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a) \(^{x^4-y^4}\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left[\left(x-y\right).\left(x+y\right)\right].\left(x^2-y^2\right)\)
\(=\left(x-y\right).\left(x+y\right).\left(x^2-y^2\right)\)
c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)
\(=\left[\left(3x-2y\right)+\left(2x-3y\right)\right].\left[\left(3x-2y\right)-\left(2x-3y\right)\right]\)
\(=\left(3x-2y+2x-3y\right)\left(3x-2y-2x+3y\right)\)
b) \(x^2-3y^2\)
\(=\left(x-3y\right)\left(x+3y\right)\)
d) \(9\left(x-y\right)^2-4\left(x+y\right)^2\)
\(=9\left(x-y\right)^2+4\left(x-y\right)^2\)
\(=\left(x-y\right).\left(9+4\right)\)
\(=\left(x-y\right).13\)
\(=13\left(x-y\right)\)
f) \(x^3+27\)
\(=x^3+3^3\)
\(=\left(x+3\right)\left(x^2-x.3+3^2\right)\)
h) \(125x^3-1\)
\(=\left(5x\right)^3-1^3\)
\(=\left(5x-1\right)\left(5x^2+5x.1+1^2\right)\)
\(=\left(5x-1\right)\left(5x^2+5x+1\right)\)
\(a,x^4-y^4=\left(x^2+y^2\right)\left(x^2-y^2\right)=\left(x^2+y^2\right)\left(x+y\right)\left(x-y\right)\)
\(b,x^2-3y^2=\left(x+\sqrt{3}y\right)\left(x-\sqrt{3}y\right)\)
cn lại tg tự nha bn
\(a,y^4-14y^2+49\)
\(\left(y^2-7\right)^2\)
\(b,x^2-2\)
\(x^2-\left(\sqrt{2}\right)^2=\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\)
\(c,y^2-13\)
\(y^2-\left(\sqrt{13}\right)^2=\left(y-\sqrt{13}\right)\left(y+\sqrt{13}\right)\)
\(d,-4x^2+9y^2\)
\(\left(3y\right)^2-\left(2x\right)^2\)
\(\left(3y-2x\right)\left(3y+2x\right)\)
\(3x\left(x-5\right)-x\left(4+3x\right)=43\)
\(\Leftrightarrow3x^2-15x-4x-3x^2=43\)
\(\Leftrightarrow-19x=43\)
\(\Leftrightarrow x=\frac{-43}{19}\)
a) Đặt t = x2
bthuc <=> t2 - 7t + 16
Từ đây ta không thể phân tích được :)
b) x3 - 2x2 + 5x - 4
= x3 - x2 - x2 + x + 4x - 4
= x2( x - 1 ) - x( x - 1 ) + 4( x - 1 )
= ( x - 1 )( x2 - x + 4 )
c) x3 - 2x2 + x - 3 ( phân tích hổng ra :)) )
d) 3x3 - 4x2 + 12x - 4 ( phân tích hổng ra p2 :)) )
e) 6x3 + x2 + x + 1
= 6x3 + 3x2 - 2x2 - x + 2x + 1
= 3x2( 2x + 1 ) - x( 2x - 1 ) + ( 2x + 1 )
= ( 2x + 1 )( 3x2 - x + 1 )
f) 4x3 + 6x2 + 4x + 1
= 4x3 + 2x2 + 4x2 + 2x + 2x + 1
= 2x2( 2x + 1 ) + 2x( 2x + 1 ) + ( 2x + 1 )
= ( 2x + 1 )( 2x2 + 2x + 1 )
a) \(x^3+x^2-2x-8\)
\(=\left(x^3-2x^2\right)+\left(3x^2-6x\right)+\left(4x-8\right)\)
\(=x^2\left(x-2\right)+3x\left(x-2\right)+4\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+3x+4\right)\)
b) \(125x^3-10x^2+2x-1\)
\(=\left(125x^3-25x^2\right)+\left(15x^2-3x\right)+\left(5x-1\right)\)
\(=25x^2\left(5x-1\right)+3x\left(5x-1\right)+\left(5x-1\right)\)
\(=\left(5x-1\right)\left(25x^2+3x+1\right)\)
c) \(x^3-4x^2+12x-27\)
\(=\left(x^3-3x^2\right)-\left(x^2-3x\right)+\left(9x-27\right)\)
\(=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
d) Đề sai sai, nghiệm ra khá xấu nên bạn xem lại nhé
e) \(x^3-3x^2-3x+1\)
\(=\left(x^3+x^2\right)-\left(4x^2+4x\right)+\left(x+1\right)\)
\(=x^2\left(x+1\right)-4x\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
1.a) 2x4-4x3+2x2
=2x2(x2-2x+1)
=2x2(x-1)2
b) 2x2-2xy+5x-5y
=2x(x-y)+5(x-y)
=(2x+5)(x-y)
2.
a) 4x(x-3)-x+3=0
=>4x(x-3)-(x-3)=0
=>(4x-1)(x-3)=0
=> 2 TH:
*4x-1=0 *x-3=0
=>4x=0+1 =>x=0+3
=>4x=1 =>x=3
=>x=1/4
vậy x=1/4 hoặc x=3
b) (2x-3)^2-(x+1)^2=0
=> (2x-3-x-1).(2x-3+x+1)=0
=>(x-4).(3x-2)=0
=> 2 TH
*x-4=0
=> x=0+4
=> x=4
*3x-2=0
=>3x=0-2
=>3x=-2
=>x=-2/3
vậy x=4 hoặc x=-2/3
a) \(x^2-xy+4x-2y+4\)
\(=\left(x^2+4x+4\right)-\left(xy+2y\right)\\ =\left(x+2\right)^2-y.\left(x+2\right)\)
\(=\left(x+2\right).\left(x+2-y\right)\)
b) \(2x^2-5x-3\)
\(=2x^2+x-6x-3\)
\(=\left(2x^2+x\right)-\left(6x+3\right)=x\left(2x+1\right)-3\left(2x+1\right)\)
\(=\left(2x+1\right).\left(x-3\right)\)
c)\(\)
c);d);e) tạm thời tớ chưa nghĩ ra-.-"
tham khả tạm 2 câu ạ, chúc học tốt'.'
a) \(x^2+4x+3\)
\(=x^2+3x+x+3\)
\(=x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
1: \(4x^2+4x+1=\left(2x+1\right)^2\)
2: \(x^2-20x+100=\left(x-10\right)^2\)
3: \(y^4-14y^2+49=\left(y^2-7\right)^2\)
4: \(125x^3-64y^3=\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)\)