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a) \(^{x^4-y^4}\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left[\left(x-y\right).\left(x+y\right)\right].\left(x^2-y^2\right)\)
\(=\left(x-y\right).\left(x+y\right).\left(x^2-y^2\right)\)
c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)
\(=\left[\left(3x-2y\right)+\left(2x-3y\right)\right].\left[\left(3x-2y\right)-\left(2x-3y\right)\right]\)
\(=\left(3x-2y+2x-3y\right)\left(3x-2y-2x+3y\right)\)
b) \(x^2-3y^2\)
\(=\left(x-3y\right)\left(x+3y\right)\)
d) \(9\left(x-y\right)^2-4\left(x+y\right)^2\)
\(=9\left(x-y\right)^2+4\left(x-y\right)^2\)
\(=\left(x-y\right).\left(9+4\right)\)
\(=\left(x-y\right).13\)
\(=13\left(x-y\right)\)
f) \(x^3+27\)
\(=x^3+3^3\)
\(=\left(x+3\right)\left(x^2-x.3+3^2\right)\)
h) \(125x^3-1\)
\(=\left(5x\right)^3-1^3\)
\(=\left(5x-1\right)\left(5x^2+5x.1+1^2\right)\)
\(=\left(5x-1\right)\left(5x^2+5x+1\right)\)
\(a,x^4-y^4=\left(x^2+y^2\right)\left(x^2-y^2\right)=\left(x^2+y^2\right)\left(x+y\right)\left(x-y\right)\)
\(b,x^2-3y^2=\left(x+\sqrt{3}y\right)\left(x-\sqrt{3}y\right)\)
cn lại tg tự nha bn
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) Đặt t = x2
bthuc <=> t2 - 7t + 16
Từ đây ta không thể phân tích được :)
b) x3 - 2x2 + 5x - 4
= x3 - x2 - x2 + x + 4x - 4
= x2( x - 1 ) - x( x - 1 ) + 4( x - 1 )
= ( x - 1 )( x2 - x + 4 )
c) x3 - 2x2 + x - 3 ( phân tích hổng ra :)) )
d) 3x3 - 4x2 + 12x - 4 ( phân tích hổng ra p2 :)) )
e) 6x3 + x2 + x + 1
= 6x3 + 3x2 - 2x2 - x + 2x + 1
= 3x2( 2x + 1 ) - x( 2x - 1 ) + ( 2x + 1 )
= ( 2x + 1 )( 3x2 - x + 1 )
f) 4x3 + 6x2 + 4x + 1
= 4x3 + 2x2 + 4x2 + 2x + 2x + 1
= 2x2( 2x + 1 ) + 2x( 2x + 1 ) + ( 2x + 1 )
= ( 2x + 1 )( 2x2 + 2x + 1 )
a) ( 3x - 1 )2 - 16 = ( 3x - 1 )2 - 42 = ( 3x - 1 - 4 )( 3x - 1 + 4 ) = ( 3x - 5 )( 3x + 3 ) = 3( 3x - 5 )( x + 1 )
b) ( 5x - 4 )2 - 49x2 = ( 5x - 4 )2 - ( 7x )2 = ( 5x - 4 - 7x )( 5x - 4 + 7x ) = ( -2x - 4 )( 12x - 4 ) = -2( x + 2 ).4( 3x - 1 ) = -8( x + 2 )( 3x - 1 )
c) ( 2x + 5 )2 - ( x - 9 )2 = [ ( 2x + 5 ) - ( x - 9 ) ][ ( 2x + 5 ) + ( x - 9 ) ] = ( 2x + 5 - x + 9 )( 2x + 5 + x - 9 ) = ( x + 14 )( 3x - 4 )
d) ( 3x + 1 )2 - 4( x - 2 )2 = ( 3x + 1 )2 - 22( x - 2 )2 = ( 3x + 1 )2 - [ 2( x - 2 ) ]2 = ( 3x + 1 )2 - ( 2x - 4 )2 = [ ( 3x + 1 ) - ( 2x - 4 ) ][ ( 3x + 1 ) + ( 2x - 4 ) ] = ( 3x + 1 - 2x + 4 )( 3x + 1 + 2x - 4 ) = ( x + 5 )( 5x - 3 )
e) 9( 2x + 3 )2 - 4( x + 1 )2 = 32( 2x + 3 )2 - 22( x + 1 )2 = [ 3( 2x + 3 ) ]2 - [ 2( x + 1 ) ]2 = ( 6x + 9 )2 - ( 2x + 2 )2 = [ ( 6x + 9 ) - ( 2x + 2 ) ][ ( 6x + 9 ) + ( 2x + 2 ) ] = ( 6x + 9 - 2x - 2 )( 6x + 9 + 2x + 2 ) = ( 4x + 7 )( 8x + 11 )
f) 4b2c2 - ( b2 + c2 - a2 )2 = ( 2bc )2 - ( b2 + c2 - a2 )2 = [ 2bc - ( b2 + c2 - a2 ) ][ 2bc + ( b2 + c2 - a2 ] = ( 2bc - b2 - c2 + a2 )( 2bc + b2+ c2 - a2 ) = [ a2 - ( b2 - 2bc + c2 ) ][ ( b2 + 2bc + c2 ) - a2 ] = [ a2 - ( b - c )2 ][ ( b + c )2 - a2 ] = ( a - b + c )( a + b - c )( b + c - a )( b + c + a )
g) ( ax + by )2 - ( ay + bx )2
= [ ( ax + by ) - ( ay + bx ) ][ ( ax + by ) + ( ay + bx ) ]
= ( ax + by - ay - bx )( ax + by + ay + bx )
= [ a( x - y ) - b( x - y ) ][ a( x + y ) + b( x + y ) ]
= ( a - b )( x - y )( x + y )( a + b )
h) ( a2 + b2 - 5 )2 - 4( ab + 2 )2
= ( a2 + b2 - 5 )2 - 22( ab + 2 )2
= ( a2 + b2 - 5 )2 - [ 2( ab + 2 ) ]2
= ( a2 + b2 - 5 )2 - ( 2ab + 4 )2
= [ ( a2 + b2 - 5 ) - ( 2ab + 4 ) ][ ( a2 + b2 - 5 ) + ( 2ab + 4 ) ]
= ( a2 + b2 - 5 - 2ab - 4 )( a2 + b2 - 5 + 2ab + 4 )
= [ ( a2 - 2ab + b2 ) - 9 ][ ( a2 + 2ab + b2 ) - 1 ]
= [ ( a - b )2 - 32 ][ ( a + b )2 - 12 ]
= ( a - b - 3 )( a - b + 3 )( a + b - 1 )( a + b + 1 )
i) ( 4x2 - 3x - 18 )2 - ( 4x2 + 3x )2
= [ ( 4x2 - 3x - 18 ) - ( 4x2 + 3x ) ][ ( 4x2 - 3x - 18 ) + ( 4x2 + 3x ) ]
= ( 4x2 - 3x - 18 - 4x2 - 3x )( 4x2 - 3x - 18 + 4x2 + 3x )
= ( -6x - 18 )( 8x2 - 18 )
= -6( x + 3 ).2( 4x2 - 9 )
= -12( x + 3 )( 2x - 3 )( 2x + 3 )
k) 9( x + y - 1 )2 - 4( 2x + 3y + 1 )2
= 32( x + y - 1 )2 - 22( 2x + 3y + 1 )2
= [ 3( x + y - 1 ) ]2 - [ 2( 2x + 3y + 1 ) ]2
= ( 3x + 3y - 3 )2 - ( 4x + 6y + 2 )2
= [ ( 3x + 3y - 3 ) - ( 4x + 6y + 2 ) ][ ( 3x + 3y - 3 ) + ( 4x + 6y + 2 ) ]
= ( 3x + 3y - 3 - 4x - 6y - 2 )( 3x + 3y - 3 + 4x + 6y + 2 )
= ( -x - 3y - 5 )( 7x + 9y - 1 )
l) -4x2 + 12xy - 9y2 + 25
= 25 - ( 4x2 - 12xy + 9y2 )
= 52 - ( 2x - 3y )2
= ( 5 - 2x + 3y )( 5 + 2x - 3y )
m) x2 - 2xy + y2 - 4m2 + 4mn - n2
= ( x2 - 2xy + y2 ) - ( 4m2 - 4mn + n2 )
= ( x - y )2 - ( 2m - n )2
= ( x - y - 2m + n )( x - y + 2m - n )
a) ( 4x2 - 3x - 18 )2 - ( 4x2 + 3x )2
= [ ( 4x2 - 3x - 18 ) - ( 4x2 + 3x ) ][ ( 4x2 - 3x - 18 ) + ( 4x2 + 3x ) ]
= ( 4x2 - 3x - 18 - 4x2 - 3x )( 4x2 - 3x - 18 + 4x2 + 3x )
= ( -6x - 18 )( 8x2 - 18 )
= -6( x + 3 ).2( 4x2 - 9 )
= -12( x + 3 )( 2x - 3 )( 2x + 3 )
b) 9( x + y - 1 )2 - 4( 2x + 3y + 1 )2
= 32( x + y - 1 )2 - 22( 2x + 3y + 1 )2
= [ 3( x + y - 1 ) ]2 - [ 2( 2x + 3y + 1 ) ]2
= ( 3x + 3y - 3 )2 - ( 4x + 6y + 2 )2
= [ ( 3x + 3y - 3 ) - ( 4x + 6y + 2 ) ][ ( 3x + 3y - 3 ) + ( 4x + 6y + 2 ) ]
= ( 3x + 3y - 3 - 4x - 6y - 2 )( 3x + 3y - 3 + 4x + 6y + 2 )
= ( -x - 3y - 5 )( 7x + 9y - 1 )
c) -4x2 + 12xy - 9y2 + 25
= 25 - ( 4x2 - 12xy + 9y2 )
= 52 - ( 2x - 3y )2
= [ 5 - ( 2x - 3y ) ][ 5 + ( 2x - 3y ) ]
= ( 5 - 2x + 3y )( 5 + 2x - 3y )
d) x2 - 2xy + y2 - 4m2 + 4mn - n2
= ( x2 - 2xy + y2 ) - ( 4m2 - 4mn + n2 )
= ( x - y )2 - ( 2m - n )2
= [ ( x - y ) - ( 2m - n ) ][ ( x - y ) + ( 2m - n ) ]
= ( x - y - 2m + n )( x - y + 2m - n )
a) \(\left(2x+3\right)^2-3\left(x-4\right)\left(x+4\right)=\left(x-2\right)^2+1\)
\(\Leftrightarrow4x^2+12x+9-3\left(x^2-16\right)=x^2-4x+4+1\)
\(\Leftrightarrow4x^2+12x+9-3x^2+48=x^2-4x+5\)
\(\Leftrightarrow x^2+12x+57=x^2-4x+5\)
\(\Leftrightarrow16x+52=0\)
\(\Leftrightarrow x=-\frac{13}{4}\)
b) \(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(\Leftrightarrow\)Xem lại đề !
c) \(x\left(x-1\right)-\left(x-3\right)\left(x+4\right)=5x\)
\(\Leftrightarrow x^2-x-x^2-x+12=5x\)
\(\Leftrightarrow-2x+12=5x\)
\(\Leftrightarrow7x-12=0\)
\(\Leftrightarrow x=\frac{12}{7}\)
d) \(\left(2x+1\right)\left(2x-1\right)=4x\left(x-7\right)-3x\)
\(\Leftrightarrow4x^2-1=4x^2-28x-3x\)
\(\Leftrightarrow28x+3x-1=0\)
\(\Leftrightarrow31x-1=0\)
\(\Leftrightarrow x=\frac{1}{31}\)
a) \(4x^2-4x+1=\left(2x-1\right)^2\)
\(3x\left(x-5\right)-x\left(4+3x\right)=43\)
\(\Leftrightarrow3x^2-15x-4x-3x^2=43\)
\(\Leftrightarrow-19x=43\)
\(\Leftrightarrow x=\frac{-43}{19}\)