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\(\left(x^2+x\right)^2-2x^2-2x-15\)
\(=\left(x^2+x\right)^2-\left(2x^2+2x+15\right)\)
\(=\left(x^2+x\right)^2-\left[\left(2x^2+2x\right)+15\right]\)
\(=\left(x^2+x\right)^2-\left[2.\left(x^2+x\right)+15\right]\)
\(=\left(x^2+x\right)^2-2\left(x^2+x\right)-15\) \(\left(1\right)\)
đặt \(x^2+x=t\)
\(\left(1\right)\)\(=\) \(t^2-2t-15\)
\(=\left(t-1\right)^2-16\)
\(=\left(t-1-4\right)\left(t-1+4\right)\)
\(=\left(t-5\right)\left(t+3\right)\)
thay \(t=x^2+x\) ta có
\(\left(1\right)=\left(x^2+x-5\right)\left(x^2+x+3\right)\)
các câu còn lại tương tự nha
học tốt
a) 1/2(x3+8)=1/2(x+2)(x2-2x+4)
b) x4(x-y)+2x3(x-y)=x3(x+2)(x-y)
c) x2-(y2-6y+9)=x2-(y-3)2=(x-y+3)(x+y-3)
d) xy(x3+y3)=xy(x+y)(x2-xy+y2)
e)3x2(x2-25y2)=3x2(x-5y)(x+5y)
f) 4x4+4x2y2+y4-4x2y2= (2x2+y2)2-(2xy)2=(2x2-2xy+y2)(2x2+2xy+y2)
a) \(\frac{1}{2}x^3+4=\frac{1}{2}\left(x^3+8\right)=\frac{1}{2}\left(x+2\right)\left(x^2-2x+4\right)\)
b) \(x^5-x^4y+2x^4-2x^3y=x^3\left(x^2-xy+2x-2y\right)=x^3\left[x\left(x-y\right)+2\left(x-y\right)\right]=x^2\left(x-y\right)\left(x+2\right)\)
c) \(x^2-y^2+6y-9=x^2-\left(y-3\right)^2=\left(x+y-3\right)\left(x-y+3\right)\)
d) \(x^4y+xy^4=xy\left(x^3+y^3\right)=xy\left(x+y\right)\left(x^2-xy+y^2\right)\)
e) \(3x^4-75x^2y^2=3x^2\left(x^2-25y^2\right)=3x^2\left(x+5y\right)\left(x-5y\right)\).
f) \(4x^4+y^4=\left(2x^2+y^2\right)^2-\left(2xy\right)^2=\left(2x^2+y^2+2xy\right)\left(2x^2-y^2-2xy\right)\)
a)x7+x5+1=x7+x6-x6+2x5-x5+x4-x4+x3-x3+x2-x2+1
=x7-x6+x5-x3+x2+x6-x5+x4-x2+x+x5-x4+x3-x+1
=x2(x5-x4+x3-x+1)+x(x5-x4+x3-x+1)+1(x5-x4+x3-x+1)
=(x2+x+1)(x5-x4+x3-x+1)
b)4x4-32x2+1=4x4+12x3+2x2-12x3-36x2-6x+2x2+6x+1
=2x2(2x2+6x+1)-6x(2x2+6x+1)+1(2x2+6x+1)
=(2x2-6x+1)(2x2+6x+1)
c)x6+27=(x2+3)(x2-3x+3)(x2+3x+3)
d)3(x4+x2+1)-(x2+x+1)
=3x4-3x3+2x2+3x3-3x2+2x+3x2-3x+2
=x2(3x2-3x+2)+x(3x2-3x+2)+1(3x2-3x+2)
=(x2+x+1)(3x2-3x+2)
e)bạn tự làm nhé
1/(x+2)2 -(3x-1)2=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x2+12x
2/(x4+x2)(-2x3-2x)=x2(x2+1)-2x(x2+1)=(x2+1)(x2-2x)
Bài 2:
\(\left(5x+1\right)^2-\left(2xy-3\right)^2\)
\(=25x^2+10x+1-\left(2xy-3\right)^2\)
\(=25x^2+10x+1\left(4x^2y^2-12xy+9\right)\)
\(=25x^2+10x+1-4x^2y^2+12xy-9\)
\(=25x^2-4x^2y^2+10x+12xy-8\)
Bài 2:
\(\left(x-1\right)\left(x^2+x+1\right)=x^2\left(x-9\right)+2x+6\)
\(=x^3-1=x^3-9x^2+2x+6\)
\(=x^3-9x^2+2x+6=x^3-1\)
\(=x^3-9x^2+2x+6+1=x^3-1+1\)
\(=x^3-9x^2+2x+7=x^3\)
\(=x^3-9x^2+2x+7-x^3=x^3-x^3\)
\(=-9x^2+2x+7=0\)
\(\Rightarrow x=-\frac{7}{9};x=1\)
Bài 1.
a) -2x( -3x + 2 ) - ( x + 2 )2
= 6x2 - 4x - ( x2 + 4x + 4 )
= 6x2 - 4x - x2 - 4x - 4
= 5x2 - 8x - 4
b) ( x + 2 )( x2 - 2x + 4 ) - 2( x + 1 )( 1 - x )
= x3 + 8 + 2( x + 1 )( x - 1 )
= x3 + 8 + 2( x2 - 1 )
= x3 + 8 + 2x2 - 2
= x3 + 2x2 + 6
c) ( 2x - 1 )2 - 2( 4x2 - 1 ) + ( 2x + 1 )2
= 4x2 - 4x + 1 - 8x2 + 2 + 4x2 + 4x + 1
= 4
d) x2 - 3x + xy - 3y
= x( x - 3 ) + y( x - 3 )
= ( x - 3 )( x + y )
Bài 2.
a) 4x2 - 4xy + y2 = ( 2x - y )2
b) 9x3 - 9x2y - 4x + 4y
= 9x2( x - y ) - 4( x - y )
= ( x - y )( 9x2 - 4 )
= ( x - y )( 3x - 2 )( 3x + 2 )
c) x3 + 2 + 3( x3 - 2 )
= x3 + 2 + 3x3 - 6
= 4x3 - 4
= 4( x3 - 1 )
= 4( x - 1 )( x2 + x + 1 )
Bài 3.
2( x - 2 ) = x2 - 4x + 4
⇔ ( x - 2 )2 - 2( x - 2 ) = 0
⇔ ( x - 2 )( x - 2 - 2 ) = 0
⇔ ( x - 2 )( x - 4 ) = 0
⇔ x = 2 hoặc x = 4
a) \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)-4=\left(x^2+6x+5\right)\left(x^2+6x+8\right)-4\)
Đặt \(t=x^2+6x+5\)
\(PT=t\left(t+3\right)-4=t^2+3t-4=\left(t-1\right)\left(t+4\right)\)
Thay t: \(PT=\left(x^2+6x+5-1\right)\left(x^2+6x+5+4\right)=\left(x^2+6x+4\right)\left(x^2+6x+9\right)=\left(x^2+6x+4\right)\left(x+3\right)^2\)
b) Đặt \(t=\left(2x+1\right)^2\)
\(PT=t^2-3t+2=\left(t^2-3t+\dfrac{9}{4}\right)-\dfrac{1}{4}=\left(t+\dfrac{3}{2}\right)^2-\dfrac{1}{4}=\left(t+1\right)\left(t+2\right)\)
Thay t:
\(PT=\left[\left(2x+1\right)^2+1\right]\left[\left(2x+1\right)^2+2\right]=\left[4x^2+4x+2\right]\left[4x^2+4x+3\right]=2\left[2x^2+2x+1\right]\left[4x^2+4x+3\right]\)
a: =(x^2-1)^2-2x(x^2-1)+x(x^2-1)-2x^2
=(x^2-1)(x^2-1-2x)+x(x^2-1-2x)
=(x^2-2x-1)(x^2+x-1)
b: \(=\left(x^2+1\right)^2+x\left(x^2+1\right)+2x\left(x^2+1\right)+2x^2\)
\(=\left(x^2+1\right)\left(x^2+x+1\right)+2x\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2+2x+1\right)\)
\(=\left(x+1\right)^2\cdot\left(x^2+x+1\right)\)