2H₂+O₂-to>2H₂O

0,5----0,25------0,5 mol

n H2=\(\dfrac{11,2}{22,4}\)=0.5 mol

n O2=\(\dfrac{8}{22,4}\)=\(\dfrac{5}{14}\) mol

=>O2 dư

=>VH2O=0,5.18.1=9 l

=>m O2 dư =(\(\dfrac{5}{14}\)-0,25).32=3,428g

\(A=-\left(x^2-4x+4\right)-\left(y^2+4y+4\right)+10\\ A=-\left(x-2\right)^2-\left(y+2\right)^2+10\le10\\ A_{max}=10\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)

Nhìu thiệt á, tách bớt ra đi nhé!

e: Ta có: \(\left(x+1\right)\left(x+2\right)=444222\)

\(\Leftrightarrow x^2+3x-444220=0\)

\(\text{Δ}=3^2-4\cdot1\cdot\left(-444220\right)=1776889\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là 

\(\left\{{}\begin{matrix}x_1=\dfrac{-3-1333}{2}=-668\\x_2=\dfrac{-3+1333}{2}=665\end{matrix}\right.\)

IX
2. j
3. i
4. f
5. c
6. a
7. h
8. e
9. g
10. d

XI
2. part => parts
3. a => an
4. a => an
5. a => the
6. are => will be (không chắc lắm)
7. taking => take
8. are => is

C. 
Bài 1
1. C
2. B
3. C
4. B

(Nên double-check trước khi chép)
 

A = \(\dfrac{4\sqrt{x}+9}{2\sqrt{x}+1}\)

Mà \(4\sqrt{x}+9>0\)

\(2\sqrt{x}+1>0\)

=> A > 0

A = \(\dfrac{2\left(2\sqrt{x}+1\right)+7}{2\sqrt{x}+1}\) = \(2+\dfrac{7}{2\sqrt{x}+1}\)

Mà \(2\sqrt{x}+1\ge1< =>\dfrac{7}{2\sqrt{x}+1}\le7\)

<=> \(A\le9\)

<=> 0 < A \(\le9\)

Mà A thuộc Z

<=> A \(\in\){1;2;3;4;5;6;7;8;9}

Đến đây bn thay A vào để tìm x nhé

A = \(\dfrac{2\left(2\sqrt{x}+1\right)+7}{2\sqrt{x}+1}=2+\dfrac{7}{2\sqrt{x}+1}\)

Mà \(2\sqrt{x}+1>0< =>\dfrac{7}{2\sqrt{x}+1}>0\)

<=> A > 2

Có \(2\sqrt{x}+1\ge1< =>\dfrac{7}{2\sqrt{x}+1}\le7\)

<=> \(A\le9\)

<=> 2 < A \(\le9\)

Mà A thuộc Z

<=> \(A\in\left\{3;4;5;6;7;8;9\right\}\)

Đến đây bn thay A vào để tìm x nhé

A = \(\dfrac{6\sqrt{x}+8}{3\sqrt{x}+2}=2+\dfrac{4}{3\sqrt{x}+2}\)

Có \(3\sqrt{x}+2>0< =>\dfrac{4}{3\sqrt{x}+2}>0\) <=> A > 2

Có: \(3\sqrt{x}+2\ge2< =>\dfrac{4}{3\sqrt{x}+2}\le2\) <=> A \(\le4\)

<=> 2 < A \(\le4\)

Mà A nguyên

<=> \(\left[{}\begin{matrix}A=3\\A=4\end{matrix}\right.\)

TH1: A = 3

<=> \(\dfrac{4}{3\sqrt{x}+2}=1\)

<=> \(3\sqrt{x}+2=4< =>x=\dfrac{4}{9}\)

TH2: A = 4

<=> \(\dfrac{4}{3\sqrt{x}+2}=2< =>3\sqrt{x}+2=2< =>x=0\)

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