Ta có x³ - y³ + z³ + 3xyz

= (x - y)³ + 3xy(x - y) + z³ + 3xyz

= [(x - y)³ + z³] + [3xy(x - y) + 3xyz]

= (x - y + z)[(x - y)² - (x - y)z + z²] + 3xy(x - y + z)

= (x - y + z)[x² - 2xy + y² - xz + yz + z²] + 3xy(x - y + z)

= (x - y + z)(x² + y² + z² + xy - xz + yz) 

= 2(x² + y² + z² + xy - xz + yz) (vì x - y+ z = 2)

Lại có (x + y)² + (y + z)² + (z - x)²

= x² + 2xy + y² + y² + 2yz + z² + z² - 2xz + z²

= 2x² + 2y² + 2z² + 2xy + 2yz - 2xz

= 2(x² + y² + z² + xy - xz + yz)

Khi đó P = \(\frac{2\left(x^2+y^2+z^2+xy-xz+yz\right)}{2\left(x^2+y^2+z^2+xy-xz+yz\right)}=1\)

\(ĐXKĐ:\hept{\begin{cases}x\ne0\\y\ne0\end{cases}}\)

\(x^2+\frac{1}{x^2}+y^2+\frac{1}{y^2}=4\)

\(\Leftrightarrow x^2+\frac{1}{x^2}+y^2+\frac{1}{y^2}-4=0\)

\(\Leftrightarrow\left(x^2-2+\frac{1}{x^2}\right)+\left(y^2-2+\frac{1}{y^2}\right)=0\)

\(\Leftrightarrow\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2=0\)

Vì \(\left(x-\frac{1}{x}\right)^2\ge0\), \(\left(y-\frac{1}{y}\right)^2\ge0\)

\(\Rightarrow\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2\ge0\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{x}=0\\y-\frac{1}{y}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{x}\\y=\frac{1}{y}\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2=1\\y^2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\pm1\\y=\pm1\end{cases}}\)

Vậy nghiệm của phương trình là \(\left(x;y\right)=\left(-1;-1\right),\left(-1;1\right),\left(1;-1\right),\left(1;1\right)\)

Áp dụng BĐT AM - GM cho 2 số 

\(x^2+\frac{1}{x^2}\ge2\sqrt{x^2\frac{1}{x^2}}=2\)

\(y^2+\frac{1}{y^2}\ge2\sqrt{y^2\frac{1}{y^2}}=2\)

Cộng vế với vế của BĐT : 

\(x^2+\frac{1}{x^2}+y^2+\frac{1}{y^2}\ge4\)

Dấu ''='' xxảy ra <=> \(x=y=\pm1\)

P/s : AM - GM cho 4 số đều được =)) 

ĐKXĐ : x² - 6x + 9 \(\ne\)0

<=> x \(\ne\)3

a) A = 0

=> 3x² - 11x + 6 = 0

<=> 3x² - 9x - 2x + 6 = 0

<=> 3x(x - 3) - 2(x - 3) = 0

<=> (3x - 2)(x - 3) = 0

<=> \(\orbr{\begin{cases}x=\frac{2}{3}\left(tm\right)\\x=3\left(\text{loại}\right)\end{cases}}\)

Vậy x = 2/3 thì A = 0

b) Ta có A = \(\frac{3x^2-11x+6}{x^2-6x+9}=3+\frac{7x-21}{x^2-6x+9}=3+\frac{7}{x-3}\)

Để : A \(\inℤ\Leftrightarrow7⋮x-3\Leftrightarrow x-3\inƯ\left(7\right)\Leftrightarrow x-3\in\left\{1;7;-1;-7\right\}\)

Lập bảng xét các trường hợp 

Vậy \(x\in\left\{4;10;2;-4\right\}\)thì A \(\inℤ\)

10 chất khí ở nhiệt độ phòng là :

- Hydro \(H_2\)

- Oxi \(O_2\)

- Ozon \(O_3\)

- Nitơ \(N_2\)

- Cacbon monoxit \(CO\)

- Cacbon đioxit \(CO_2\)

- Metan \(CH_4\)

- Etilen \(C_2H_4\)

- Axetilen \(C_2H_2\)

- Bis(triflometyl)peroxit \(C_2F_6O_2\)

Hydro

Oxi

Nitơ

Clo

Flo

Neon

Argon

cacbon dioxit

cacbon monoxit

Chào em, em tham khảo nhé!

1. She always (remember) remembers my birthday.

2. How long (know) have you known Simon?

3. Can you hear those girls? What (they/talk) are they talking this exam?

4. How many times (you/talk) have you talked this exam?

5. He (eat) eats six bars of chocolate today!

6. When I'm in Paris, I (usually/stay) usually stay in Hotel du Pont, but this time I (stay) am staying in the more expensive Hotel Notre Dame.

7. (They/arrive) Have they arrived already?

8. My father (be) is an engineer, but he (not/work) is not working right now.

9. My parents (live) live in Sydney. Where (your parents/live) do your parent live?

Chúc em học tốt và có những trải nghiệm tuyệt vời tại olm.vn!

Put the verbs in brackets in correct form.

1. She always (remember) .....remembers......... my birthday.

2. How long (know) .......have you known..... Simon?

3. Can you hear those girls? What (they/talk) ........are they talking..... this exam?

4. How many times (you/talk) ......do you talk....... this exam?

5. He (eat) ate six bars of chcolates today!

6. When I'm in Paris, I (usually/stay) ......usually stayed....... in Hotel du Pont, but this time I (stay) .......am staying.... in the more expensive Hotel Notre Dame.

7. (They/arrive) ......Have they arrived........ already?

8. My father (be) .......is....... an engineer, but he (not/work) ......isn't working......... right now.

9. My parents (live) ........live...... in Sydney. Where (your parents/live) .....do your parents live............?

-Gọi công thức là C_xH_y

-Ta cos:

x:y=$\frac{m_C}{12}:\frac{m_H}{1}=\frac{4}{12}:\frac{1}{1}=1:3$

$\to$Công thức nguyên (CH₃)_n

-Ta có: 15n=30$\to$n=2$\to$CTPT: C₂H₆

Chọn A nha bn

#H

Đáp án A nha !

too bad : quá tệ

( bạn nên viết là) => very bad : rất tệ 

Như thế này : Their consciousness is very bad .

k nha FPT

Dịch sang tiếng anh : Ý thức của họ rất tệ.

Đáp án : Their consciousness is too bad   

Áp dụng định lý PYTAGO vào tam giác ABC có

BC^2=AB^2+AC^2= 9^2+12^2=225

=>BC= 15

Sabc= 1/2.AB.AC = 54 mà Sabc = 1/2.AH.BC 

=>1/2.AH = Sabc: BC = 3.6=> AH =7,2

mình ghi thiếu đề ạ:

Cho P=...  

\(P=\left(\frac{x^2+x-4}{x^2-2x-3}\right):\left(1-\frac{x-3}{x-2}\right)\)

\(=\frac{x^2+x-4}{\left(x+1\right)\left(x-3\right)}:\left(\frac{x-2-x+3}{x-2}\right)\)

\(=\frac{x^2+x-4}{\left(x+1\right)\left(x-3\right)}.\frac{x-2}{1}=\frac{\left(x^2+x-4\right)\left(x-2\right)}{\left(x+1\right)\left(x-3\right)}\)