PT bảo toàn khối lượng : 

m_a + m_b = m_c + m_d

\(A=3-x^2+2x-\left|y-3\right|=-\left(x^2-2x+1\right)+4-\left|y-3\right|=-\left[\left(x-1\right)^2+\left|y-3\right|\right]+4\)

Mà ta luôn có : \(\begin{cases}\left(x-1\right)^2\ge0\\\left|y-3\right|\ge0\end{cases}\) \(\Rightarrow\left(x-1\right)^2+\left|y-3\right|\ge0\)

\(\Rightarrow-\left[\left(x-1\right)^2+\left|y-3\right|\right]\le0\Rightarrow A\le4\)

Vậy Max A = 4 <=> x = 1 , y = 3

 ĐỒNG Ý VỚI Đinh Tuấn Việt

1/ Ta có \(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)

\(\Leftrightarrow\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}=\frac{abz-acy+bcx-abz+acy-bcx}{a^2+b^2+c^2}=0\)

\(\Rightarrow bz-cy=cx-az=ay-bx=0\Leftrightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)

2/ Giả sử \(a>b\Rightarrow\frac{a}{b}>1\)

Ta sẽ chứng minh \(\frac{a}{b}>\frac{a+2017}{b+2017}\)  . Thật vậy : \(\frac{a}{b}>\frac{a+2017}{b+2017}\Leftrightarrow ab+2017a>ab+2017b\Leftrightarrow a>b\) luôn đúng

Giả sử \(a< b\) thì \(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+2017}{b+2017}\) . Thật vậy : 

\(\frac{a}{b}< \frac{a+2017}{b+2017}\Rightarrow ab+2017a< ab+2017b\Leftrightarrow a< b\) luôn đúng

Giả sử \(a=b\Leftrightarrow\frac{a}{b}=1=\frac{2017}{2017}=\frac{a+2017}{b+2017}\)

1. That's my dog. Its name's Lulu

2. He's a doctor. His name's Tom

3. We are in our living room now

4. Are these your books? No. Those are their books

5. Are they Tom and Mary? No, they aren't. Their names are John and Jane

6. Is this your eraser, Lan? Yes, it's my eraser

7. This is my sister. Her name is Minh

8. Lan is here. This is her bag

nMg = 2,4/24 = 0,1 (mol)

a/                    Mg + H2SO4 ------> MgSO4 + H2

b/   Từ PTHH ta suy ra nH2 = nMg = 0,1 (mol)

Suy ra \(V_{H_2}=22,4\times0,1=2,24\left(l\right)\)

c/ Từ PTHH suy ra nH2SO4 = nMg = 0,1 mol

Suy ra \(C_{M_{H2SO4}}=\frac{0,1}{\frac{200}{1000}}=0,5M\)

nếu dịch chuyển dấu phẩy của số trừ sang phải 1 bên thì số đó tăng lên 10 lần

khi đó hiệu giảm xuống 9 lần số trừ

9 lần số trừ là:70,78-30,64=40,14

số trừ là:40,14:9=4,46

số bị trừ là:70,78+4,46=75,24

Ta chứng minh được \(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n}\) với mọi n là số tự nhiên lớn hơn 0

Đặt \(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}\)

Ta có \(2A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{79}+\sqrt{80}}>\)

\(>\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\)

\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\sqrt{5}-\sqrt{4}+...+\sqrt{80}-\sqrt{79}+\sqrt{81}-\sqrt{80}\)

\(=\sqrt{81}-\sqrt{1}=8\)

\(\Rightarrow2A>8\Rightarrow A>4\)

a/ Giả sử \(x^4+2x^3+3x^2+ax+b=\left(x^2+cx+d\right)^2\)

\(\Leftrightarrow x^4+2x^3+3x^2+ax+b=x^4+c^2x^2+d^2+2x^3c+2xcd+2dx^2\)

\(\Leftrightarrow x^3\left(2-2c\right)+x^2\left(3-c^2-2d\right)+x\left(a-2cd\right)+\left(b-d^2\right)=0\)

Áp dụng hệ số bất định, ta có : 

\(\begin{cases}2-2c=0\\3-c^2-2d=0\\a-2cd=0\\b-d^2=0\end{cases}\) \(\Leftrightarrow\begin{cases}a=2\\b=1\\c=1\\d=1\end{cases}\)

Vậy : \(x^4+2x^3+3x^2+2x+1=\left(x^2+x+1\right)^2\)

b/ Tương tự

1/ +) They were happy yesterday

-) They weren't happy yesterday

?) Were they happy yesterday?

2/ +) We went swimming last summer

-) We didn't go swimming last summer

?) Did you go swimming last summer?

3/ +) The students were in the school yard 2 hours ago

-) The students weren't in the school yard 2 hours ago

?) Were the students in the school yard 2 hours ago?

4/ +) Lan bought a new lantern last night

-) Lan didn't buy a new lantern last night

?) Did Lan buy a new lantern last night?