$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
$n_{CO_2} = n_{CaCO_3} = \dfrac{15}{100} = 0,15(mol)$
$n_{NaOH} = \dfrac{80.12,5\%}{40} = 0,25(mol)$

Ta thấy \(1< \dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,25}{0,15}=1,67< 2\) nên X gồm $Na_2CO_3(a\ mol);NaHCO_3( b\ mol)$

$2NaOH + CO_2 \to Na_2CO_3 + H_2O$
$NaOH + CO_2 \to NaHCO_3$

Ta có : $2a + b = 0,25$ và $a + b = 0,15$
Suy ra : a = 0,1 ; b = 0,05

$m_{dd\ X} = m_{CO_2} + m_{dd\ NaOH} = 86,6(gam)$
$C\%_{Na_2CO_3} = \dfrac{0,1.106}{86,6}.100\% = 12,24\%$

$C\%_{NaHCO_3} = \dfrac{0,05.84}{86,6}.100\% = 4,85\%$

Giải bài này giúp mk ik!!!!!!

\(\sqrt{x+1}\) + \(\sqrt{x+6}\) = 5

\(\sqrt{\left(x+1\right)\left(x-1\right)}\) + \(\sqrt{\left(x+6\right)\left(x-6\right)}\) = 5

x - 1 + x - 6 = 5

2x - 7 = 5

x = 6

\(ĐKXĐ:\left\{{}\begin{matrix}x+1\ge0\\x+6\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ge-6\end{matrix}\right.\Leftrightarrow x\ge-1\)

- Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{x+6}=b\end{matrix}\right.\left(a\ge0,b\ge\sqrt{5}\right)\)

\(\Rightarrow b^2-a^2=\left(\sqrt{x+6}\right)^2-\left(\sqrt{x+1}\right)=x+6-\left(x+1\right)=5\)

- Ta có hệ phương trình:

\(\left\{{}\begin{matrix}b+a=5\\b^2-a^2=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b+a=5\\\left(b+a\right)\left(b-a\right)=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b+a=5\\5\left(b-a\right)=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b+a=5\\b-a=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b+a=5\\2b=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3+a=5\\b=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x+1}=2\\\sqrt{x+6}=3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+1=4\\x+6=9\end{matrix}\right.\)

\(\Leftrightarrow x=3\left(nhận\right)\)

- Vậy tập nghiệm của pt trên là \(S=\left\{3\right\}\)

BH=CH=2,3 (trong tg cân đường cao đồng thời là đường trung trực) \(\Rightarrow BC=2BH=4,6\)

\(\widehat{BAH}=\widehat{CAH}=\dfrac{\widehat{A}}{2}=25^o\) (trong tg cân đường cao đồng thời là đường phân giác)

\(AB=AC\) (cạnh bên tg cân)

Xét tg vuông ABH có

\(\sin\widehat{BAH}=\dfrac{BH}{AC}\Rightarrow AB=AC=\dfrac{BH}{\sin\widehat{BAH}}=\dfrac{2,3}{\sin25^o}\)

\(C_{ABC}=AB+AC+BC=\dfrac{4,6}{\sin25^o}+4,6\)

ĐKXĐ: x>=-1

Ta có: \(\sqrt{x+1}+\sqrt{x+6}=5\)

=>\(\sqrt{x+1}-2+\sqrt{x+6}-3=5-5=0\)

=>\(\frac{x+1-4}{\sqrt{x+1}+2}+\frac{x+6-9}{\sqrt{x+6}+9}=0\)

=>\(\left(x-3\right)\left(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+9}\right)=0\)

=>x-3=0

=>x=3(nhận)

ai giải hộ mk cái đi!!

Đặt \(A=\sqrt{2003}+\sqrt{2005};B=2\sqrt{2004}\)

\(A^2=2003+2005+2\sqrt{\left(2003.2005\right)}\)

\(=4008+2\sqrt{\left[2004-1\left(2004+1\right)\right]}\)

\(=3008+2\sqrt{\left(2004^2-1\right)}< 2.2004+2\sqrt{\left(2004^2\right)}=4.2004=B^2\\ \Rightarrow A< B\)

\(\dfrac{x-1}{5}+\dfrac{x+1}{7}+\left(x-1\right)\left(x+1\right)=\left(x+1\right)^2\)

\(\Leftrightarrow\dfrac{7\left(x-1\right)+5\left(x+1\right)}{35}+x^2-1=x^2+2x+1\)

\(\Leftrightarrow\dfrac{12x-2}{35}=2x+2\)

\(\Leftrightarrow\dfrac{6x-1}{35}=x+1\)

\(\Leftrightarrow35x+35=6x-1\)

\(\Leftrightarrow x=-\dfrac{36}{29}\)

Ta có: \(\frac{x-1}{5}+\frac{x+1}{7}+\left(x-1\right)\left(x+1\right)=\left(x+1\right)^2\)

=>\(\frac{7\cdot\left(x-1\right)+5\left(x+1\right)}{35}+x^2-1-x^2-2x-1=0\)

=>\(\frac{12x-2}{35}-2x-2=0\)

=>12x-2+35(-2x-2)=0

=>12x-2-70x-70=0

=>-58x=72

=>\(x=-\frac{72}{58}=-\frac{36}{29}\)

b) Before Sue left the house, she had checked that she had her keys.

c) While Mark was parking his car, he noticed the wing-mirror was broken.

d) After Julia had cleaned the house, she fell asleep on the sofa.

e) Before Brian bought a new television, he had checked all the prices.

f) While Alan was skiing in Switzerland, she met his old friend, Ken.

g) After Kate had taken 2 aspirins, she felt a lot better.

h) Before Sheila went out for the evening, she had washed her hair.

mik cũng ko chắc đâu nếu sai thì thôi nhé:
 

ĐKXĐ : $-x^2+6x-9\ge 0$

$\iff$$-\left(-x^2+6x-9\right)\le 0$

$\iff$$x^2-6x+9\le 0$

$\iff$${\left(x-3\right)}^2\le 0$

Mà $\left(x-3\right)\ge 0$

Suy ra : ${\left(x-3\right)}^2=0$

$\iff$

ĐKXĐ: \(^{x^2}\)- 6x + 9 ≥0 với mọi x