Câu 1: 

a: ĐKXĐ: \(2cos2x+1< >0\)

=>cos2x<>-1/2

\(\Leftrightarrow\left\{{}\begin{matrix}2x\ne\dfrac{2}{3}\Pi+k2\Pi\\2x\ne-\dfrac{2}{3}\Pi+k2\Pi\end{matrix}\right.\Leftrightarrow x\ne\pm\dfrac{1}{3}\Pi+k\Pi\)

b: ĐKXĐ: \(cotx< >1\)

=>x<>pi/4+kpi

1: =-3x-2x=-5x

2: \(=x-4+\left|x-4\right|=x-4+x-4=2x-8\)

3: \(=\dfrac{3}{2\left(2x-1\right)}\cdot\left(2x-1\right)=\dfrac{3}{2}\)

4: \(=\dfrac{2}{2a-1}\cdot\left(1-2a\right)=-2\)

a) Gọi n là hóa trị của M

$2M +2nHCl \to 2MCl_n + nH_2$

Theo PTHH : $n_M = \dfrac{2}{n}n_{H_2} = \dfrac{1}{n}(mol)$

$\Rightarrow \dfrac{1}{n}.M = 9$

$\Rightarrow M = 9n$

Với n = 3 thì M = 27(Al)

b) $n_{HCl} = 2n_{H_2} = 1(mol)$
$C_{M_{HCl}} = \dfrac{1}{0,5} = 2M$

$C_{M_{AlCl_3}} = \dfrac{\dfrac{9}{27}}{0,5} = 0,67M$

c) $AlCl_3 + 3NaOH \to Al(OH)_3 + 3NaCl$
$n_{NaOH} = 3n_{AlCl_3} = 1(mol)$
$V_{dd\ NaOH} = \dfrac{1}{0,5} = 2(lít)$

a) $A(OH)_2 + 2HCl \to ACl_2 + 2H_2O$

$n_{HCl} = \dfrac{146.10\%}{36,5} = 0,4(mol)$

$n_{A(OH)_2} = \dfrac{1}{2}n_{HCl} = 0,2(mol)$
$\Rightarrow M_{A(OH)_2} = A + 3 4= \dfrac{18}{0,2} = 90$

$\Rightarrow A = 56(Fe)$

Vậy CT của hidroxit kim loại A là $Fe(OH)_2$
b) $m_{dd\ sau\ pư} = 18 + 146 =164(gam)$
$C\%_{FeCl_2} = \dfrac{0,2.127}{164}.100\% = 15,49\%$

1: =>(5x-3)(5x+3)=0

=>x=3/5 hoặc x=-3/5

2: =>x²+8x+16-x²+1=16

=>8x+17=16

=>x=-1/8

1, \(25x^2-9=0\)

\(\Leftrightarrow25x^2=9\) \(\Leftrightarrow\left(5x\right)^2=3^2=\left(-3\right)^2\)\(\Leftrightarrow\left[{}\begin{matrix}5x=3\\5x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

2, \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(\Leftrightarrow\left(x+4\right)^2-\left(x^2-1\right)=16\)

\(\Leftrightarrow x^2+8x+16-x^2+1=16\)

\(\Leftrightarrow8x+17=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)

3, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

\(\Leftrightarrow5x^2+2x+10-5x^2+245=0\)

\(\Leftrightarrow2x+10+245=0\Leftrightarrow2x=-255\Leftrightarrow x=-\dfrac{255}{2}\)