Rightarrow0\le x;y;z\le1\)

\(\Rightarrow\left\{{}\begin{matrix}1-x\ge0\\1-y\ge0\\1-z\ge0\end{matrix}\right.\) (1)

Lại có:

\(x^{2016}+y^{2016}+z^{2016}=x^{2017}+y^{2017}+z^{2017}\)

\(\Leftrightarrow x^{2016}\left(1-x\right)+y^{2016}\left(1-y\right)+z^{2016}\left(1-z\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}1-x=0\\1-y=0\\1-z=0\end{matrix}\right.\) (2)

(1);(2) và đề`=>` 1 trong 3 số x,y,z có 2 số bằng 0, 1 số bằng 1

`=>`\(P=1\)

\(\)

\(\dfrac{1}{2}-\dfrac{43}{101}+\left(-\dfrac{1}{3}\right)-\dfrac{1}{6}\)

\(=\dfrac{1}{2}-\dfrac{43}{101}-\dfrac{1}{3}-\dfrac{1}{6}\)

\(=-\dfrac{43}{11}+0\)

\(=-\dfrac{43}{101}\)

`7/11-x=3/11`

`=>x=7/11-3/11

`=>x=4/11`

21. \(6x^2-7x-3=\left(2x-3\right)\left(3x+1\right)\)

22.\(-12x^2+11x-2=\left(3x-2\right)\left(-4x+1\right)\)

23.\(-x^2+2x+3=\left(x+1\right)\left(-x+3\right)\)

24.\(2x^2+5x-3=\left(2x-1\right)\left(x+3\right)\)

xem lại bài!