12 This matter to which you should pay attention is of great importance

13 The woman to whom you gave place on the bus lives next door to me

Bài 14:
a) xx' // yy' vì :
Góc x'An = góc ABy = 60 độ ( 2 góc so le trong)
=> xx' // yy'
b) Ax // By vì:
Góc zBy = góc zAx = 50 độ ( 2 góc đồng vị )
=> Ax // By

1: \(x^2-\left(m+1\right)x-2023=0\)

a=1; b=-(m+1); c=-2023

Vì \(a\cdot c=-2023< 0\)

nên phương trình luôn có hai nghiệm phân biệt

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-\left[-\left(m+1\right)\right]}{1}=m+1\\x_2\cdot x_1=\dfrac{c}{a}=-\dfrac{2023}{1}=-2023\end{matrix}\right.\)

\(\dfrac{1}{x_1-2023}+\dfrac{1}{x_2-2023}=1\)

=>\(\dfrac{x_2-2023+x_1-2023}{\left(x_1-2023\right)\left(x_2-2023\right)}=1\)

=>\(x_2+x_1-4046=\left(x_1-2023\right)\left(x_2-2023\right)\)

=>\(m+1-4046=x_1x_2-2023\left(x_1+x_2\right)+2023^2\)

=>\(m-4045=-2023-2023\left(m+1\right)+2023^2\)

=>\(m-4045=-2023-2023m-2023+2023^2\)

=>\(2024m=4092528\)

=>\(m=\dfrac{4092528}{2024}=2022\)

a) Gọi $n_{Mg} = x(mol) ; n_{Al} = y(mol) \Rightarrow 24x + 27y = 25,5(1)$

$n_{N_2} = \dfrac{5,6}{22,4} = 0,25(mol)$

Bảo toàn electron : $2n_{Mg} + 3n_{Al} = 10n_{N_2} \Rightarrow 2x + 3y = 0,25.10(2)$

Từ (1)(2) suy ra :  x = 0,5 ; y = 0,5

$m_{Mg} = 0,5.24 = 12(gam)$
$m_{Al} = 0,5.27 = 13,5(gam)$

b) $n_{Mg(OH)_2} = x = 0,5(mol) ; n_{Al(OH)_3} = y = 0,5(mol)$

$\Rightarrow m_{Kết\ tủa} = 0,5.58 + 0,5.78 = 68(gam)$