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a) \(9x^2-12x+4\)
\(=9x^2-6x-6x+4\)
\(=3x\left(3x-2\right)-2\left(3x-2\right)\)
\(=\left(3x-2\right)^2\)
b) \(2xy+16-x^2-y^2\)
\(=-\left(x^2-2xy+y^2-16\right)\)
\(=-\left(x-y\right)^2+16\)
\(=\left(4-x+y\right)\left(4+x-y\right)\)
c) \(3x+2x^2-2\)
\(=2x^2+4x-x-2\)
\(=2x\left(x+2\right)-\left(x+2\right)=\left(x+2\right)\left(2x-1\right)\)
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
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) \(3x^2-3y^2-12x+12y\)
\(=\left(3x^2-3y^2\right)-\left(12x-12y\right)\)
\(=3\left(x^2-y^2\right)-12\left(x-y\right)\)
\(=3\left(x-y\right)\left(x+y\right)-12\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-3y-12\right)\)
\(=\left(x-y\right).3.\left(x-y-4\right)\)
b) \(4x^3+4xy^2+8x^2y-16x\)
\(=\left(4x^3-16x\right)+\left(4xy^2+8x^2y\right)\)
\(=4x\left(x^2-4\right)+4xy\left(y+2x\right)\)
c) \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=\left(x^4-x^2\right)-\left(4x^2-4\right)\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-4\right)\left(x^2-1\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)
a/ \(12x^2+5x-12y^2+12y-10xy-3.\)
\(=12x^2+9x-4x-12y^2+6y+6y-18xy+8xy-3.\)
\(=\left(12x^2-18xy+9x\right)-\left(4x-6y+3\right)+\left(8xy-12y^2+6y\right)\)
\(=3x\left(4x-6y+3\right)-\left(4x-6y+3\right)+2y\left(4x-6y+3\right)\)
\(=\left(4x-6y+3\right)\left(3x-1+2y\right)\)
2/ \(2x^2+y^2+3x-2y-3xy+1\)
\(=\left(y^2-2y+1\right)+\left(3x-3xy\right)+2x^2\)
\(=\left(y-1\right)^2+3x\left(1-y\right)+2x^2\)
\(=\left(y-1\right)^2-3x\left(y-1\right)+2x^2\)
a: \(=\left(x-1\right)^2-4y^2=\left(x-1-2y\right)\left(x-1+2y\right)\)
b: \(=3\left(x^2-4x+4\right)=3\left(x-2\right)^2\)
c: \(=4x^2y^3\left(3y^2-8xyz+7x^2\right)\)
d: \(=x^4+16x^2+64-16x^2=\left(x^2+8\right)^2-16x^2\)
\(=\left(x^2+8-4x\right)\left(x^2+8+4x\right)\)
c: \(=\left(x-y\right)\left(x+y\right)-2z\left(x-y\right)=\left(x-y\right)\left(x+y-2z\right)\)
a) \(3x^2-6xy+3y^2-12x^2=3\left(x^2-2xy+y^2\right)-12x^2=3\left(x-y\right)^2-12x^2=3\left[\left(x-y\right)^2-4x^2\right]=3\left(x-y-2x\right)\left(x-y+2x\right)=3\left(-x-y\right)\left(3x-y\right)\)
b)\(3x^2y^2-6x^2y^3+12x^2y^2=3x^2y^2\left(1-2y+4\right)=3x^2y^2\left(5-2y\right)\)
c) \(3x^2-3y^2+12x-12y=3\left(x^2-y^2\right)+12\left(x-y\right)=3\left(x-y\right)\left(x+y+4\right)\)
a: \(3x^2-6xy+3y^2-12x^2\)
\(=3\left(x^2-2xy+y^2-4x^2\right)\)
\(=3\left[\left(x-y\right)^2-4x^2\right]\)
\(=3\left(x-y-2x\right)\left(x-y+2x\right)\)
\(=3\left(-x-y\right)\left(3x-y\right)\)
b: \(3x^2y^2-6x^2y^3+12x^2y^2\)
\(=3x^2y^2\left(1-2y+4\right)\)
\(=3x^2y^2\left(-2y+5\right)\)
c: Ta có: \(3x^2-3y^2+12x-12y\)
\(=3\left(x-y\right)\left(x+y\right)+12\left(x-y\right)\)
\(=3\left(x-y\right)\left(x+y+4\right)\)