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a) Đăt \(x^2+x=t\) khi đó bt trở thành:
\(t^2-2t-15=t^2+3t-5t-15=t\left(t+3\right)-5\left(t+3\right)\\ =\left(t+3\right)\left(1-5\right)=\left(x^2+x+3\right)\left(x^2+x-5\right)\)
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)18x2-12x
=3x(6x-4)
b)3x2-11x+6
=x(3x-11+6)
=x(3x-5)
c)x3+6x2+11x+6
=x2(x+23
\(18x^2-12x\)
\(=6x\left(3x-2\right)\)
\(3x^2-11x+6\)
\(=3x^2-9x-2x+6\)
\(=3x\left(x-3\right)-2\left(x-3\right)\)
\(=\left(x-3\right)\left(3x-2\right)\)
a) x2 - 3x + 2 = x2 - x - 2x + 2 = x( x - 1 ) - 2( x - 1 ) = ( x - 1 )( x - 2 )
b) 2x2 - x - 6 = 2x2 - 4x + 3x - 6 = 2x( x - 2 ) + 3( x - 2 ) = ( x - 2 )( 2x + 3 )
c) x2 - 5x - 6 = x2 + x - 6x - 6 = x( x + 1 ) - 6( x + 1 ) = ( x + 1 )( x - 6 )
d) x2 + 8x + 7 = x2 + x + 7x + 7 = x( x + 1 ) + 7( x + 1 ) = ( x + 1 )( x + 7 )
e) 3x2 + 2x - 5 = 3x2 - 3x + 5x - 5 = 3x( x - 1 ) + 5( x - 1 ) = ( x - 1 )( 3x + 5 )
f) 4x2 - 3x - 1 = 4x2 - 4x + x - 1 = 4x( x - 1 ) + ( x - 1 ) = ( x - 1 )( 4x + 1 )
a \(x^2-3x+2=x^2-x-2x+2=\left(x-1\right)\left(x-2\right)\)
b, \(2x^2-x-6=2x^2-4x+3x-6=\left(x-2\right)\left(2x+3\right)\)
c, \(x^2-5x-6=x^2+x-6x-6=\left(x+1\right)\left(x-6\right)\)
d, \(x^2+8x+7=x^2+x+7x+7=\left(x+1\right)\left(x+7\right)\)
e, \(3x^2+2x-5=3x^2-3x+5x-5=\left(x-1\right)\left(3x+5\right)\)
f, \(4x^2-3x-1=4x^2-4x+x-1=\left(x-1\right)\left(4x+1\right)\)
\(2x^2+3x-27=2x^2-6x+9x-27=2x\left(x-3\right)+9\left(x-3\right)=\left(2x+9\right)\left(x-3\right)\)
\(x^3-7x+6=x^3-x-6x+6=x\left(x^2-1\right)-6\left(x-1\right)=x\left(x-1\right)\left(x+1\right)-6\left(x-1\right)=\left(x-1\right)\left(x^2+x-6\right)\)
\(x^3+5x^2+8x+4=x^3+x^2+4x^2+8x+4=x^2\left(x+1\right)+4\left(x^2+2x+1\right)=x^2\left(x+1\right)+4\left(x+1\right)^2\)
\(=\left(x+1\right)\left(x^2+4x+4\right)=\left(x+1\right)\left(x+2\right)^2\)
\(27x^3-27x^2+18x-4=27x^3-9x^2-18x^2+6x+12x-4\)
\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)=\left(3x-1\right)\left(9x^2-6x+4\right)\)
a) ( x2 + x )2 - 2( x2 + x ) - 15 (*)
Đặt t = x2 + x
(*) <=> t2 - 2t - 15
= t2 + 3t - 5t - 15
= t( t + 3 ) - 5( t + 3 )
= ( t + 3 )( t - 5 )
= ( x2 + x + 3 )( x2 + x - 5 )
b) ( x2 + 2x )2 + 9x2 + 18x + 20
= ( x2 + 2x )2 + 9( x2 + 2x ) + 20 (*)
Đặt t = x2 + 2x
(*) <=> t2 + 9t + 20
= t2 + 4t + 5t + 20
= t( t + 4 ) + 5( t + 4 )
= ( t + 4 )( t + 5 )
= ( x2 + x + 4 )( x2 + x + 5 )
c) ( x2 + 3x + 1 )( x2 + 3x + 2 ) - 6 (*)
Đặt t = x2 + 3x + 1
(*) <=> t( t + 1 ) - 6
= t2 + t - 6
= t2 - 2t + 3t - 6
= t( t - 2 ) + 3( t - 2 )
= ( t - 2 )( t + 3 )
= ( x2 + 3x + 1 - 2 )( x2 + 3x + 1 + 3 )
= ( x2 + 3x - 1 )( x2 + 3x + 4 )
d) ( x2 + 8x + 7 )( x + 3 )( x + 5 ) + 15
= ( x2 + 8x + 7 )( x2 + 8x + 15 ) + 15 (*)
Đặt t = x2 + 8x + 7
(*) <=> t( t + 8 ) + 15
= t2 + 8t + 15
= t2 + 3t + 5t + 15
= t( t + 3 ) + 5( t + 3 )
= ( t + 3 )( t + 5 )
= ( x2 + 8x + 7 + 3 )( x2 + 8x + 7 + 5 )
= ( x2 + 8x + 10 )( x2 + 8x + 12 )