1. I have never talked to a foreigner before.
2. Long was sleepy, but he still stayed up late to watch the end of the game on TV.
3. Living in the countryside is what she likes.
4. Do you want a cup of coffee?
5. You should play sports.
6. You shouldn't play cards all night.
7. How many people are there in Vietnam?

I have never talked to foreigner before.

Long was sleepy, but he still stayed up late to watch the end of the game onTV.

Do you like a cup of coffe? 

you should go to play sport 

-you shoundn't play cards all night

- How many population in viet nam? 

**I'm planning to buy a new house. I 1) have been looking (look) for one for two months now. So far I 2) have looked (look) at ten houses, but I 3) have not found (not/find) one I like.

**My Spanish lessons are going very well. I 1) have been learning (learn) Spanish for five months now and I love it. 2) have already learned (already/learn) a lot.

**John 1) has been (be) very busy recently. He 2) has been painting (paint) the living room and the bedrooms, but he 3) has not started (not/start) painting the kitchen yet.

A

**I'm planning to buy a new house. I 1) have been looking (look) for one for two months now. So far I 2) have looked (look) at ten houses, but I 3) have not found (not/find) one I like.

B

**My Spanish lessons are going very well. I 1) have been learning (learn) Spanish for five months now and I love it. 2) have already learned (already/learn) a lot.

C

**John 1) has been (be) very busy recently. He 2) has been painting (paint) the living room and the bedrooms, but he 3) has not started (not/start) painting the kitchen yet.

Ta có: \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)

PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

____0,2__________0,1 (mol)

\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)

0,1_________________0,2 (mol)

\(\Rightarrow C_{M_{H_3PO_4}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)

Ta có: \(m_{ddH_3PO_4}=400.1,15=460\left(g\right)\)

\(\Rightarrow C\%_{H_3PO_4}=\dfrac{0,2.98}{460}.100\%\approx4,26\%\)

\(n\left(n+1\right)\left(2n+1\right)\\ =n\left(n+1\right)\left[2n-2+3\right]\\ =n\left(n+1\right)\left(2n-2\right)+3n\left(n+1\right)\\ =2\left(n-1\right)n\left(n+1\right)+3n\left(n+1\right)\)

Ta có:

`+)(n-1)n(n+1)` là tích của 3 số tự nhiên liên tiếp `=>(n-1)n(n+1)` chia hết cho 3 

`=>2(n-1)n(n+1)` chia hết cho 6 (1) 

`+)n(n+1)` là tích của 2 số tự nhiên liên tiếp `=>n(n+1)` chia hết cho 2

`=>3n(n+1)` chia hết cho 6 (2)

Từ (1) và (2) => `n(n+1)(2n+1)` chia hết cho 6 

\(a,\dfrac{xy^2}{xy+y}=\dfrac{xy^2}{y\left(x+1\right)}=\dfrac{xy}{x+1}\\ b,\dfrac{xy-y}{x}\ne\dfrac{xy-x}{y}\\ c,\dfrac{3ac}{a^3b}=\dfrac{3c}{a^2b}=\dfrac{6c}{2a^2b}\\ d,\dfrac{3ab-3b^2}{6b^2}=\dfrac{3b\left(a-b\right)}{6b^2}=\dfrac{a-b}{2b}\\ e,\dfrac{3x\left(x-y\right)^2}{9x^2\left(x-y\right)}=\dfrac{3x\left(x-y\right)}{9x^2}=\dfrac{x-y}{3x}\\ f,\dfrac{8-x^3}{x\left(x^2+2x+4\right)}=\dfrac{-\left(x^3-8\right)}{x\left(x^2+2x+4\right)}=\dfrac{-\left(x-2\right)\left(x^2+2x+4\right)}{x\left(x^2+2x+4\right)}=\dfrac{-\left(x-2\right)}{x}=\dfrac{x-2}{-x}\)

ĐK: `x>=0` 

Ta có:

\(B=\dfrac{5\sqrt{x}-1}{\sqrt{x}+1}\\ =\dfrac{\left(5\sqrt{x}+5\right)-6}{\sqrt{x}+1}=\dfrac{5\left(\sqrt{x}+1\right)-6}{\sqrt{x}+1}\\ =\dfrac{5\left(\sqrt{x}+1\right)}{\sqrt{x}+1}-\dfrac{6}{\sqrt{x}+1}\\ =5-\dfrac{6}{\sqrt{x}+1}\)

Vì: \(\sqrt{x}\ge0\forall x\)

\(=>\sqrt{x}+1\ge1\forall x=>\dfrac{6}{\sqrt{x}+1}\le6\\ =>5-\dfrac{6}{\sqrt{x}+1}\ge5-6=-1\)

Dấu "=" xảy ra: `x=0` 

Ta có pt hoành độ giao điểm là: 

\(-x^2=\left(2-m\right)x+m-3\\ \Leftrightarrow x^2+\left(2-m\right)x+m-3=0\)

Để pt có nghiệm phân biệt thì: 

\(\Delta=\left(2-m\right)^2-4\cdot1\cdot\left(m-3\right)\\ =4-4m+m^2-4m+12=m^2-8m+16=\left(m-4\right)^2>0\) 

`=>m-4<>0<=>m<>4` 

Ta có: `a+b+c=1+(2-m)+(m-3)=0`

\(=>x_1=1\)
Theo vi-ét ta có: \(x_1+x_2=m-2=>x_2=m-2-x_2=m-2-1=m-3\) 

\(\left|x_1\right|+x_2^2=2\\ =>1+\left(m-3\right)^2=2\\< =>\left(m-3\right)^2=2-1=1\\ < =>\left[{}\begin{matrix}m-3=1\\m-3=-1\end{matrix}\right.\\ < =>\left[{}\begin{matrix}m=1+3=4\left(ktm\right)\\m=-1+3=2\left(tm\right)\end{matrix}\right.\)

Vậy: ... 

Bài 22: 

\(a^6+b^6\\ =\left(a^2\right)^3+\left(b^2\right)^3\\ =\left(a^2+b^2\right)\left[\left(a^2\right)^2-a^2b^2+\left(b^2\right)^2\right]\\ =\left(a^2+b^2\right)\left[\left(a^4+2a^2b^2+b^4\right)-3a^2b^2\right]\\ =\left(a^2+b^2\right)\left[\left(a^2+b^2\right)^2-3a^2b^2\right]\) 

Bài 24: 

a) Ta có:

`(a+b)^2=2(a^2+b^2)`

`<=>a^2+2ab+b^2=2a^2+2b^2`

`<=>a^2-2ab+b^2=0`

`<=>(a-b)^2=0`

`<=>a-b=0`

`<=>a=b`

b) Ta có: 

`a^2+b^2+c^2=ab+bc+ca`

`<=>2a^2+2b^2+2c^2=2ab+2bc+2ca`

`<=>(a^2-2ab+b^2)+(a^2-2ca+c^2)+(b^2-2bc+c^2)=0`

`<=>(a-b)^2+(a-c)^2+(b-c)^2=0`

`<=>a-b=0` và `a-c=0` và `b-c=0`

`<=>a=b=c` 

c) Ta có:

`(a+b+c)^2=3(ab+bc+bc)`

`<=>a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca`

`<=>a^2+b^2+c^2=ab+bc+ca`

`<=>(a-b)^2+(b-c)^2+(a-c)^2=0`

`<=>a=b=c`

ΔAHB vuông tại H

=>\(HA^2+HB^2=AB^2\)

=>\(HB=\sqrt{15^2-9^2}=12\left(cm\right)\)

ΔABC cân tại A

mà AH là đường cao

nên H là trung điểm của BC

=>\(BC=2\cdot BH=24\left(cm\right)\)

Xét ΔABC có \(cosBAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\dfrac{15^2+15^2-24^2}{2\cdot15\cdot15}=\dfrac{-7}{25}\)

=>\(sinBAC=\sqrt{1-\left(-\dfrac{7}{25}\right)^2}=\sqrt{1-\dfrac{49}{625}}=\dfrac{24}{25}\)

Xét ΔABC có \(\dfrac{BC}{sinBAC}=2R\)

=>\(2R=24:\dfrac{24}{25}=25\)

=>R=12,5(cm)

\(\left[19\left(4\cdot2^3+18\right)-9\cdot50\right]:5^2\\ =\left[19\left(4\cdot8+18\right)-450\right]:25\\ =\left[19\cdot50-450\right]:25\\ =\left[950-450\right]:25\\ =500:25\\ =20\)

\(\left[19\left(4\cdot2^3+18\right)-9\cdot50\right]:5^2\\ =\left[19\left(4\cdot8+18\right)-9\cdot50\right]:25\\ =\left[19\left(32+18\right)-9\cdot50\right]:25\\ =\left(19\cdot50-9\cdot50\right):25\\ =50\cdot\left(19-9\right):25\\ =\left(50:25\right)\cdot10\\ =2\cdot10\\ =20\)