1) 

a = 100 - 8 = 92 (g)

\(\%_{NaOH}=\dfrac{8}{100}.100\%=8\%\)

2) \(m_{dd.HCl}=1,2.150=180\left(g\right)\)

n_HCl = 0,15.4 = 0,6 (mol)

\(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)

PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,6}{1}\) => NaOH hết, HCl dư

PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)

               0,2----->0,2----->0,2

=> \(\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{0,2.58,5}{100+180}.100\%=4,18\%\\C\%_{HCl\left(dư\right)}=\dfrac{\left(0,6-0,2\right).36,5}{100+180}.100\%=5,21\%\end{matrix}\right.\)

Ta có \(\dfrac{1}{\sqrt{n}+\sqrt{n+2}}=\dfrac{\sqrt{n+2}-\sqrt{n}}{\left(\sqrt{n+2}+\sqrt{n}\right)\left(\sqrt{n+2}-\sqrt{n}\right)}\) \(=\dfrac{\sqrt{n+2}-\sqrt{n}}{\left(\sqrt{n+2}\right)^2-\left(\sqrt{n}\right)^2}\) \(=\dfrac{\sqrt{n+2}-\sqrt{n}}{2}\)

Như vậy, ta có \(\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}+...+\dfrac{1}{\sqrt{23}+\sqrt{25}}\) 

\(=\dfrac{\sqrt{3}-1}{2}+\dfrac{\sqrt{5}-\sqrt{3}}{2}+\dfrac{\sqrt{7}-\sqrt{5}}{2}+...+\dfrac{\sqrt{25}-\sqrt{23}}{2}\)

\(=\dfrac{\sqrt{3}-1+\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+...+\sqrt{25}-\sqrt{23}}{2}\) 

\(=\dfrac{\sqrt{25}-1}{2}=\dfrac{5-1}{2}=2\)

\(A=\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}+...+\dfrac{1}{\sqrt{23}+\sqrt{25}}\)

\(=\dfrac{\sqrt{3}-\sqrt{1}}{\left(\sqrt{3}-\sqrt{1}\right)\left(\sqrt{3}+1\right)}+\dfrac{\sqrt{5}-\sqrt{3}}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}+...+\dfrac{\sqrt{25}-\sqrt{23}}{\left(\sqrt{25}-\sqrt{23}\right)\left(\sqrt{25}+\sqrt{23}\right)}\)

\(=\dfrac{\sqrt{3}-1}{2}+\dfrac{\sqrt{5}-\sqrt{3}}{2}+...+\dfrac{\sqrt{25}-\sqrt{23}}{2}\)

\(=-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{5}}{2}-...-\dfrac{\sqrt{23}}{2}+\dfrac{\sqrt{25}}{2}\\ =\dfrac{1}{2}+\dfrac{\sqrt{25}}{2}=\dfrac{1}{2}+\dfrac{5}{2}=3\)

a) ĐK: \(x\ge0\)

\(\sqrt{x}+9\le31\Leftrightarrow\sqrt{x}\le22\Leftrightarrow x\le484\)

Kết hợp với đk ta được \(0\le x\le484\)

b) ĐK: \(2x-1\ge0\Leftrightarrow x\ge\dfrac{1}{2}\)

Ta có \(\sqrt{2x-1}>6\Leftrightarrow2x-1>36\Leftrightarrow x>\dfrac{37}{2}\) (TM)

Vậy \(x>\dfrac{37}{2}\)

c) ĐK: \(x+3\ge0\Leftrightarrow x\ge-3\)

Ta có \(\sqrt{x+3}\ge5\Leftrightarrow x+3\ge25\Leftrightarrow x\ge22\) (TM)

Vậy \(x\ge22\)

d) ĐK: \(2x-1\ge0\Leftrightarrow x\ge\dfrac{1}{2}\)

Ta có \(\sqrt{2x-1}+5< 2\Leftrightarrow\sqrt{2x-1}< -3\) (vô lí)

Vậy không có \(x\) thỏa mãn.

giúp mình với ạ

a: \(-xy\cdot\sqrt{\frac{y}{x}}=-xy\cdot\frac{\sqrt{y}}{\sqrt{x}}=-\sqrt{x}\cdot y\sqrt{y}=-y\cdot\sqrt{xy}\)

b: \(\sqrt{\frac{-3x^3}{35}}=\sqrt{\frac{-3x^3\cdot35}{35\cdot35}}=\sqrt{-\frac{105x^3}{35^2}}=\sqrt{\frac{x^2}{35^2}}\cdot\sqrt{-105x}=-\frac{x}{35}\cdot\sqrt{-105x}\) (Vì x<0)

c: \(\sqrt{\frac{5a^3}{49b}}=\sqrt{\frac{5a^3b}{49b^2}}\)

\(=\sqrt{\frac{a^2}{49b^2}\cdot5ab}=\frac{a}{7b}\cdot\sqrt{5ab}\)

d: \(-7xy\cdot\sqrt{\frac{3}{xy}}=-7\cdot\sqrt{3xy}\)

                (phương pháp phản chứng )

giả sử x + \(\dfrac{1}{x}\) ϵ Q ⇔ x + \(\dfrac{1}{x}\) = \(\dfrac{a}{b}\) (a,b ϵN, b#0)

⇔ x = \(\dfrac{a}{b}\) - \(\dfrac{1}{x}\)⇔ x - \(\dfrac{1}{x}\) = \(\dfrac{a}{b}\) - \(\dfrac{1}{x}\) - \(\dfrac{1}{x}\) ⇔ x - \(\dfrac{1}{x}\) = \(\dfrac{a}{b}\)- \(\dfrac{2}{x}\)

nếu x = 2 ta có x - \(\dfrac{1}{x}\) = 2 - \(\dfrac{1}{2}\) = \(\dfrac{3}{4}\) (loại vì \(\dfrac{3}{4}\) không thuộc Z)

nếu \(\dfrac{a}{b}\)= \(\dfrac{2}{x}\) ⇔ x - \(\dfrac{1}{x}\) = 0 ⇔ x = +- 1 (loại) ⇔ \(\dfrac{a}{b}\) # \(\dfrac{2}{x}\)

vậy với x # +-1

⇔ x - \(\dfrac{1}{x}\)= \(\dfrac{a}{b}\) - \(\dfrac{2}{x}\)  \(\notin\)  Z ⇔ x + \(\dfrac{1}{x}\) \(\notin\) Q ⇔ x + \(\dfrac{1}{x}\) \(\in\) I (đpcm)

ĐKXĐ: \(x\ge-\dfrac{9}{2};x\ne0\)

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

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

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

\(\Rightarrow\left(3+\sqrt{9+2x}\right)^2=2x+18\)

Đặt \(\sqrt{2x+9}=t\ge0;t\ne3\)

\(\Rightarrow\left(3+t\right)^2=t^2+9\Rightarrow t=0\)

\(\Rightarrow\sqrt{2x+9}=0\Rightarrow x=-\dfrac{9}{2}\)

a.

Để cho biểu thức \(\sqrt{\dfrac{2}{x-1}}\) xác định thì

\(\Leftrightarrow\dfrac{2}{x-1}\ge0\)

\(\Leftrightarrow x-1>0\)

\(\Leftrightarrow x>1\)

b.

Để cho biểu thức \(\sqrt{1+x^2}\) xác định thì

\(\Leftrightarrow1+x^2\ge0\) (Luôn đúng với mọi \(x\))

Do \(x^2\ge\forall x\)

\(\Rightarrow x^2+1>0\)

c.

Để cho biểu thức \(\sqrt{\dfrac{x-1}{2x-4}}\) xác định thì 

\(\Leftrightarrow\dfrac{x-1}{2x-4}\ge0\)

Trường hợp 1: \(\left\{{}\begin{matrix}x-1\ge0\\2x-4>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x>2\end{matrix}\right.\Leftrightarrow x>2\)

Trường hợp 2: \(\left\{{}\begin{matrix}x-1\le0\\2x-4< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x< 2\end{matrix}\right.\Leftrightarrow x\le1\)

a: \(\sqrt2<\sqrt3;\sqrt{11}<\sqrt{25}=5\)

Do đó: \(\sqrt2+\sqrt{11}<\sqrt3+5\)

b: \(1\frac12\cdot\sqrt{\frac{17}{2}}=\sqrt{\frac94\cdot\frac{17}{2}}=\sqrt{\frac{153}{8}}\)

\(\frac13\sqrt{19}=\sqrt{19\cdot\frac19}=\sqrt{\frac{19}{9}}\)

mà \(\frac{153}{8}>10>\frac{19}{9}\)

nên \(1\frac12\cdot\sqrt{\frac{17}{2}}>\frac13\sqrt{19}\)

c: \(\sqrt{21}>\sqrt{20};-\sqrt5>-\sqrt6\left(5<6\right)\)

Do đó: \(\sqrt{21}-\sqrt5>\sqrt{20}-\sqrt6\)

d: \(\frac14\cdot\sqrt{82}=\sqrt{\frac{82}{16}}\)

mà \(\frac{82}{16}>1>\frac17\)

nên \(\frac14\sqrt{82}>\sqrt{\frac17}\)