\(\dfrac{x+\sqrt{11}}{x^2+2x\sqrt{11}+11}\) = \(\dfrac{x+\sqrt{11}}{(x+\sqrt{11)}^2}\) =  x +\(\sqrt{11}\)

$\frac{x+\sqrt{11}}{x^2+2x\sqrt{11}+11}$ = $\frac{x+\sqrt{11}}{(x+{\sqrt{11)}}^2}$ =  x +$\sqrt{11}$

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

\(\dfrac{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}{x-\sqrt{3}}=x+\sqrt{3}\)

đề bài có sai không nhỉ sao biểu thức không đối xứng

I'm don't know

14 balls are not blue.    => yellow + red + pink = 14
16 balls are not yellow. => blue + red + pink = 16
24 balls are not red.      => blue + yellow + pink = 24
12 balls are not pink.    => blue + yellow + red = 12

====> 3 yellow + 3 red + 3 blue + 3 pink = 14+16+24+12
    3(yellow + red + blue + pink) = 66
    yellow + red + blue + pink = 66:3 =22
 

\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ n_{NaOH}=0,5.0,1=0,05\left(mol\right)\\ n_{Na_2CO_3}=1,5.0,1=0,15\left(mol\right)\)

Xét \(T=\dfrac{0,05}{1}=\dfrac{1}{2}\) => Tạo muối NaHCO₃ 

PTHH:   \(NaOH+CO_2\rightarrow NaHCO_3\)

               0,05---->0,05----->0,05

               \(Na_2CO_3+CO_2+H_2O\rightarrow2NaHCO_3\)

bđ            0,15           0,05

sau pư     0,1               0                       0,1

Vậy ddX gồm \(\left\{{}\begin{matrix}Na_2CO_3:n_{Na_2CO_3}=0,1\left(mol\right)\\NaHCO_3:n_{NaHCO_3}=0,05+0,1=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{CO_3^{2-}}=0,1\left(mol\right)\\n_{HCO_3^-}=0,15\left(mol\right)\end{matrix}\right.\)

\(n_{BaCO_3\downarrow}=\dfrac{43,34}{197}=0,22\left(mol\right)\)

Xét \(0,22< 0,1+0,15=0,25\)

=> Trong dd có chứa muối \(HCO_3^-\)

+) TH₁: Muối đó là NaHCO₃ dư

\(n_{Ba^{2+}}=n_{BaCO_3}=0,22\left(mol\right)\\ n_{BaCl_2}=2.0,1=0,2\left(mol\right)\\ \xrightarrow[]{\text{BTNT Ba}}n_{Ba\left(OH\right)_2}=0,22-0,2=0,02\left(mol\right)\\ \rightarrow a=C_{M\left(Ba\left(OH\right)_2\right)}=\dfrac{0,02}{0,1}=0,2M\)

+) TH₂: Muối đó là Ba(HCO₃)₂

\(\xrightarrow[\text{BTNT C}]{}n_{Ba\left(HCO_3\right)_2}=\dfrac{1}{2}n_{HCO_3^-}=\dfrac{1}{2}.\left(0,25-0,22\right)=0,015\left(mol\right)\\ \sum n_{Ba^{2+}}=n_{Ba\left(HCO_3\right)_2}+n_{BaCO_3}=0,015+0,22=0,235\left(mol\right)\\ \xrightarrow[\text{BTNT Ba}]{}n_{Ba\left(OH\right)_2}=0,235-0,2=0,035\left(mol\right)\\ \rightarrow a=C_{M\left(Ba\left(OH\right)_2\right)}=\dfrac{0,03}{0,1}=0,35M\)

Vậy \(0,2\le a\le0,35\)

hnhu thay a = 0,2 với 0,35 vào đều ra khác với đề hsy :) ?