Gọi CTHH : C_xH_y 

\(n_{CO_2}=0,3\left(mol\right);n_{H_2O}=0.4\left(mol\right)\)

PT : 4C_xH_y + (4x + y)O₂ \(\rightarrow\) 4xCO₂ + yH₂O 

\(\Rightarrow\dfrac{4x}{2y}=\dfrac{0,3}{0,4}\Leftrightarrow\dfrac{x}{y}=\dfrac{3}{8}\Rightarrow x=3;y=8\) =

CTHH C₃H₈

b)Công thức tổng quát hợp chất : C_nH_{2n + 2}

\(\Rightarrow\) Phản ứng cộng Cl 

C₃H₈ + Cl₂ \(\Rightarrow C_3H_7Cl+HCl\)

1) will arrive

2) has been sung

3) turning 

4) spelt 

5) have made

6) increased

7) are planted

8) wrote 

9) knocked

10) stop 

1 . with

2) into 

3) from

4) up 

5) of 

6)  from

Ta có x³ + y³ \(\le8-6xy\)

<=> (x + y)³ - 8 - 3xy(x + y) + 6xy \(\le0\)

<=> (x + y - 2)[(x + y)² + 2(x + y) + 4)] - 3xy(x + y - 2) \(\le0\)

<=> (x + y - 2)(x² - xy + y² + 2x + 2y + 4) \(\le0\)

<=> x + y - 2 \(\le\) 0 (Vì x² - xy + y² + 2x + 2y + 4 > 0 \(\forall xy\)) 

<=> x + y \(\le2\)

Khi đó 

P = \(\dfrac{1}{x^2+y^2}+\dfrac{2}{xy}=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{3}{2xy}\ge\dfrac{4}{x^2+y^2+2xy}+\dfrac{3}{2xy}\)

\(\dfrac{4}{\left(x+y\right)^2}+\dfrac{3}{2xy}\ge\dfrac{4}{\left(x+y\right)^2}+\dfrac{6}{\left(x+y\right)^2}=\dfrac{10}{\left(x+y\right)^2}\ge\dfrac{10}{2^2}=\dfrac{5}{2}\)

Dấu "=" xảy ra <=> x = y = 1 

ĐKXĐ : \(\left\{{}\begin{matrix}x\ge2\\y\ge-\dfrac{1}{2}\end{matrix}\right.\)

Ta có \(\left\{{}\begin{matrix}x+2y-1-2\sqrt{2xy+x-4y-2}=0\\\sqrt{x-2}+3\sqrt{2y+1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)+\left(2y+1\right)-2\sqrt{\left(x-2\right)\left(2y+1\right)}=0\\\sqrt{x-2}+3\sqrt{2y+1}=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x-2}-\sqrt{2y+1}\right)^2=0\\\sqrt{x-2}+3\sqrt{2y+1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=\sqrt{2y+1}\\4\sqrt{2y+1}=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=\sqrt{2y+1}\\2y+1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=\sqrt{2y+1}\\y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=0\end{matrix}\right.\)

Gọi số học sinh nam x ; số học sinh nữ y 

Theo đề ra => HPT : \(\left\{{}\begin{matrix}x+y=24\\2x+3y=62\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=14\end{matrix}\right.\)

(p) đi qua A(-1;2) 

=> 2 = (m - 2).(-1)² 

<=> m - 2 = 2

<=> m = 4

Vậy m = 4 thì (p) đi qua A(-1 ; 2) 

a) Vì khối lượng nước đun như nhau ; không tính hao phí 

=> Q_i = Q_tp không đổi 

=> Q₁ = Q₂ = Q_nt = Q_ss 

=> Q₁ = Q₂ 

<=> \(\dfrac{U^2}{R_1}.t_1=\dfrac{U_2}{R_2}.t_2\Leftrightarrow t_2=\dfrac{t_1}{R_1}.R_2=\dfrac{10}{4}.6=15\left(\text{phút}\right)\)

b) R_tđ = R₁ + R₂ = 4 + 6 = 10 \(\Omega\)

Ta có Q₁ = Q_nt 

=> \(\dfrac{U^2}{R_1}.t_1=\dfrac{U^2}{R_{tđ}}.t_{nt}\Leftrightarrow t_{nt}=\dfrac{t_1.R_{tđ}}{R_1}=\dfrac{10.10}{4}=25\left(\text{phút}\right)\)

c) \(R_{tđ}=\dfrac{R_1.R_2}{R_1+R_2}=\dfrac{4.6}{4+6}=2,4\left(\Omega\right)\)

mà Q₁ = Q_ss 

<=> \(\dfrac{U^2}{R_1}.t_1=\dfrac{U^2}{R_{tđ}}.t_{ss}\Leftrightarrow t_{ss}=\dfrac{t_1.R_{tđ}}{R_1}=\dfrac{10.2,4}{4}=6\left(\text{phút}\right)\)

ĐKXĐ : x \(\ge\dfrac{2}{3}\)

Ta có \(\left(x-1\right)\left(x-2\right)=\sqrt{3x-2}\left(1-x\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\sqrt{3x-2}=2-x\end{matrix}\right.\)

Khi x - 1 = 0 <=> x = 1 (tm)

Khi \(\sqrt{3x-2}=2-x\)

<=> \(\left\{{}\begin{matrix}3x-2=x^2-4x+4\\\dfrac{2}{3}\le x\le2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-7x+6=0\\\dfrac{2}{3}\le x\le2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(x-6\right)=0\\\dfrac{2}{3}\le x\le2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=6\end{matrix}\right.\\\dfrac{2}{3}\le x\le2\end{matrix}\right.\Leftrightarrow x=1\)

Vậy phương trình 1 nghiêm \(x=1\)

S = 1² + 2² + 3² + ... + 2018²

= 1.2 + 2.3 + 3.4 + ... + 2018.2019 - (1 + 2 + 3 + 4 + ... + 2018) 

= \(\dfrac{2018.2019.2020}{3}-\dfrac{2018.2019}{2}=1009.673.367.11\)

=> S không là số chính phươn