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a) \(\left(x^2-2x+1\right)-\left(y^2+2y+1\right)\)
\(=\left(x-1\right)^2-\left(y+1\right)^2\)
\(=\left(x-y-2\right)\left(x+y\right)\)
1:
a) \(x^3+2x^2+x=x\left(x^2+2x+1\right)=x\left(x+1\right)^2\)
b) \(25-x^2+4xy-4y^2=25-\left(x-2y\right)^2=\left(5-x+2y\right)\left(5+x-2y\right)\)
2
\(-2x^2-4x+6=0\)
\(\Leftrightarrow-2\left(x^2+2x-3\right)=0\)
\(\Leftrightarrow x^2-x+3x-3=0\)
\(\Leftrightarrow x\left(x-1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-3\end{array}\right.\)
1,
a) x( x2 + 2x +1) = x(x+1)2
b)25 - (x-2y)2 = (5-x+2y)(5+x-2y)
2,
(x-1)(x+3)=0
<=>x=1 hoặc x=-3
a) \(-5x^2+16x-3=-5x^2+15x+x-3=-5x\left(x-3\right)+x-3=\left(x-3\right)\left(1-5x\right).\)
b) \(x^4+64=x^4+16x^2+64-16x^2=\left(x^2+8\right)^2-\left(4x\right)^2=\left(x^2+4x+8\right)\left(x^2-4x+8\right).\)
c) \(64x^2+4y^4=4\left(16x^2+y^4\right)\)
d) \(x^5+x-1\)đa thức này có nghiệm vô tỷ. Mik ko phân tích được.
Bài 1:
a) 2x^2 -3x + 1 = 2x^2 -2x -x +1 = 2x.(x-1) - (x-1) = (x-1).(2x-1)
b) 2x^3y - 2xy^3 - 4xy^2 - 2xy = 2xy.(x^2 - y^2 - 2y -1) = 2xy.[ x^2 - (y^2 + 2y+1)] = 2xy.[x^2 - (y+1)^2]
= 2xy.(x-y-1).(x+y+1)
c) (x^2 + x+3).(x^2 + x +5) - 8 = (x^2+x+4-1).(x^2+x+4+1) - 8 = (x^2+x+4)^2 - 1 - 8 = (x^2+x+4)^2 - 3^2
= (x^2+x+4-3).(x^2+x+4+3) = (x^2+x+1).(x^2+x+7)
Bài 2:
a) (x+2).(x^2-2x+4) - (x^3+2x) = 0
x^3 + 8 - x^3 - 2x = 0
8 - 2x = 0
x = 4
b) x^2 - 2x - 8 = 0
x^2 +2x - 4x - 8 = 0
x.(x+2) - 4.(x+2) = 0
(x+2).(x-4) = 0
...
bn tự làm tiếp nha
a. 2x-1-x2= -(x2-2x+1)=-(x-1)2
b. 8x3+y6=(2x)3+(y2)3
=(2x+y2)(4x2-2xy2+y4)
c. x2-16+4xy+4y2=(x2+4xy+4y2)-16
=(x+2y)2-16=(x+2y+4)(x+2y-4)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
a) \(x^3-x^2-4=x^3-2x^2+x^2-4=x^2\left(x-2\right)+\left(x-2\right)\left(x+2\right)=\left(x-2\right)\left(x^2+x+2\right)\)
b) \(x^3-5x^2+8x-4=x^3-x^2-4x^2+4x+4x-4=x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-4x+4\right)=\left(x-1\right)\left(x-2\right)^2\)
c) \(2x^3-12x^2+17x-2=2x^3-4x^2-8x^2+16x+x-2=2x^2\left(x-2\right)-8x\left(x-2\right)+\left(x-2\right).\)
\(=\left(x-2\right)\left(2x^2-8x+1\right)\)
d) \(2x^4+x^3-22x^2+15x+36=2x^4+2x^3-x^3-x^2-21x^2-21x+36x+36.\)
\(=2x^3\left(x+1\right)-x^2\left(x+1\right)-21x\left(x+1\right)+36\left(x+1\right)\)
\(=\left(x+1\right)\left(2x^3-x^2-21x+36\right)\)
1a/ x3+x2+x+1=0
x2(x+1).(x+1)=0
=> x2(x+1)=0 x =1
hoặc =>[
x+1=0 x=-1
b/(x+2)2=x+2
x2+2.x.2+22 =x+2
x+x+4x+4=x+2
6x+4=x+2
....
c/(x+1)(6x2+2x)+(x-1)(6x2+2x)=0
x2-12 + (6x2+2x)2=0
=> x2-1 = 0 x=1
hoặc => [
(6x2+2x)2=0 x= 0
a) \(x^2+2x-4y^2-4y=\left(x^2-4y^2\right)+\left(2x-4y\right)=\left(x+2y\right)\left(x-2y\right)+2\left(x-2y\right)\)
\(=\left(x-2y\right).\left(x+2y+2\right)\)
b) \(x^4-6x^3+54x-81=\left(x^4-81\right)-\left(6x^3-54x\right)=\left(x^2-9\right)\left(x^2+9\right)-6x.\left(x^2-9\right)\)
\(=\left(x^2-9\right).\left(x^2+9-6x\right)=\left(x+3\right).\left(x-3\right).\left(x-3\right)^2=\left(x+3\right).\left(x-3\right)^3\)
c) \(ax^2+ax-bx^2-bx-a+b=\left(ax^2-bx^2\right)+\left(ax-bx\right)-\left(a-b\right)\)
\(=x^2.\left(a-b\right)+x.\left(a-b\right)-\left(a-b\right)=\left(a-b\right).\left(x^2+x-1\right)\)
d) \(\left(x^2+y^2-2\right)^2-\left(2xy-2\right)^2=\left(x^2+y^2-2+2xy-2\right).\left(x^2+y^2-2-2xy+2\right)\)
\(=\left(x^2+2xy+y^2-4\right).\left(x^2+y^2-2xy\right)=\left[\left(x+y\right)^2-4\right].\left(x-y\right)^2\)
\(=\left(x+y+2\right).\left(x+y-2\right).\left(x-y\right)^2\)
Bài 1.
a) x( 8x - 2 ) - 8x2 + 12 = 0
<=> 8x2 - 2x - 8x2 + 12 = 0
<=> 12 - 2x = 0
<=> 2x = 12
<=> x = 6
b) x( 4x - 5 ) - ( 2x + 1 )2 = 0
<=> 4x2 - 5x - ( 4x2 + 4x + 1 ) = 0
<=> 4x2 - 5x - 4x2 - 4x - 1 = 0
<=> -9x - 1 = 0
<=> -9x = 1
<=> x = -1/9
c) ( 5 - 2x )( 2x + 7 ) = ( 2x - 5 )( 2x + 5 )
<=> -4x2 - 4x + 35 = 4x2 - 25
<=> -4x2 - 4x + 35 - 4x2 + 25 = 0
<=> -8x2 - 4x + 60 = 0
<=> -8x2 + 20x - 24x + 60 = 0
<=> -4x( 2x - 5 ) - 12( 2x - 5 ) = 0
<=> ( 2x - 5 )( -4x - 12 ) = 0
<=> \(\orbr{\begin{cases}2x-5=0\\-4x-12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-3\end{cases}}\)
d) 64x2 - 49 = 0
<=> ( 8x )2 - 72 = 0
<=> ( 8x - 7 )( 8x + 7 ) = 0
<=> \(\orbr{\begin{cases}8x-7=0\\8x+7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{8}\\x=-\frac{7}{8}\end{cases}}\)
e) ( x2 + 6x + 9 )( x2 + 8x + 7 ) = 0
<=> ( x + 3 )2( x2 + x + 7x + 7 ) = 0
<=> ( x + 3 )2 [ x( x + 1 ) + 7( x + 1 ) ] = 0
<=> ( x + 3 )2( x + 1 )( x + 7 ) = 0
<=> x = -3 hoặc x = -1 hoặc x = -7
g) ( x2 + 1 )( x2 - 8x + 7 ) = 0
Vì x2 + 1 ≥ 1 > 0 với mọi x
=> x2 - 8x + 7 = 0
=> x2 - x - 7x + 7 = 0
=> x( x - 1 ) - 7( x - 1 ) = 0
=> ( x - 1 )( x - 7 ) = 0
=> \(\orbr{\begin{cases}x-1=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=7\end{cases}}\)
Bài 2.
a) ( x - 1 )2 - ( x - 2 )( x + 2 )
= x2 - 2x + 1 - ( x2 - 4 )
= x2 - 2x + 1 - x2 + 4
= -2x + 5
b) ( 3x + 5 )2 + ( 26x + 10 )( 2 - 3x ) + ( 2 - 3x )2
= 9x2 + 30x + 25 - 78x2 + 22x + 20 + 9x2 - 12x + 4
= ( 9x2 - 78x2 + 9x2 ) + ( 30x + 22x - 12x ) + ( 25 + 20 + 4 )
= -60x2 + 40x2 + 49
d) ( x + y )2 - ( x + y - 2 )2
= [ x + y - ( x + y - 2 ) ][ x + y + ( x + y - 2 ) ]
= ( x + y - x - y + 2 )( x + y + x + y - 2 )
= 2( 2x + 2y - 2 )
= 4x + 4y - 4
Bài 3.
A = 3x2 + 18x + 33
= 3( x2 + 6x + 9 ) + 6
= 3( x + 3 )2 + 6 ≥ 6 ∀ x
Đẳng thức xảy ra <=> x + 3 = 0 => x = -3
=> MinA = 6 <=> x = -3
B = x2 - 6x + 10 + y2
= ( x2 - 6x + 9 ) + y2 + 1
= ( x - 3 )2 + y2 + 1 ≥ 1 ∀ x,y
Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-3=0\\y^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=0\end{cases}}\)
=> MinB = 1 <=> x = 3 ; y = 0
C = ( 2x - 1 )2 + ( x + 2 )2
= 4x2 - 4x + 1 + x2 + 4x + 4
= 5x2 + 5 ≥ 5 ∀ x
Đẳng thức xảy ra <=> 5x2 = 0 => x = 0
=> MinC = 5 <=> x = 0
D = -2/7x2 - 8x + 7 ( sửa thành tìm Max )
Để D đạt GTLN => 7x2 - 8x + 7 đạt GTNN
7x2 - 8x + 7
= 7( x2 - 8/7x + 16/49 ) + 33/7
= 7( x - 4/7 )2 + 33/7 ≥ 33/7 ∀ x
Đẳng thức xảy ra <=> x - 4/7 = 0 => x = 4/7
=> MaxC = \(\frac{-2}{\frac{33}{7}}=-\frac{14}{33}\)<=> x = 4/7
`a)x^3-8x^2+16x`
`=x(x^2-8x+16)`
`=x(x-4)^2`
`b)x^2+4y^2+2x-4y-4xy-24`
`=(x-2y)^2+2(x-2y)-24`
`=(x-2y)^2-4(x-2y)+6(x-2y)-24`
`=(x-2y-4)(x-2y+6)`
`c)x^4+x^3-x^2-2x-2`
`=x^4-2x^2+x^3-2x+x^2-2`
`=x^2(x^2-2)+x(x^2-2)+x^2-2`
`=(x^2-2)(x^2+x+1)`