Dear Tuan,

Thank you very much for the present you gave me on my birthday last week. When I opened it, I was suprised because it was an English-Vietnamese dictionary. "Great!" I shouted like that. The dictionary was nice with blue - my favorite color and it helped me a lot in learning English. You know me very well when giving me such a useful gift. Again, thank you for the present.

Now, I feel quite bored. The weather made me uncomfortable, it is hot and stuffy. Would you like to have some ice-creams with me next Sunday? I hope you will do.

If yes, please contact to me at 043890186.

I look forward to seeing from you.

Your best friend,

Trang.

đây là đoạn văn mà sao mọi người đều viết bài văn vậy?

Ta có:

\(x^2-xy=6x-5y-8\)

\(\Rightarrow x^2-6x+8=xy-5y\)

\(\Rightarrow x^2-6x+8=y\left(x-5\right)\) (1)

Nếu x = 5, thay vào phương trình ta có:

3 = 0y.(vô lí, phương trình vô nghiệm)

\(\Rightarrow\) Chia cả hai vế của (1) cho x-5 ta có:

\(y=\dfrac{x^2-6x+8}{x-5}=\dfrac{x^2-6x+5+3}{x-5}=x-1+\dfrac{3}{x-5}\)

Vì \(y\in Z\) nên ta có \(\dfrac{3}{x-5}\in Z\Rightarrow x-5\inƯ\left(3\right)\)

Từ đó ta tìm được các nghiệm (x;y) của phương trình là:(6 ; 8) ; (4 ; 0);(8 ; 8) ; (2 ; 0).

Phần lập bảng với thử chọn bạn tự làm nha!

mFe : mS : mO = 7 : 6 : 12

\(\Rightarrow\)\(\dfrac{mFe}{7}=\dfrac{mS}{6}=\dfrac{mO}{12}\)

\(\Rightarrow\dfrac{56.nFe}{7}=\dfrac{32.nS}{6}=\dfrac{16.nO}{12}\)

\(\Rightarrow\dfrac{56.nFe}{7:7.56}=\dfrac{32.nS}{6:7.56}=\dfrac{16.nO}{12:7.56}\)

\(\Rightarrow\dfrac{nFe}{1}=\dfrac{nS}{1.5}=\dfrac{nO}{6}\)\(\Rightarrow\dfrac{nFe}{2}=\dfrac{nS}{3}=\dfrac{nO}{12}\)

Vậy công thức hóa học của hợp chất là Fe₂(SO4)₃

a, \(\dfrac{x^4-4x^2+3}{x^4+6x^2-7}\)=\(\dfrac{x^4-4x^2+4-1}{x^4+7x^2-x^2-7}\)=\(\dfrac{\left(x^2-2\right)^2-1}{x^2\left(x^2+7\right)-\left(x^2+7\right)}\)

=\(\dfrac{\left(x^2-1\right)\left(x^2-3\right)}{\left(x^2-1\right)\left(x^2+7\right)}\)=\(\dfrac{x^2-3}{x^2+7}\)

b, \(\dfrac{x^4+x^3-x-1}{x^4+x^3+2x^2+x+1}\)=\(\dfrac{x^3\left(x+1\right)-\left(x+1\right)}{\left(x^2+1\right)^2+x\left(x^2+1\right)}\)=\(\dfrac{\left(x+1\right)\left(x^3-1\right)}{\left(x^2+1\right)\left(x^2+x+1\right)}\)=\(\dfrac{\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+1\right)\left(x^2+x+1\right)}\)=\(\dfrac{x^2-1}{x^2+1}\)

c, \(\dfrac{x^3+3x^2-4}{x^3-3x+2}\)=\(\dfrac{x^3+3x^2-3-1}{x^3-3x-1+3}\)=\(\dfrac{\left(x-1\right)\left(x^2+x+1\right)+3\left(x^2-1\right)}{\left(x-1\right)\left(x^2+x+1\right)-3\left(x-1\right)}\)=\(\dfrac{\left(x-1\right)\left(x^2+x+1+3x+3\right)}{\left(x-1\right)\left(x^2+x+1-3\right)}\)=\(\dfrac{x^2+4x+4}{x^2+x-2}\)=\(\dfrac{\left(x+2\right)^2}{x^2+x-2}\)

Mong là mk làm đúng. Chúc bn may mắn!

48 nha

tick cho mik với

=a, a(b²+c²)+b(a²+c²)+c(a²+b²)+2abc

= ab²+ac²+ba²+bc²+ca²+cb²+2abc

= c²(a+b)+ab(a+b)+c(a²+b²+2ab)

= c²(a+b)+ab(a+b)+c(a+b)²

= (a+b)\(\left[c^2+ab+c\left(a+b\right)\right]\)

= (a+b)(c²+ab+ca+cb)

= (a+b)\(\left[c\left(a+c\right)+b\left(a+c\right)\right]\)

=(a+b)(a+c)(b+c)

b, a(b-c)³+b(c-a)³+c(a-b)³

= a(b-c)³-b\(\left[\left(b-c\right)+\left(a-b\right)\right]\)³+c(a-b)³

= a(b-c)³-b(b-c)³-3b(b-c)²(a-b)-3b(b-c)(a-b)²-b(a-b)³+c(a-b)³

= a(b-c)³-b(b-c)³-3b(b-c)(a-b)(b-c+a-b)-b(a-b)³+c(a-b)³

= a(b-c)³-b(b-c)³-3b(b-c)(a-b)(a-c)-b(a-b)³+c(a-b)³

= (b-c)³(a-b)-3b(b-c)(a-b)(a-c)-(a-b)³(b-c)

= (b-c)(a-b)\(\left[\left(b-c\right)^2-3b\left(a-c\right)-\left(a-b\right)^2\right]\)

=(b-c)(a-b)(b²-2bc+c²-3ab+3bc-a²+2ab-b²)

= (b-c)(a-b)(c²-a²+bc-ab)

= (b-c)(a-b)\(\left[\left(c-a\right)\left(c+a\right)+b\left(c-a\right)\right]\)

= (b-c)(a-b)(c-a)(c+a+b)

c, a²b²(a-b)+b²c²(b-c)+c²a²(c-a)

= a²b²(a-b)-b²c²\(\left[\left(a-b\right)+\left(c-a\right)\right]\)+c²a²(c-a)

= a²b²(a-b)-b²c²(a-b)-b²c²(c-a)+c²a²(c-a)

= b²(a-b)(a²-c²)+c²(c-a)(a²-b²)

= b²(a-b)(a-c)(a+c)-c²(a-c)(a-b)(a+b)

= (a-c)(a-b)\(\left[b^2\left(a+c\right)-c^2\left(a+b\right)\right]\)

= (a-c)(a-b)(b²a+b²c-c²a-c²b)

= (a-c)(a-b)\(\left[a\left(b^2-c^2\right)+bc\left(b-c\right)\right]\)

= (a-c)(a-b)\(\left[a\left(b-c\right)\left(b+c\right)+bc\left(b-c\right)\right]\)

= (a-c)(a-b)(b-c)\(\left[a\left(b+c\right)+bc\right]\)

= (a-c)(a-b)(b-c)(ab+ac+bc)

d, a⁴(b-c)+b⁴(c-a)+c⁴(a-b)

= a⁴(b-c)-b⁴[(b-c)+(a-b)]+c⁴(a-b)
= (b-c)(a⁴-b⁴)+(a-b)(c⁴-b⁴)
= (b-c)(a²-b²)(a²+b²)+(a-b)(c²-b²)(c²+b²)
= (b-c)(a-b)(a+b)(a^2+b^2)-(a-b)(b-c)(b+c)(b²+c²)
= (b-c)(a-b)(a³+ab²+ba²+b³-bc²-b³-cb²-c³)

= (b-c)(a-b)(a³+ab²+ba²-bc²-c³-cb²)
= (b-c)(a-b)(a³-c³)+b²(a-c)+b(a²-c²)
= (b-c)(a-b)(a-c)(a²+ac+c²)+b²(a-c)+b(a-c)(a+c)
= (b-c)(a-b)(a-c)(a²+ac+c²+b²+ab+ac)

= (a-b)(b-c)(c-a)(a²+b²+c²+ab+bc+ca)