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a, 8x^3 - 1/8
= (2x)^3 - (1/2)^3
= ( 2x - 1/2) ( 4x^2 + x + 1/4)
b, 1/25.x^2 - 64y^2 = (1/5x)^2 - (8y)^2 = ( 1/5x - 8y)(1/5x+8y)
Đúng xho mình nha )
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)
\(x^2+6x+9=x^2+2.3x+3^2=\left(x+3\right)^2\)
\(10x-25-x^2=-\left(x^2-2.5x+5^2\right)=-\left(x-5\right)^2\)
\\(8x^3-\frac{1}{8}=\left(2x\right)^3-\left(\frac{1}{2}\right)^3=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)
\(\frac{1}{25}x^2-64y^2=\left(\frac{1}{5}x\right)^2-\left(8y\right)^2=\left(\frac{1}{5x}-8y\right)\left(\frac{1}{5x}+8y\right)\)
1/ phân tích thành nhân tử ;
= C2-( a +b )2=( c-a -b ) . ( c+a +b )
a)\(6x^2-9xy\)
\(=3x\left(2x-3y\right)\)
b)\(x^2-y^2-3x+3y\)
\(=\left(x-y\right)\left(x+y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-3\right)\)
c)\(x^4-8x^2-9\)
\(=x^4+x^2-9x^2-9\)
\(=x^2\left(x^2+1\right)-9\left(x^2+1\right)\)
\(=\left(x^2-9\right)\left(x^2+1\right)\)
\(=\left(x+3\right)\left(x-3\right)\left(x^2+1\right)\)
d)\(x^4-4\left(x^2+5\right)-25\)
\(=\left(x^2-5\right)\left(x^2+5\right)-4\left(x^2+5\right)\)
\(=\left(x^2+5\right)\left(x^2-5-4\right)\)
\(=\left(x^2+5\right)\left(x^2-9\right)\)
\(=\left(x^2+5\right)\left(x-3\right)\left(x+3\right)\)
a) 5x ( x - 2000 ) - x + 2000 = 0
5x ( x - 2000 ) - ( x - 2000 ) = 0
5x ( x - 2000 ) = 0
\(\Rightarrow\orbr{\begin{cases}5x=0\\x-2000=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2000\end{cases}}\)
Vậy ....
b) x3 - 13x = 0
x ( x2 - 13 ) = 0
x ( x - \(\sqrt{13}\)) - ( x + \(\sqrt{13}\)) = 0
\(\Rightarrow\hept{\begin{cases}x=0\\x-\sqrt{13}\\x+\sqrt{13}\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\x=\sqrt{13}\\x=\sqrt{-13}\end{cases}}\)
Vậy ....
a) x2 + 6 + 9
= x2 + 2 . 3 . x + 32
= ( x + 3 )2
b) 10x - 25 - x2
= - ( x2 - 10x + 25 )
= - ( x - 5 )2
c) 8x3 - 1/8
= ( 2x )3 - ( 1/2 )3
= ( 2x - 1/2 ) ( 4x2 + x + 1/4 )
d) 1/25 x2 - 64x2
= ( 1/5x )2 - ( 8x )2
= ( 1/5x + 8x ) ( 1/5 - 8x )
\(x^3-13x=0\)
<=> \(x\left(x^2-13\right)=0\)
<=> \(x\left(x-\sqrt{13}\right)\left(x+\sqrt{13}\right)=0\)
<=> \(x=0\)
hoặc \(x-\sqrt{13}=0\)
hoặc \(x+\sqrt{13}=0\)
<=> .....
a) x2 + 7x + 12 = x2 + 3x + 4x + 12 = x( x + 3 ) + 4( x + 3 ) = ( x + 3 )( x + 4 )
b) x2 - 10x + 16 = x2 - 2x - 8x + 16 = x( x - 2 ) - 8( x - 2 ) = ( x - 2 )( x - 8 )
c) x2 + 6x + 8 = x2 + 2x + 4x + 8 = x( x + 2 ) + 4( x + 2 ) = ( x + 2 )( x + 4 )
d) x2 - 8x + 15 = x2 - 3x - 5x + 15 = x( x - 3 ) - 5( x - 3 ) = ( x - 3 )( x - 5 )
e) x2 - 8x - 9 = x2 + x - 9x - 9 = x( x + 1 ) - 9( x + 1 ) = ( x + 1 )( x - 9 )
f) x2 + 14x + 48 = x2 + 6x + 8x + 48 = x( x + 6 ) + 8( x + 6 ) = ( x + 6 )( x + 8 )
a) \(8x^3-\frac{1}{8}\)
\(=\left(2x\right)^3-\left(\frac{1}{2}\right)^3\)
\(=\left(2x-\frac{1}{2}\right)\left(4x^2+2x+\frac{1}{4}\right)\)
b) \(\frac{1}{25}x^2-64y^2\)
\(=\left(\frac{1}{5}x\right)^2-\left(8y\right)^2\)
\(=\left(\frac{1}{5}x-8y\right)\left(\frac{1}{5}x+8y\right)\)
a) x2 + 6x + 9 = x2 + 2.3.x + 32 = (x + 3)2
b) 10x - 25 - x2 = - (x2 - 2.5.x + 52) = - (x - 5)2
c) \(8x^3-\dfrac{1}{8}=\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
d) 25x2 - 64y2 = (5x)2 - (8y)2 = (5x - 8y)(5x + 8y)
\(a,x^2+6x+9=\left(x+3\right)^2\)