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a, Ta có : \(x^2\left(3y-4\right)-4\left(3y-4\right)\)
\(=\left(3y-4\right)\left(x^2-4\right)\)
\(=\left(3y-4\right)\left(x-2\right)\left(x+2\right)\)
b, Ta có : \(x^3-x^2-5x+125\)
\(=\left(x^3+125\right)-\left(x^2+5x\right)\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
\(3x\left(x-5\right)-x\left(4+3x\right)=43\)
\(\Leftrightarrow3x^2-15x-4x-3x^2=43\)
\(\Leftrightarrow-19x=43\)
\(\Leftrightarrow x=\frac{-43}{19}\)
(X-y-4)2-(2x+3y-1)2
=(X-Y-4-2X-3Y+1)(X-Y-4+2X+3Y-1)=(-X-4Y-3)(X+2Y-5)
(2x2+1)2+6(2x2+1)+9
=(2X2+1+3)2 (dùng hằng đẳng thức a2 +2ab+b2 =(a+b)2
=(2x2+4)2=(2(x2+2))2=4(x2+2)2
\(a,\left(x-y-4\right)^2-\left(2x+3y-1\right)^2\)
\(=\left(3x-2y-5\right)\left(-x-4y-3\right)\)
\(b,\left(2x^2+1\right)^2+6\left(2x^2+1\right)+9\)
\(=\left(2x^2+4\right)^2\)
A = x8 + 2x5 - 2x4 + x2 - 2x - 100 + 10x.(x4 + x) + (5x - 1)2
A = (x8 + 2x5 + x2) - (2x4 + 2x) + 10x.(x4 + x) + (5x - 1)2 - 100
A = (x4 + x)2 - 2(x4 + x) + 10x. (x4 + x) + (5x -1)2 - 100
A = (x4 + x)2 + (x4 + x).(10x - 2) + (5x - 1)2 - 100
A = [(x4 + x)2 + 2.(x4 + x).(5x - 1) + (5x - 1)2 ] - 100
A = [x4 + x + 5x - 1]2 - 102
A = (x4 + 6x - 11).(x4 + 6x + 9)
Hok tốt ^_^
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) \(=x^2+2xy+y^2-x^2+y^2=2xy+2y^2=2y\left(x+y\right)\)
b) \(=\left(x^2-4y^2\right)-\left(2x+4y\right)=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)
c) \(=3\left[\left(x^2+2xy+y^2\right)-z^2\right]=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y+z\right)\left(x+y-z\right)\)
d) \(=\left(2xy+1+2x+y\right)\left(2xy+1-2x-y\right)\)
e) \(=\left(x-3\right)\left(x^2+3x+9\right)-2x\left(x-3\right)=\left(x-3\right)\left(x^2+x+9\right)\)
f) \(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)=\left(x+5\right)\left(x^2-6x+25\right)\)
a) \(\left(x+y\right)^2-\left(x^2-y^2\right)\)
\(=x^2+2xy+y^2-x^2+y^2\)
\(=2y^2+2xy\)
\(=2y\left(x+y\right)\)
c) \(3x^2+6xy+3y^2-3z^2\)
\(=3\left(x^2+2xy+y^2-x^2\right)\)
\(=3\left[\left(x+y\right)^2-z^2\right]\)
\(=3\left(x+y+z\right)\left(x+y-z\right)\)
d) \(\left(2xy+1\right)^2-\left(2x+y\right)^2\)
\(=\left(2xy+1+2x+y\right)\left(2xy+1-2x-y\right)\)
\(=\left[\left(2xy+2x\right)+\left(y+1\right)\right]\left[\left(2xy-2x\right)-\left(y-1\right)\right]\)
\(=\left[2x\left(y+1\right)+\left(y+1\right)\right]\left[2x\left(y-1\right)-\left(y-1\right)\right]\)
\(=\left(2x+1\right)\left(y+1\right)\left(2x-1\right)\left(y-1\right)\)
\(=\left(4x^2-1\right)\left(y^2-1\right)\)
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 =
bài 1
a) x2(3y-4)-4(3y-4)= (3y-4)(x2-4)= (3y-4)(x+2)(x-2)