b) ĐK: \(x\ge1;y\ge2;z\ge3\)

Ta có: \(x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)

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

\(\left(z-3\right)-6\sqrt{z-3}+9=0\)

<=> \(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}+3\right)^2=0\)

=> \(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{y-2}=2\\\sqrt{z-3}=3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\) (TM)

Vậy ............................................

a) ĐK: \(x\ge1\)

Ta có: \(\sqrt{4x-4}=\dfrac{x+3}{2}\)

<=> \(2\sqrt{4\left(x-1\right)}=x+3\)

<=> \(2.2\sqrt{x-1}=x+3\)

<=> \(x+3-4\sqrt{x-1}=0\)

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

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

<=> \(\sqrt{x-1}=2\)

<=> \(x-1=4\) => \(x=5\) (TM)

Vậy ............................................

Đề khá hay đấy! Nhưng lần sau đừng viết sai đề nx!

a) ĐK: \(x>4\)

b) \(P=\dfrac{\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}}{\sqrt{1-\dfrac{8}{x}+\dfrac{16}{x^2}}}\)

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

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

= \(\dfrac{\left|\sqrt{x-4}+2\right|+\left|\sqrt{x-4}-2\right|}{\left|1-\dfrac{4}{x}\right|}\)

= \(\dfrac{\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|}{1-\dfrac{4}{x}}\) = \(\left[{}\begin{matrix}\dfrac{2x\sqrt{x-4}}{x-4}khix\ge8\\\dfrac{4x}{x-4}khi4< x< 8\end{matrix}\right.\)

Xét \(P=\dfrac{2x}{\sqrt{x-4}}\left(x\ge8\right)\) thì:

Để \(P\in Z\) khi \(\dfrac{2x-8+8}{\sqrt{x-4}}\in Z\)

<=> \(2.\left(\sqrt{x-4}\right)+\dfrac{8}{\sqrt{x-4}}\in Z\)

<=> \(\left\{{}\begin{matrix}\sqrt{x-4}\in Z^+\\\sqrt{x-4}\inƯ\left(8\right)\end{matrix}\right.\)

Mà \(x\ge8\) => \(\left[{}\begin{matrix}\sqrt{x-4}=2\\\sqrt{x-4}=4\\\sqrt{x-4}=8\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=8\\x=20\\x=68\end{matrix}\right.\)

Xét \(P=\dfrac{4x}{x-4}\left(4< x< 8\right)\) thì:

Để \(P\in Z\) khi \(\dfrac{4x-16+16}{x-4}\in Z\) <=> \(4+\dfrac{16}{x-4}\in Z\)

=> \(x-4\inƯ\left(16\right)\) mà \(0< x-4< 4\)

=> \(x-4=2\) => \(x=6\)

Vậy \(x\in\left\{6;8;20;68\right\}\) thì \(P\in Z\)

P/s: Vì bài này dài nên mk lm khá tắt, ko hiểu cứ hỏi!

Bài 1:

Ta có: \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2mol\)

\(m_{NaOH}=50.20\%=10g\)

=> \(n_{NaOH}=\dfrac{10}{40}=0,25mol\)

Vì \(\dfrac{1}{2}< \dfrac{n_{CO_2}}{n_{NaOH}}=0,8< 1\) => Tạo ra 2 muối

PTHH: 2NaOH + CO₂ --> Na₂CO₃ + H₂O

(mol)............2x .............x................x

NaOH + CO₂ --> NaHCO₃

(mol)..............y.................y................y

Ta có HPT: \(\left\{{}\begin{matrix}2x+y=0,25\\x+y=0,2\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=0,05\\y=0,15\end{matrix}\right.\)

=> \(m_{Na_2CO_3}=0,05.106=5,3g\)

\(m_{NaHCO_3}=0,15.84=12,6g\)