a)

$n_R = \dfrac{3,6.10^{23}}{6.10^{23}} = 0,6(mol)$
$\Rightarrow R = \dfrac{14,4}{0,6} = 24(Mg)$

b)

$n_{X_2} = \dfrac{6,72}{22,4} = 0,3(mol)$
$\RIghtarrow M_{X_2} =2X = \dfrac{21,3}{0,3} = 71$

Vậy X là $Cl_2$

\(\left(x+4\right)^2-81=0\Leftrightarrow\left(x+4\right)^2-9^2=0\)

\(\Leftrightarrow\left(x+4+9\right)\times\left(x+4-9\right)=0\)

\(\Leftrightarrow\left(x+13\right)\times\left(x-5\right)=0\)

\(\left[{}\begin{matrix}x+13=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=5\end{matrix}\right.\)

a, `5:y = 0,8`

`=> y = 5 : 0.8`

`=> y = 6,25`.

b, `72,9 xx 99 + 72 + 0,9`

`= 72,9 xx 99 + 72,9`

`= 72,9(99+1)`

`= 72,9 xx 100`

`= 7290`.

34,8 + 5 : y = 35,6  

5 : y = 35,6  - 34,8

5 : y = 0.8

y = 5 : 0.8

y = 6.25

72,9 x 99 + 72 + 0,9

=72,9 x 99 + (72 + 0,9)

=72,9 x 99 + 72,9

=72,9 x 99 + 72,9 x 1

=72,9 x  (99 + 1)

=7290

a) \(=63,52\)

b) \(=168,25\)

c) \(=83,835\)

d) \(=32,5\)

a= 63,52 b= 193,25 c= 83,835 d= 32,5

Bài 1:

a: ĐKXĐ: x<>0; x<>3; x<>1

\(A=\left(\frac{x-3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\right):\frac{2x-2}{x}\)

\(=\frac{\left(x-3\right)^2-x^2+9}{x\left(x-3\right)}:\frac{2\left(x-1\right)}{x}\)

\(=\frac{x^2-6x+9-x^2+9}{x\cdot\left(x-3\right)}:\frac{2\left(x-1\right)}{x}=\frac{-6\left(x-3\right)}{x\left(x-3\right)}:\frac{2\left(x-1\right)}{x}=\frac{-6}{x}\cdot\frac{x}{2\left(x-1\right)}=\frac{-3}{x-1}\)

b: Để A là số nguyên thì -3⋮x-1

=>x-1∈{1;-1;3;-3}

=>x∈{2;0;4;-2}

Kết hợp ĐKXĐ, ta được: x∈{2;4;-2}

Bài 2:

a: \(x^3-2x^2=x^2\cdot x-x^2\cdot2=x^2\left(x-2\right)\)

b: \(y^2-2y-x^2+1\)

\(=\left(y^2-2y+1\right)-x^2\)

\(=\left(y-1\right)^2-x^2=\left(y-1-x\right)\left(y-1+x\right)\)

c: \(\left(x+1\right)^2-25\)

=(x+1+5)(x+1-5)

=(x-4)(x+6)

a: ĐKXĐ: x>=0; x<>1

\(A=\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}}{\sqrt{x}-1}\right):\frac{2}{\sqrt{x}+1}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{\sqrt{x}+1}{2}\)

\(=\frac{x-\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}-1\right)}\cdot\frac12=\frac{-2\sqrt{x}}{2\left(\sqrt{x}-1\right)}=\frac{-\sqrt{x}}{\sqrt{x}-1}\)

b: A=2

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

=>\(2\sqrt{x}-2+\sqrt{x}=0\)

=>\(3\sqrt{x}-2=0\)

=>\(\sqrt{x}=\frac23\)

=>\(x=\frac49\) (nhận)

1111111

a )\(A=\frac{x^2+4x+4}{x^2-4}=\frac{\left(x+2\right)^2}{x^2-2^2}=\frac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}=\frac{x+2}{x-2}=\frac{5}{3}\)

<=> (x + 2).3 = (x - 2).5

<=> 3x + 6 = 5x - 10

<=> 3x - 5x = - 10 - 6

<=> - 2x = - 16

=> x = 8

b ) \(\frac{x+2}{x-2}=\frac{\left(x-2\right)+4}{x-2}=1+\frac{4}{x-2}\)

đến đây tự tìm đc 

Bài 2 lớp 8 ko làm đc thì đi chết đi