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a) \(x^3+x^2-2x-8\)
\(=\left(x^3-2x^2\right)+\left(3x^2-6x\right)+\left(4x-8\right)\)
\(=x^2\left(x-2\right)+3x\left(x-2\right)+4\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+3x+4\right)\)
b) \(125x^3-10x^2+2x-1\)
\(=\left(125x^3-25x^2\right)+\left(15x^2-3x\right)+\left(5x-1\right)\)
\(=25x^2\left(5x-1\right)+3x\left(5x-1\right)+\left(5x-1\right)\)
\(=\left(5x-1\right)\left(25x^2+3x+1\right)\)
c) \(x^3-4x^2+12x-27\)
\(=\left(x^3-3x^2\right)-\left(x^2-3x\right)+\left(9x-27\right)\)
\(=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
d) Đề sai sai, nghiệm ra khá xấu nên bạn xem lại nhé
e) \(x^3-3x^2-3x+1\)
\(=\left(x^3+x^2\right)-\left(4x^2+4x\right)+\left(x+1\right)\)
\(=x^2\left(x+1\right)-4x\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
\(3x\left(x-5\right)-x\left(4+3x\right)=43\)
\(\Leftrightarrow3x^2-15x-4x-3x^2=43\)
\(\Leftrightarrow-19x=43\)
\(\Leftrightarrow x=\frac{-43}{19}\)
\(a,xy+1-x-y\)
\(=\left(xy-y\right)+\left(1-x\right)\)
\(=y\left(x-1\right)- \left(x-1\right)\)
\(=\left(x-1\right)\left(y-1\right)\)
\(b,ax+ay-3x-3y\)
\(=a\left(x+y\right)-3\left(x+y\right)\)
\(=\left(x+y\right)\left(a-3\right)\)
\(c,x^3-2x^2+2x-4\)
\(=x^2\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x^2+2\right)\left(x-2\right)\)
\(d,x^2+ab+ax+bx\)
\(=\left(x^2+ax\right)+\left(ab+bx\right)\)
\(=x\left(a+x\right)+b\left(a+x\right)\)
\(=\left(a+x\right)\left(b+x\right)\)
\(e,16-x^2+2xy-y^2\)
\(=4^2-\left(x^2-2xy+y^2\right)\)
\(=4^2-\left(x-y\right)^2\)
\(=\left(4-x+y\right)\left(4+x-y\right)\)
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
1) (3x+4)(x+1) = 3x2+7x+4 đặt là a
(6x+7)2= 36x2+84x+49 = 12a+1
=> a(12a+1)- 6 = 12a2 -a -6 = (3a+2)(4a-3) = (9x2+21x+14)(12x2+28x+13)
2) (x-2)2=x2-4x+4 đặt là a
(2x-5)(2x-3)= 4x2-16x+15 =4a-1
=> a(4a-1)-5 = 4a2-a-5 = (4a-5)(a+1) = ( 4x2-16x+11)(x2-4x+5)
3) đặt (x+3)2 =a ta làm tương tự
4) (x-2)(x-10)(x-4)(x-5) = (x2-12x+20)(x2-9x+20)
đặt x2+20=a => (a-12x)(a-9x)-54x2 = a2-21ax+54x2 = (a-18x)(a-3x) = (x2-18x+20)(x2-3x+20)
a) \(x^3+2x^2+2x+1\)
\(=\left(x^3+x^2\right)+\left(x^2+x\right)+\left(x+1\right)\)
\(=x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+x+1\right)\)
b) \(x^3+x^2-4x-4\)
\(=x^2\left(x+1\right)-4\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-4\right)\)
\(=\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
c) \(x^4-2x^3+2x-1\)
\(=\left(x^4-x^3\right)-\left(x^3-x^2\right)-\left(x^2-x\right)+\left(x-1\right)\)
\(=x^3\left(x-1\right)-x^2\left(x-1\right)-x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3-x^2-x+1\right)\)
\(=\left(x-1\right)\left[x^2\left(x-1\right)-\left(x-1\right)\right]\)
\(=\left(x-1\right)\left(x-1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)^3\left(x+1\right)\)
d) \(x^5-x^4+x^3-x^2\)
\(=x^4\left(x-1\right)+x^2\left(x-1\right)\)
\(=x^2\left(x-1\right)\left(x^2+1\right)\)