CMR : f(x) = x6n - x3n +1 không chia hết cho g(x) = x2 + x + 1 ( n ϵ N , n ≠ 0 ) mấy bn giúp vs
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\(\left(x-2\right)^3-\left(x-2\right)\left(x^2+2x+4\right)=0\\ \Leftrightarrow\left(x-2\right)\left[\left(x-2\right)^2-\left(x^2+2x+4\right)\right]=0\\ \Leftrightarrow\left(x-2\right)\left(x^2-4x+4-x^2-2x-4\right)=0\\ \Leftrightarrow\left(x-2\right).\left(-6x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\-6x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
Cách làm khác:
\(\left(x-2\right)^3-\left(x-2\right)\left(x^2+2x+4\right)=0\\ \Leftrightarrow x^3-3.x^2.2+3.x.2^2-2^3-\left(x^3-2^3\right)=0\\ \Leftrightarrow x^3-6x^2+12x-8-x^3+8=0\\ \Leftrightarrow-6x^2+12x=0\\ \Leftrightarrow-6x\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
ĐK: \(\left\{{}\begin{matrix}x\ne2\\x\ne-1\end{matrix}\right.\)
PT trở thành:
\(\dfrac{x}{x-2}=\dfrac{1}{x+1}+\dfrac{x+1}{x+1}\\ \Leftrightarrow\dfrac{x}{x-2}=\dfrac{x+2}{x+1}\\ \Leftrightarrow x\left(x+1\right)=x^2-4\\ \Leftrightarrow x^2+x-x^2+4=0\\ \Leftrightarrow x+4=0\Leftrightarrow x=-4\left(tm\right)\)
[Không biết mình làm có đúng không ạ!]
1- London is visited by thousands of people every year.
2- The thief is arrested by police cars.
3- CO2 is absorbed by trees.
4- Photos are taken by Jane.
5- A lot of waste is produced by people.
6- The animals are fed by the farmers.
7- Films are watched by Mr. Jones.
8- English is spoken by the people.
9- Comics are read by him.
10- Volleyball is played by us.
11- The song is sung by them.
12- Photos are taken by me.
13- Poems are written by the writer.
14- The children are helped by the police.
15- The flowers are watered by mother.
16- The car is washed by my husband.
17- Halloween is loved by people around the world.
18- Blue shoes are worn by the teachers.
19- The school is cleaned by the janitor.
20- The house is cleaned by my mother everyday.
Xét ΔABC vuông tại A có AH là đường cao
nên \(AH^2=HB\cdot HC\)
=>\(HB=\dfrac{12^2}{16}=9\left(cm\right)\)
BC=BH+CH=9+16=25(cm)
Xét ΔABC vuông tại A có AH là đường cao
nên \(\left\{{}\begin{matrix}AB^2=BH\cdot BC\\AC^2=CH\cdot BC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}AB=\sqrt{9\cdot25}=15\left(cm\right)\\AC=\sqrt{16\cdot25}=20\left(cm\right)\end{matrix}\right.\)
Lời giải:
$f(x)=x^{6n}-x^{3n}+1=x^{3n}(x^{3n}-1)+1$
$=x^{3n}[(x^3)^n-1^n]+1$
$=x^{3n}(x^3-1)[(x^3)^{n-1}+(x^3)^{n-2}+...+1]+1$
$=x^{3n}(x^2+x+1)(x-1)[(x^3)^{n-1}+(x^3)^{n-2}+...+1]+1$
$\Rightarrow f(x)$ chia $x^2+x+1$ dư $1$
$\Rightarrow f(x)$ không chia hết cho $g(x)$