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a)5x.(x-2y)+2.(2y-x)2 = 5x.(x-2y)+2.(x-2y)2 = (x-2y)[ 5x+2(x-2y)] = (x-2y)( 5x + 2x - 4y) = (x-2y)(7x-4y)
b) tương tự như trước, do là (A-B)^2 = a^2 - 2ab + b^2 = b^2 - 2ab + a^2 = ( b-a)^2
c)100x2-(x2+25)2= (10x)^2 -(x2+25)2= (10x + x^2 + 25)( 10x-x^2-25) = (x^2+10x+25)[-(x^2-10x+25)] =-1(x+5)^2 (x-5)^2
d)x2-xz-9y2+3y2= ( từ để suy nghĩ đã -_-)
e)x3-x2.5x+125 =
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)
a)
\(4x^2-9y^2+6x-9y=\left(2x-3y\right)\left(2x+3\right)+3\left(2x-3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y+3\right)\)
b)
\(1-2x+2yz+x^2-y^2-z^2=\left(x^2-2x+1\right)-\left(y^2-2yz+z^2\right)\) (đổi dấu)
\(=\left(x-1\right)^2-\left(y-z\right)^2\)
c)
\(x^3-1+5x^2-5+3x-3=\left(x-1\right)\left(x^2+x+1\right)+5\left(x-1\right)\left(x+1\right)+3\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1+5\left(x+1\right)+3\right)\)
\(=\left(x-1\right)\left(x^2+x+1+5x+5+3\right)\)
\(=\left(x-1\right)\left(x^2+6x+9\right)=\left(x-1\right)\left(x+3\right)^2\)
\(\left(x-2\right)^3-1=\left(x-2\right)\left[\left(x-3\right)^2+x-2\right]=\left(x-2\right)\left(x^2+5x+7\right)\)
\(\left(x+3y\right)^2-9y^2=x\left(x+6y\right)\)
\(\left(x+3\right)^2-\left(x-1\right)^2=4\left(2x+4\right)=8\left(x+2\right)\)
a) \(\left(x-2\right)^3-1=\left(x-2\right)^3-1^3=\left(x-2-1\right)\left[\left(x-2\right)^2+\left(x-2\right)\cdot1+1^2\right]\)\(=\left(x-3\right)\left(x^2-4x+4+x-2+1\right)\)
\(=\left(x-3\right)\left(x^2-3x+3\right)\)
b) \(\left(x+3y\right)^2-9y^2\)
\(=\left(x+3y\right)^2-\left(3y\right)^2\)
\(=\left(x+3y+3y\right)\left(x+3y-3y\right)\)
\(=x\left(x+6y\right)\)
c) \(\left(x+3\right)^2-\left(x-1\right)^2\)
\(=\left(x+3-x+1\right)\left(x+3+x-1\right)\)
\(=4\left(2x+2\right)\)
\(=8\left(x+1\right)\)
a) \(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
b) \(5x^2-5xy-3x+3y\)
\(=5x\left(x-y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(5x-3\right)\)
c) \(x^2-2x-4y^2+1\)
\(=\left(x-1\right)^2-4y^2\)
\(=\left(x-2y-1\right)\left(x+2y-1\right)\)
a) ( x-3y ) ( x + 1 )
b) ( x+y+5 ) ( x+y-5 )
c) ( x-5 ) ( x+2 )
Hk tốt
a) Hạng tử z là thừa nhé!
b) \(x^3-x^2-5x+125\)
\(=\left(x^3+5^3\right)-\left(x^2+5x\right)\)
\(=\left(x+5\right)\left(x^2+25-5x\right)-\left(x+5\right)x\)
\(=\left(x+5\right)\left(x^2+25-6x\right)\)
a ) \(x^3-x^2-5x+125\)
\(=\left(x^3+125\right)-\left(x^2+5x\right)\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
b ) \(x^3+2x^2-6x-27\)
\(=\left(x^3-27\right)+\left(2x^2-6x\right)\)
\(=\left(x-3\right)\left(x^2+3x+9\right)+2x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+3x+9+2x\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
a) x3 - x2 - 5x + 125
=(x3-6x2+25x)+(5x2-30x+125)
=x(x2-6x+25)+5(x2-6x+25)
=(x+5)(x2-6x+25)
b) x3 + 2x2 - 6x - 27
=x3+5x2+9-3x2-15x-27
=x(x2+5x+9)-3(x2+5x+9)
=(x-3)(x2+5x+9)
câu b nè : ta có x3 - x2 - 5x + 125 = x ^3 + 5* x^2 - 6x^2 - 30x + 25 x + 125 = x^2 (x+5 ) - 6x (x + 5) + 25 (x+5)
= (x+ 5 ) (x^2 - 6x + 25)
nha bạn
câu a nè bạn :
có cl í tin người vcl !!!