a, Ta có : \(P_{\left(x\right)}=x^2+5x^4-3x^3+x^2+4x^4+3x^2-x+5\)

\(=9x^4-3x^3+5x^2-x+5\)

Ta có : \(Q_{\left(x\right)}=x-5x^3-x^2-x^4+4x^3-x^2+3x-1\)

\(Q_{\left(x\right)}=-x^4-x^3-2x^2+4x-1\)

b, Ta có : \(P_{\left(x\right)}+Q_{\left(x\right)}=9x^4-3x^3+5x^2-x+5-x^4-x^3-2x^2+4x-1\)

\(=8x^4-4x^3+3x^2+3x+4\)

Ta có : \(P_{\left(x\right)}-Q_{\left(x\right)}=9x^4-3x^3+5x^2-x+5+x^4+x^3+2x^2-4x+1\)

\(=10x^4-2x^3+7x^2-5x+6\)

Ta có : \(n_{O2}=\dfrac{V}{22,4}=0,05\left(mol\right)\)

\(\Rightarrow O2=n.A=3.10^{22}\) ( phân tử )

Ta có : \(n_{SO3}=\dfrac{V}{22,4}=0,1\left(mol\right)\)

\(\Rightarrow SO3=n.A=6.10^{22}\) ( phân tử )

Ta có : \(n_{NaOH}=\dfrac{m}{M}=0,4\left(mol\right)\)

\(\Rightarrow NaOH=n.A=2,4.10^{23}\) ( phân tử )

Ta có : \(n_{SO3}=\dfrac{m}{M}=0,405\left(mol\right)\)

\(\Rightarrow SO3=n.A=2,4381.10^{23}\) ( phân tử )

\(4FeS_2+11O_2\rightarrow2Fe_2O_3+8SO_2\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\)

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

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

Bài 2 mình viết gọn chút nha .

- Qùy tím

Na2SO4 không màu

HCl H2SO4 Đỏ => Nhóm 1

Ba(OH)2 KOH Xanh => Nhóm 2

- Cho Na2SO4 vào nhóm 2

=> Ba(OH)2 tạo kt

- Cho Ba(OH)2 và Nhóm 1

=> H2SO4 tạo kết tủa

Sai chỗ nào v ?

\(a,A=-x^2+2x+10=-x^2+2x-1+11=-\left(x^2-2x+1\right)+11\)

\(=11-\left(x-1\right)^2\)

- Thấy : \(\left(x-1\right)^2\ge0\forall x\in R\)

\(\Rightarrow A=11-\left(x-1\right)^2\le11\)

Vậy MaxA = 11 <=> x = 1 .

\(b,B=-2x^2+4x-2+10=-2\left(x^2-2x+1\right)+10=10-2\left(x-1\right)^2\)

- Thấy : \(\left(x-1\right)^2\ge0\forall x\in R\)

\(\Rightarrow B=10-2\left(x-1\right)^2\le10\)

Vậy MaxB = 10 <=> x = 1 .

\(c,C=-x^2-\dfrac{1}{2}.2.x-\dfrac{1}{4}+\dfrac{5}{4}=\dfrac{5}{4}-\left(x+\dfrac{1}{2}\right)^2\)

- Thấy : \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\in R\)

\(\Rightarrow C=\dfrac{5}{4}-\left(x+\dfrac{1}{2}\right)^2\le\dfrac{5}{4}\)

Vậy MaxC = 5/4 <=> x = -1/2 .

\(d,D=-4x^2+6x+3=-4x^2+2x.2.\dfrac{6}{4}-\dfrac{9}{4}+\dfrac{21}{4}=-\left(4x^2-6x+\dfrac{9}{4}\right)+\dfrac{21}{4}\)

\(=\dfrac{21}{4}-\left(2x-\dfrac{3}{2}\right)^2\)

- Thấy : \(\left(2x-\dfrac{3}{2}\right)^2\ge0\forall x\in R\)

\(\Rightarrow A=\dfrac{21}{4}-\left(2x-\dfrac{3}{2}\right)^2\le\dfrac{21}{4}\)

Vậy MaxD=21/4 <=> x = 3/4 .

\(n_{Na}=\dfrac{m}{M}=0,2mol\)

\(n_{Na2O}=\dfrac{m}{M}=0,1\left(mol\right)\)

\(BTNT\left(Na\right):n_{NaOH}=0,2+2.0,1=0,4\left(mol\right)\)

\(\Rightarrow m_{NaOH}=n.M=16g\)

\(PTHH:Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)

...............0,2...................................0,1.

\(\Rightarrow m_{dd}=m_{KL}+m_{H2O}-m_{H2}=110,6g\)

\(\Rightarrow C\%_{NaOH}=\dfrac{m}{m_{dd}}.100\%=14,46\%\)