ĐKXĐ : \(x\ge-4\)

(2x + 1)² = (x + 4)\(\sqrt{4x^2+1}\)

<=> \(4x^2+1+4x=x\sqrt{4x^2+1}+4\sqrt{4x^2+1}\)

<=> \(\left(\sqrt{4x^2+1}-4\right)\left(\sqrt{4x^2+1}-x\right)=0\)

<=> \(\left[{}\begin{matrix}\sqrt{4x^2+1}=4\\\sqrt{4x^2+1}=x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x^2=15\\\left\{{}\begin{matrix}3x^2+1=0\\x\ge0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\sqrt{15}}{2}\left(tm\right)\\∄x\end{matrix}\right.\Leftrightarrow x=\pm\dfrac{\sqrt{15}}{2}\)

\(\sqrt{\dfrac{3x^2-3x-2x+2}{4x^2-12x+9}}=\sqrt{\dfrac{\left(3x-2\right)\left(x-1\right)}{\left(2x-3\right)^2}}\)

đk \(\left\{{}\begin{matrix}\left(3x-2\right)\left(x-1\right)>0\\2x-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>1\\x< \dfrac{2}{3}\\x\ne\dfrac{3}{2}\end{matrix}\right.\)

PTHH: \(K_2O+H_2O\rightarrow2KOH\)

Ta có: \(n_{K_2O}=\dfrac{x}{94}\left(mol\right)\) \(\Rightarrow n_{KOH\left(thêm\right)}=\dfrac{x}{47}\left(mol\right)\) \(\Rightarrow m_{KOH\left(thêm\right)}=\dfrac{56x}{47}\left(g\right)\)

Theo đề bài: \(\Sigma m_{KOH}=120\cdot51\%=61,2\left(g\right)\)

Ta lập được hệ phương trình: \(\left\{{}\begin{matrix}\dfrac{56x}{47}+12\%\cdot y=61,2\\x+y=120\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\approx43,68\\y\approx76,32\end{matrix}\right.\)

bạn xem lại số liệu nhé chứ số như kia xấu quá :((

\(=\sqrt{2}-1-\sqrt{2}-1-2=-4\)

Gọi CTHH của oxit là R₂O_n 

m_{dd sau phản ứng} = 211,75 + 38,25 = 250 (g)

=> \(m_{R\left(OH\right)_n}=17,1\%.250=42,75\left(g\right)\)

=> \(n_{R\left(OH\right)_n}=\dfrac{42,75}{M_R+17n}\left(mol\right)\)

\(n_{R_2O_n}=\dfrac{38,25}{2M_R+16n}\left(mol\right)\)

PTHH: \(R_2O_n+nH_2O\rightarrow2R\left(OH\right)_n\)

        \(\dfrac{38,25}{2M_R+16n}\)---------->\(\dfrac{38,25}{M_R+8n}\)

=> \(\dfrac{38,25}{M_R+8n}=\dfrac{42,75}{M_R+17n}\)

=> M_R = 68,5n (g/mol)

n = 2 => M_R = 68,5.2 = 137 (h/mol)

=> R là Ba

CTHH của oxit là BaO

Gọi công thức của oxit bazơ là R₂O_x  

PTHH: \(R_2O_x+xH_2O\rightarrow2R\left(OH\right)_x\)

Ta có: \(m_{dd}=m_{Oxit}+m_{H_2O}=38,25+211,75=250\left(g\right)\)

\(\Rightarrow m_{R\left(OH\right)_x}=250\cdot17,1\%=42,75\left(g\right)\)

Theo PTHH: \(n_{R\left(OH\right)_x}=2n_{R_2O_x}\) \(\Rightarrow2\cdot\dfrac{38,25}{2R+16x}=\dfrac{42,75}{R+17x}\)

Ta thấy với \(x=2\) thì \(R=137\)  (Bari)

    Vậy CTHH là BaO

\(ĐKXĐ:x\ge0\)

\(4x-2\sqrt{x}=0\)

\(\Leftrightarrow2\sqrt{x}\left(2\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\2\sqrt{x}-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{1}{4}\left(tm\right)\end{matrix}\right.\)

đk x >= 0 

\(2\sqrt{x}\left(2\sqrt{x}-1\right)=0\Leftrightarrow x=0;x=\dfrac{1}{4}\left(tm\right)\)