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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)
F=x2+2xy+y2-x-y-12
= (x + y)^2 - (x + y) - 12
= (x + y)(x + y - 1) - 12
đặt x + y = t
F = t(t - 1) - 12
= t^2 - t - 12
= (t - 4)(t + 3)
G=(x2-3x-1)2-12(x2-3x-1)+27
đăth x^2 - 3x - 1 = t
G = t^2 - 12t + 27
= (t - 3)(t - 9)
có t = x^2 - 3x - 1
thay vào
Câu F ( kiểm tra lại đề )
Câu G . Đặt x^2 -3x-1=t
t^2 -12t+27 ( thực hiện pp tách)
\(x^3-4x-12+3x^2=x\left(x^2-2^2\right)+3\left(x^2-2^2\right)=\left(x-2\right)\left(x+2\right)\left(x+3\right)\)
\(x^2+2xy-15y^2=x^2+2xy+y^2-16y^2=\left(x+y\right)^2-\left(4y\right)^2=\left(x-3y\right)\left(x+5y\right)\)
\(\left(x-y\right)^2-6\left(x-y\right)-16=\left(x-y\right)^2-2\times\left(x-y\right)\times3+9-25=\left(x-y-3\right)^2-5^2=\left(x-y-8\right)\left(x-y+2\right)\)
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
Áp dụng HĐT a2 - b2 = ( a - b )( a + b )
và tính chất an.bn = ( a.b )n ( với n ∈ N* )
a) ( 3x + 1 )2 - ( x + 1 )2
= [ ( 3x + 1 ) - ( x + 1 ) ][ ( 3x + 1 ) + ( x + 1 ) ]
= ( 3x + 1 - x - 1 )( 3x + 1 + x + 1 )
= 2x( 4x + 2 )
= 2x.2( 2x + 1 )
= 4x( 2x + 1 )
b) ( x + y )2 - ( x - y )2
= [ ( x + y ) - ( x - y ) ][ ( x + y ) + ( x - y ) ]
= ( x + y - x + y )( x + y + x - y )
= 2y.2x = 4xy
c) ( 2xy + 1 )2 - ( 2x + y )2
= [ ( 2xy + 1 ) - ( 2x + y ) ][ ( 2xy + 1 ) + ( 2x + y ) ]
= ( 2xy + 1 - 2x - y )( 2xy + 1 + 2x + y )
= [ ( 2xy - 2x ) - ( y - 1 ) ][ ( 2xy + 2x ) + ( y + 1 ) ]
= [ 2x( y - 1 ) - ( y - 1 ) ][ 2x( y + 1 ) + ( y + 1 ) ]
= ( y - 1 )( 2x - 1 )9 y + 1 )( 2x + 1 )
d) 9( x - y )2 - 4( x + y )2
= 32( x - y )2 - 22( x + y )2
= [ 3( x - y ) ]2 - [ 2( x + y ) ]2
= ( 3x - 3y )2 - ( 2x + 2y )2
= [ ( 3x - 3y ) - ( 2x + 2y ) ][ ( 3x - 3y ) + ( 2x + 2y ) ]
= ( 3x - 3y - 2x - 2y )( 3x - 3y + 2x + 2y )
= ( x - 5y )( 5x - y )
e) ( 3x - 2y )2 - ( 2x - 3y )2
= [ ( 3x - 2y ) - ( 2x - 3y ) ][ ( 3x - 2y ) + ( 2x - 3y ) ]
= ( 3x - 2y - 2x + 3y )( 3x - 2y + 2x - 3y )
= ( x + y )( 5x - 5y )
= ( x + y )5( x - y )
f) ( 4x2 - 4x + 1 ) - ( x + 1 )2
= ( 2x - 1 )2 - ( x + 1 )2
= [ ( 2x - 1 ) - ( x + 1 ) ][ ( 2x - 1 ) + ( x + 1 ) ]
= ( 2x - 1 - x - 1 )( 2x - 1 + x + 1 )
= 3x( x - 2 )
\(x^2-3x+xy-3y\)
\(=\left(x^2+xy\right)-\left(3x+3y\right)\)
\(=x.\left(x+y\right)-3.\left(x+y\right)\)
\(=\left(x-3\right).\left(x+y\right)\)
\(2x^2-x+2xy-y\)
\(=2x^2-\left(x-2xy+y\right)\)
\(=2x^2-\left(x-y\right)^2\)
\(=\left(\sqrt{2}x\right)^2-\left(x-y\right)^2\)
\(=\left(\sqrt{2}x-x+y\right).\left(\sqrt{2}x+x-y\right)\)
\(x^4+x^3+2x^2+x+1\)
\(=\left(x^4+2x^2+1\right)+\left(x^3+x\right)\)
\(=\left(x^2+1\right)^2+x.\left(x^2+1\right)\)
\(=\left(x^2+1\right).\left(x^2+1+x\right)\)
\(16+2xy-x^2-y^2\)
\(=16-x^2+2xy-y^2\)
\(=16-\left(x^2-2xy+y^2\right)\)
\(=4^2-\left(x-y\right)^2\)
\(=[4-\left(x-y\right)].[4+\left(x-y\right)]\)
\(=\left(4-x+y\right).\left(4+x-y\right)\)
a) =(x2+x-1)2 -16=(x2+x-5)(x2+x+3)
b) Đặt a=x2+3x
=(a+1)(a+2)-6=a2+3a-4=(a-1)(a+4)=(x2+3x-1)(x2+3x+4)
c) =(x+y)2-(x+y)-12=(x+y)(x+y+1)-12
Đặt a=x+y
=a(a+1)-12=a2+a-12=(a-3)(a+4)=(x+y-3)(x+y+4)
d) Đặt a=x2+x
=(a+1)(a+2)-12=a2+3a-10=(a-2)(a+5)=(x2+x-2)(x2+x+5)