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Bài làm :
\(a,\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)+10\)
\(=8x+16-5x^2-10x+\left(4x-8\right)\left(x+1\right)+2\left(x^2-2^2\right)+10\)
\(=8x+16-5x^2-10x+4x^2+4x-8x-8+2x^2-8+10\)
\(=\left(8x-10x+4x-8x\right)+\left(-5x^2+4x^2+2x^2\right)+\left(16-8-8+10\right)\)
\(=-6x+x^2+10\)
a)\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)+10\)\(=8x+16-5x^2-2+4x-8x-8+2x-4x-4+10\)\(=\left(8x+4x-8x+2x-4x\right)+\left(16-2-8-4+10\right)+5x^2\)
\(=2x+12+5x^2\)
b)\(4\left(x-1\right)\left(x+5\right)-\left(x+2\right)\left(x+5\right)-3\left(x-1\right)\left(x+2\right)\)
\(=4x-4x-20-\left[x^2+5x+2x+10\right]-3\left[x^2+2x-1x-2\right]\)
\(=4x-4x-20-x^2-5x-2x-10-3x^2-6x+3x+6\)
\(=\left(4x-4x-5x-2x-6x+3x\right)+\left(-20-10+6\right)+\left(-x^2-3x^2\right)\)
\(=-10x-24-4x^2\)
c)\(\left(x^{2n}+x^ny^n+y^{2n}\right)\left(x^n-y^n\right)\left(x^{3n}+y^{3n}\right)\)
Xét tích \(\left(x^{2n}+x^ny^n+y^{2n}\right)\left(x^n-y^n\right)\Leftrightarrow\left(x^n\right)^3-\left(y^n\right)^3=x^{3n}-y^{3n}\)
Thay vào bt đã cho ta có \(\left(x^{3n}-y^{3n}\right)\left(x^{3n}+y^{3n}\right)\)
\(\Leftrightarrow\left(x^{3n}\right)^2-\left(y^{3n}\right)^2=x^{6n}-y^{6n}\)
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.\(4\left(x-1\right)\left(x+5\right)-\left(x-2\right)\left(x+5\right)-3\left(x-1\right)\left(x+2\right)\)
\(=4\left(x^2+4x-5\right)-\left(x^2+3x-10\right)-3\left(x^2+x-2\right)\)
\(=4x^2+16x-20-x^2-3x+10-3x^2-3x+6\)
\(=10x-4\)
\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)
\(8x+16-5x^2-10x+4\left(x^2+x-2x-2\right)+2\left(x^2+2x-2x-4\right)=0\)
\(-2x+16-5x^2+4x^2-4x-8+2x^2-8=0\)
\(x^2-6x=0\)
\(x\left(x-6\right)=0\)
\(\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)
a: \(4x^2\left(3x^{n+1}-2x^n\right)\)
\(=4x^2\cdot3x^{n+1}-4x^2\cdot2x^n\)
\(=12x^{n+3}-8x^{n+2}\)
b: \(2\left(x^{2n}+2x^ny^n+y^{2n}\right)-y^n\left(4x^n+2y^n\right)\)
\(=2x^{2n}+4x^ny^n+2y^{2n}-4x^ny^n-2y^{2n}\)
\(=2x^{2n}\)
c: \(=\left(x^{3n}-y^{3n}\right)\left(x^{3n}+y^{3n}\right)\)
\(=x^{6n}-y^{6n}\)
d: \(=4^n\cdot4-3\cdot4^n=4^n\)
a: 4x2(3xn+1−2xn)4x2(3xn+1−2xn)
=4x2⋅3xn+1−4x2⋅2xn=4x2⋅3xn+1−4x2⋅2xn
=12xn+3−8xn+2=12xn+3−8xn+2
b: 2(x2n+2xnyn+y2n)−yn(4xn+2yn)2(x2n+2xnyn+y2n)−yn(4xn+2yn)
=2x2n+4xnyn+2y2n−4xnyn−2y2n=2x2n+4xnyn+2y2n−4xnyn−2y2n
=2x2n=2x2n
c: =(x3n−y3n)(x3n+y3n)=(x3n−y3n)(x3n+y3n)
=x6n−y6n=x6n−y6n
d: =4n⋅4−3⋅4n=4n
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: \(=4x^2y^2-2x^2y+2xy^2-xy-4x^2y^2+xy\)
\(=-2\left(x^2y-xy^2\right)⋮2\)
Nhận xét chút nè:))
Ở chỗ 3.(x - 1).(x + 1) thì bạn có thể áp dụng HĐT để ra lun là:
= 3. (x2 - 12)
= ...................
Như thế sẽ giảm thiểu khả năng sai sót hơn!
Oke cảm ơn bạn nhiều nha ^^