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a: \(=n^3+2n^2+3n^2+6n-n-2-n^3+5\)
\(=5n^2+5n+3⋮̸5\)
b:\(=6n^2+30n+n+5-6n^2+3n-10n+5\)
\(=24n+10=2\left(12n+5\right)⋮2\)
d: \(=4x^2y^2-2x^2y+2xy^2-xy-4x^2y^2+xy\)
\(=-2\left(x^2y-xy^2\right)⋮2\)
\(x^2-x-6=x^2-3x+2x-6=x\left(x-3\right)+2\left(x-3\right)=\left(x-3\right)\left(x+2\right)\)
\(x^4+x^2+1=x^4+2x^2+1-x^2=\left(x^2+1\right)-x^2=\left(x^2+x+1\right)\left(x^2-x+1\right)\)\(x^3-19x-30=\left(x^3+8\right)-\left(19x-38\right)=\left(x+2\right)\left(x^2-2x+4\right)-19\left(x+2\right)=\left(x+2\right)\left(x^2-2x-15\right)=\left(x+2\right)\left(x^2-5x+3x-15\right)=\left(x+2\right)\left(x-5\right)\left(x+3\right)\)
\(x^4+4x^2-5=x^4+4x^2+4-9=\left(x^2+2\right)^2-9=\left(x^2+5\right)\left(x^2-1\right)=\left(x^2+5\right)\left(x-1\right)\left(x+1\right)\)
\(x^3-7x-6=0\Leftrightarrow\left(x^3+1\right)-\left(7x+7\right)=0\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)-7\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(x^2-x-6\right)=0\Leftrightarrow\left(x+1\right)\left(x^2-3x+2x-6\right)=0\Leftrightarrow\left(x+1\right)\left(x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\x=-1\end{matrix}\right.\)
\(x^3-3x^2-16x+48=x^2\left(x-3\right)-16\left(x-3\right)=\left(x^2-16\right)\left(x-3\right)=\left(x-4\right)\left(x+4\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\\x=-4\end{matrix}\right.\)
Bài 1.
a) 2x2 + 3( x - 1 )( x + 1 ) - 5x( x + 1 )
= 2x2 + 3( x2 - 1 ) - 5x2 - 5x
= 2x2 + 3x2 - 3 - 5x2 - 5x
= -5x - 3
b) 4( x - 1 )( x + 5 ) - ( x - 2 )( x + 5 ) - 3( x - 1 )( x + 2 )
= 4( x2 + 4x - 5 ) - ( x2 + 3x - 10 ) - 3( x2 + x - 2 )
= 4x2 + 16x - 20 - x2 - 3x + 10 - 3x2 - 3x + 6
= 10x - 4
Bài 2.
a) ( 8 - 5x )( x + 2 ) + 4( x - 2 )( x + 1 ) + 2( x - 2 )( x + 2 ) = 0
<=> -5x2 - 2x + 16 + 4( x2 - x - 2 ) + 2( x2 - 4 ) = 0
<=> -5x2 - 2x + 16 + 4x2 - 4x - 8 + 2x2 - 8 = 0
<=> x2 - 6x = 0
<=> x( x - 6 ) = 0
<=> x = 0 hoặc x = 6
b) ( x + 3 )( x + 2 ) - ( x - 2 )( x + 5 ) = 0
<=> x2 + 5x + 6 - ( x2 + 3x - 10 ) = 0
<=> x2 + 5x + 6 - x2 - 3x + 10 = 0
<=> 2x + 16 = 0
<=> 2x = -16
<=> x = -8
Bài 3.
A = ( n2 + 3n - 1 )( n + 2 ) - n3 + 2
= n3 + 2n2 + 3n2 + 6n - n - 2 - n3 + 2
= 5n2 + 5n
= 5n( n + 1 ) chia hết cho 5 ( đpcm )
B = ( 6n + 1 )( n + 5 ) - ( 3n + 5 )( 2n - 1 )
= 6n2 + 30n + n + 5 - ( 6n2 - 3n + 10n - 5 )
= 6n2 + 31n + 5 - 6n2 - 7n + 5
= 24n + 10
= 2( 12n + 5 ) chia hết cho 2 ( đpcm )
bài 1:a,\(2x^2+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)\)
\(=2x^2+3x^2-3-5x^2-5x\)
\(=-3-5x\)
b.\(4\left(x-1\right)\left(x+5\right)-\left(x-2\right)\left(x+5\right)-3\left(x-1\right)\left(x+2\right)\)
\(=4\left(x^2+4x-5\right)-\left(x^2+3x-10\right)-3\left(x^2+x-2\right)\)
\(=4x^2+16x-20-x^2-3x+10-3x^2-3x+6\)
\(=10x-4\)
\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)
\(8x+16-5x^2-10x+4\left(x^2+x-2x-2\right)+2\left(x^2+2x-2x-4\right)=0\)
\(-2x+16-5x^2+4x^2-4x-8+2x^2-8=0\)
\(x^2-6x=0\)
\(x\left(x-6\right)=0\)
\(\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)
Bài 3:
a: \(n\left(2n-3\right)-2n\left(n+1\right)\)
\(=2n^2-3n-2n^2-2n\)
=-5n chia hết cho 5
b: \(\left(n-1\right)\left(n+4\right)-\left(n-4\right)\left(n+1\right)\)
\(=n^2+4n-n-4-\left(n^2+n-4n-4\right)\)
\(=n^2+3n-4-\left(n^2-3n-4\right)\)
\(=6n⋮6\)
Câu 1 cậu lên mạng tra định ký Bơdu nhé
Câu 2 ta lấy f(x) chia cho g(x) đc x + 6 dư n+12
Để f(x) chia hết cho g(x) thì ta cho n+12=0
=> n= -12
Vậy để f(x) chia hết cho g(x) <=> n = -12
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 $g(x)$