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a) ĐKXĐ: \(\left\{{}\begin{matrix}3x\left(x+2\right)\ne0\\x+1\ne0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}3x\ne0\\x+2\ne0\\x+1\ne0\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x\ne0\\x\ne-2\\x\ne-1\end{matrix}\right.\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}x^2-x+1\ne0\\2x\ne0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}\left(x-1\right)^2\ne0\\x\ne0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x-1\ne0\\x\ne0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x\ne1\\x\ne0\end{matrix}\right.\)
bn ơi cho mk hỏi bn lm tiếng anh hay toán mà chủ đề là tiếng anh mà bài lại là toán vậy alo????
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a .
a. =x3 -x2-4x2+4x+4x-4=(x-1)(x2-4x+4)=(x-1)(x-2)2
b. =x3+x2-6x2-6x+9x+9=(x+1)(x-3)2
c. =x3+x2+7x2+7x+10x+10=(x+1)(x+2)(X+5)
d. =x4+x3+x3+x2+x+1=x3(x+1)+x2(x+1)+x+1=(x+1)(x3+x2+x)=x(x+1)(x2+x+1).Ok
\(4x^2-6x-16⋮x-3\)
\(\Leftrightarrow4x^2-12x+6x-18+2⋮x-3\)
\(\Leftrightarrow4x\left(x-3\right)+6\left(x-3\right)+2⋮x-3\)
\(\Leftrightarrow\left(x-3\right)\left(4x+6\right)+2⋮x-3\)
Mà \(\left(x-3\right)\left(4x+6\right)⋮x-3\)
\(\Rightarrow2⋮x-3\)
\(\Rightarrow x-3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
làm nốt
cách 2:
4x^2-6x-16 x-3 4x+6 4x^2-12x - 6x-12 6x-18 - 2
Để \(4x^2-6x-16\)chia hết cho x-3
\(\Leftrightarrow2⋮x-3\)
\(\Leftrightarrow x-3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Làm nốt
1.We are now planning a holiday for next travel.
2.Our holiday will begin on July 15th and we will return on July 19th.
3.My daughter dislikes traveling by coach, so we wish to go by air.
4. We expect to stay at a hotel on the East coast.
5.We would be grateful if you could send us some information with details of charges.
1.We are now planning a holiday for next summer.
2.Our holiday will begin on July 15th and we return on July 19th.
3.My daughter dislikes traveling by coach' so we wish to go by air.
4.We expect to stay at a hotel on the East coast.
5.We would be grateful if you could send us some information with details charges.
\(Đặt A=(n^4-3n^3+n^2-3n+1):(n^2+1) \)
\(=(n^4+n^2-3n^3+n^2-3n+10):(n^2+1)\)
\(=[n^2(n^2+1)-3n(n^2+1)+1]:(n^2+1)\)
\(=[(n^2+1)(n^2-3n)+1]:(n^2+1)\)
Để A thuộc Z thì tử phải chia hết cho mẫu mà\((n^2+1)(n^2-3n) \) chia hết cho \(n^2+1\)
=> 1 chia hết cho \(n^2+1\)
=> \(n^2+1\) thuộc Ư(1)
mà \(n^2+1>=1\) (với mọi n)
=>\(n^2+1=1\)
=>n=0
Vậy....................