Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
\(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)
\(=\left(x-y+4+2x+3y-1\right)\left(x-y+4-2x-3y+1\right)\)
\(=\left(3x+2y+3\right)\left(-x-4y+5\right)\)
\(49\left(y-4\right)^2-9y^2-36y-36\)
\(=49\left(y-4\right)^2-\left(9y^2+36y+36\right)\)
\(=49\left(y-4\right)^2-\left(3y+6\right)^2\)
\(=[7\left(y-4\right)]^2-\left(3y+6\right)^2\)
\(=\left(7y-28\right)^2-\left(3y+6\right)^2\)
\(=\left(7y-28+3y+6\right)\left(7y-28-3y-6\right)\)
\(=\left(10y-22\right)\left(4y-34\right)\)
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
a) 36 - 4a2 + 20ab - 25b2 = 36 - ( 4a2 - 20ab + 25b2 ) = 62 - ( 2a - 5b )2 = ( 6 - 2a + 5b )( 6 + 2a - 5b )
b) ( xy + 4 )2 - 4( x + y )2 = ( xy + 4 )2 - 22( x + y )2 = ( xy + 4 )2 - [ 2( x + y ) ]2
= ( xy + 4 )2 - ( 2x + 2y )2 = ( xy + 4 - 2x - 2y )( xy + 4 + 2x + 2y )
= [ x( y - 2 ) - 2( y - 2 ) ][ x( y + 2 ) + 2( y + 2 ) ]
= ( y - 2 )( x - 2 )( y + 2 )( x + 2 )
c) x2 + y2 - x2y2 + xy - x - y
= ( x2 - x2y2 ) + ( y2 - y ) + ( xy - x )
= x2( 1 - y2 ) + y( y - 1 ) + x( y - 1 )
= x2( 1 - y )( 1 + y ) - y( 1 - y ) - x( 1 - y )
= ( 1 - y )[ x2( 1 + y ) - y - x ) ]
= ( 1 - y )( x2 + x2y - y - x )
= ( 1 - y )[ ( x2 - x ) + ( x2y - y ) ]
= ( 1 - y )[ x( x - 1 ) + y( x2 - 1 ) ]
= ( 1 - y )[ x( x - 1 ) + y( x - 1 )( x + 1 ) ]
= ( 1 - y )( x - 1 )[ x + y( x + 1 ) ]
= ( 1 - y )( x - 1 )( x + xy + y )
d) 3x + 3y - x2 - 2xy - y2
= 3( x + y ) - ( x2 + 2xy + y2 )
= 3( x + y ) - ( x + y )2
= ( x + y )( 3 - x - y )
e) ( 2xy + 1 )2 - ( 2x + y )2
= ( 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 )( y + 1 )( 2x + 1 )
a) \(36-4a^2+20ab-25b^2\)
\(=36-\left(4a^2-20ab+25b^2\right)\)
\(=36-\left(2a-5b\right)^2\)
\(=\left(6-2a+5b\right)\left(6+2a-5b\right)\)
b) \(\left(xy+4\right)^2-4\left(x+y\right)^2\)
\(=\left(xy+4-2x-2y\right)\left(xy+4+2x+2y\right)\)
\(=\left[x\left(y-2\right)-2\left(y-2\right)\right]\left[x\left(y+2\right)+2\left(y+2\right)\right]\)
\(=\left(x+2\right)\left(x-2\right)\left(y+2\right)\left(y-2\right)\)
c) \(x^2+y^2-x^2y^2+xy-x-y\)
\(=-\left(x^2y^2-x^2\right)+\left(y^2-y\right)+\left(xy-x\right)\)
\(=-x^2\left(y-1\right)\left(y+1\right)+y\left(y-1\right)+x\left(y-1\right)\)
\(=\left(y-1\right)\left(-x^2y-x^2+y+x\right)\)
\(=\left(1-y\right)\left[\left(x^2y-y\right)+\left(x^2-x\right)\right]\)
\(=\left(1-y\right)\left(x-1\right)\left(xy+y+x\right)\)
\(16x^4\left(x-y\right)-x+y\)
\(=\left(x-y\right)\left(16x^2-1\right)\)
\(=\left(x-y\right)\left(4x+1\right)\left(4x-1\right)\)