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a) x3 + 2x2 + x
= x3 + x2 + x2 + x
= x2 ( x + 1 ) + x ( x + 1 )
= ( x2 + x ) ( x + 1 )
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
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2)
a)x2-y2=(x+y).(x-y)=(87+13).(87-13)=100.74=7400
b)x3-3x2+3x-1=(x-1)3=(101-1)3=1003=1000000
c)x3+9x2+27x+27=(x+3)3=(97+3)3=1003=1000000
4)
a)x2-6x+10=x2-6x+9+1=(x-3)2+1>=1>0 voi moi x
b)4x-x2-5= -(x2-4x+5)= -(x2-4x+4+1)= -(x-2)2 - 1<0 voi moi x
Bài 1 :
a ) \(x^2-6x-y^2+9=\left(x^2-6x+9\right)-y^2=\left(x-3\right)^2-y^2=\left(x-3+y\right)\left(x-3-y\right)\)
b) \(25-4x^2-4xy-y^2=5^2-\left(4x^2+4xy+y^2\right)=5^2-\left(2x+y\right)^2=\left(5+2x+y\right)\left(5-2x-y\right)\)
c) \(x^2+2xy+y^2-xz-yz=\left(x+y\right)^2-z.\left(x+y\right)=\left(x+y\right)\left(x+y-z\right)\)
d) \(x^2-4xy+4y^2-z^2+4tz-4t^2=\left(x^2-4xy+4y^2\right)-\left(z^2-4tz+4t^2\right)\)
\(=\left(x-2y\right)^2-\left(z-2t\right)^2=\left(x-2y+z-2t\right).\left(x-2y-z+2t\right)\)
BÀi 2 :
a) \(ax^2+cx^2-ay+ay^2-cy+cy^2=\left(ax^2+cx^2\right)-\left(ay+cy\right)+\left(ay^2+cy^2\right)\)
\(=x^2.\left(a+c\right)-y\left(a+c\right)+y^2.\left(a+c\right)=\left(a+c\right).\left(x^2-y+y^2\right)\)
b) \(ax^2+ay^2-bx^2-by^2+b-a=\left(ax^2-bx^2\right)+\left(ay^2-by^2\right)-\left(a-b\right)\)
\(=x^2.\left(a-b\right)+y^2.\left(a-b\right)-\left(a-b\right)=\left(a-b\right)\left(x^2+y^2-1\right)\)
c) \(ac^2-ad-bc^2+cd+bd-c^3=\left(ac^2-ad\right)+\left(cd+bd\right)-\left(bc^2+c^3\right)\)
\(=-a.\left(d-c^2\right)+d.\left(b+c\right)-c^2.\left(b+c\right)=\left(b+c\right).\left(d-c^2\right)-a\left(d-c^2\right)\)
\(=\left(b+c-a\right)\left(d-c^2\right)\)
BÀi 3 :
a) \(x.\left(x-5\right)-4x+20=0\) \(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\) \(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x-5=0\\x-4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\\x=4\end{cases}}}\)
b) \(x.\left(x+6\right)-7x-42=0\)\(\Leftrightarrow x.\left(x+6\right)-7.\left(x+6\right)=0\) \(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x+6=0\\x-7=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-6\\x=7\end{cases}}}\)
c) \(x^3-5x^2+x-5=0\) \(\Leftrightarrow x^2.\left(x-5\right)+\left(x-5\right)=0\) \(\Leftrightarrow\left(x-5\right)\left(x^2+1\right)\)
\(\Leftrightarrow\hept{\begin{cases}x^2+1=0\\x-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2=-1\left(KTM\right)\\x=5\end{cases}}}\)
d) \(x^4-2x^3+10x^2-20x=0\) \(\Leftrightarrow x.\left(x^3-2x^2+10x-20\right)=0\)\(\Leftrightarrow x.\left[x^2.\left(x-2\right)+10.\left(x-2\right)\right]=0\) \(\Leftrightarrow x.\left(x-2\right)\left(x^2+10=0\right)\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x-2=0\\x^2+10=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\left(KTM\right)\end{cases}}}\)
kết quả thôi nha
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