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\(a^3+a^2c-abc+b^2c+b^3\)
\(=\left(a^3+b^3\right)+\left(a^2c+b^2c-abc\right)\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)+\)\(c\left(a^2+b^2-ab\right)\)
\(=\left(a^2+b^2-ab\right)\left(a+b+c\right)\)
mk chỉnh lại đề nha:
\(x^2-x-6\)
\(=x^2-3x+2x-6\)
\(=x\left(x-3\right)+2\left(x-3\right)\)
\(=\left(x-3\right)\left(x+2\right)\)
a) x^3 - 7x - 6
= x^3 - x - 6x - 6
= x(x^2 - 1 ) - 6 (x + 1 )
= x(x-1)(x+1) - 6 ( x + 1 )
= ( x+ 1 ) [ x(x-1) - 6 ]
= ( x + 1 )(x^2 - x - 6 )
= ( x+ 1 ) ( x^2 - 3x + 2x - 6 )
= ( x+ 1 ) [ x(x-3) + 2 ( x- 3 )]
=(x+1)(x+2)(x-3)
b) x^3 - x^2 - 14x + 24
= x^3 - 3x^2 + 2x^2 - 6x - 8x + 24
= x^2 ( x - 3 ) + 2x(x-3) - 8 ( x- 3 )
= ( x - 3 )( x^2 + 2 x - 8 )
= ( x- 3 ) [ x^2 + 4x - 2x - 8 )]
= ( x- 3 )( [ x( x + 4 ) - 2 ( x+ 4) ]
= ( x - 3 )( x+ 4 )( x- 2 )
c) x^5 + x + 1
= x^5 - x^2 + x^2 + x + 1
= x^2(x^3 - 1 ) + x^2 + x + 1
= x^2 ( x- 1 )(x^2 + x + 1 ) + x^2 + x+ 1
= ( x^2 + x + 1 )( x^3 - x^2 ) + x^2 + x + 1
=( x^2 + x + 1 )( X^3 - x^2 + 1 )
x2+7x+6
giúp mình nha (phân tích thành nhân tử )
nhanh mình k cho
a: \(2x^3+x^2-13x+6\)
\(=2x^3-4x^2+5x^2-10x-3x+6\)
\(=\left(x-2\right)\left(2x^2+5x-3\right)\)
\(=\left(x-2\right)\left(2x^2+6x-x-3\right)\)
\(=\left(x-2\right)\left(x+3\right)\left(2x-1\right)\)
b: \(2x^2+y^2-6x+2xy-2y+5=0\)
\(\Leftrightarrow x^2+2xy+y^2+x^2-4x+4-2x-2y+1=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x-2\right)^2-2\left(x+y\right)+1=0\)
\(\Leftrightarrow\left(x-2\right)^2+\left(x+y-1\right)^2=0\)
=>x-2=0 và x+y-1=0
=>x=2 và y=-1
a, \(x^3-6x^2+9x\)
\(=x\left(x^2-6x+9\right)\)
\(=x \left(x-3\right)\)
Câu b, c cũng tượng tự nha bn , dễ mà
#hoc_tot#
b) \(x^2-2xy+3x-6y=x\left(x-2y\right)+3\left(x-2y\right)=\left(x-2y\right)\left(x+3\right)\)
c)\(x^2-8x+7=x^2-x-7x+7=x\left(x-1\right)-7\left(x-1\right)=\left(x-1\right)\left(x-7\right)\)
a)\(x^3-6x^2+9x=x\left(x^2-2\cdot x\cdot3+3^2\right)=x\left(x-3\right)^2\)
~ Chúc bạn học tốt ~
Bài 1 :
a ) \(x^2-6x-y^2+9=\left(x^2-6x+9\right)-y^2=\left(x-3\right)^2-y^2=\left(x-3+y\right)\left(x-3-y\right)\)
b) \(25-4x^2-4xy-y^2=5^2-\left(4x^2+4xy+y^2\right)=5^2-\left(2x+y\right)^2=\left(5+2x+y\right)\left(5-2x-y\right)\)
c) \(x^2+2xy+y^2-xz-yz=\left(x+y\right)^2-z.\left(x+y\right)=\left(x+y\right)\left(x+y-z\right)\)
d) \(x^2-4xy+4y^2-z^2+4tz-4t^2=\left(x^2-4xy+4y^2\right)-\left(z^2-4tz+4t^2\right)\)
\(=\left(x-2y\right)^2-\left(z-2t\right)^2=\left(x-2y+z-2t\right).\left(x-2y-z+2t\right)\)
BÀi 2 :
a) \(ax^2+cx^2-ay+ay^2-cy+cy^2=\left(ax^2+cx^2\right)-\left(ay+cy\right)+\left(ay^2+cy^2\right)\)
\(=x^2.\left(a+c\right)-y\left(a+c\right)+y^2.\left(a+c\right)=\left(a+c\right).\left(x^2-y+y^2\right)\)
b) \(ax^2+ay^2-bx^2-by^2+b-a=\left(ax^2-bx^2\right)+\left(ay^2-by^2\right)-\left(a-b\right)\)
\(=x^2.\left(a-b\right)+y^2.\left(a-b\right)-\left(a-b\right)=\left(a-b\right)\left(x^2+y^2-1\right)\)
c) \(ac^2-ad-bc^2+cd+bd-c^3=\left(ac^2-ad\right)+\left(cd+bd\right)-\left(bc^2+c^3\right)\)
\(=-a.\left(d-c^2\right)+d.\left(b+c\right)-c^2.\left(b+c\right)=\left(b+c\right).\left(d-c^2\right)-a\left(d-c^2\right)\)
\(=\left(b+c-a\right)\left(d-c^2\right)\)
BÀi 3 :
a) \(x.\left(x-5\right)-4x+20=0\) \(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\) \(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x-5=0\\x-4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\\x=4\end{cases}}}\)
b) \(x.\left(x+6\right)-7x-42=0\)\(\Leftrightarrow x.\left(x+6\right)-7.\left(x+6\right)=0\) \(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x+6=0\\x-7=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-6\\x=7\end{cases}}}\)
c) \(x^3-5x^2+x-5=0\) \(\Leftrightarrow x^2.\left(x-5\right)+\left(x-5\right)=0\) \(\Leftrightarrow\left(x-5\right)\left(x^2+1\right)\)
\(\Leftrightarrow\hept{\begin{cases}x^2+1=0\\x-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2=-1\left(KTM\right)\\x=5\end{cases}}}\)
d) \(x^4-2x^3+10x^2-20x=0\) \(\Leftrightarrow x.\left(x^3-2x^2+10x-20\right)=0\)\(\Leftrightarrow x.\left[x^2.\left(x-2\right)+10.\left(x-2\right)\right]=0\) \(\Leftrightarrow x.\left(x-2\right)\left(x^2+10=0\right)\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x-2=0\\x^2+10=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\left(KTM\right)\end{cases}}}\)
1) \(\left(x^2-9\right)^2+12x\left(x-3\right)^2=\left(x-3\right)^2\left(x+3\right)^2+12x\left(x-3\right)^2\)
\(=\left(x-3\right)^2\left[\left(x+3\right)^2+12x\right]=\left(x-3\right)^2\left(x^2+6x+9+12x\right)\)
\(=\left(x-3\right)^2\left(x^2+18x+9\right)\)
\(=\left(x-3\right)^2\left[\left(x+9\right)^2-72\right]\)
\(=\left(x-3\right)^2\left(x+9-\sqrt{72}\right)\left(x+9+\sqrt{72}\right)\)
2) \(\left(a+b+c\right)^3-a^3-b^3-c^3=\left(a+b\right)^3+3c\left(a+b\right)\left(a+b+c\right)+c^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3ab\left(a+b\right)+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left[ab+c\left(a+b+c\right)\right]\)
\(=3\left(a+b\right)\left(ab+bc+ca+c^2\right)\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
a) \(x^2-2x+1-y^2=\left(x-1\right)^2-y^2=\left(x-1-y\right)\left(x-1+y\right)\)
b)\(=\left(x+y\right)^2-z^2=\left(x+y+z\right)\left(x+y-z\right)\)
mấy ý còn lại tương tự nha
a,\(x^2-y^2+1-2x\)
\(=\left(x-1\right)^2-y^2\)
\(=\left(x-1+y\right)\left(x-1-y\right)\)
\(b,x^2+2xy-z^2+y^2\)
\(=\left(x+y\right)^2-z^2\)
\(=\left(x+y+z\right)\left(x+y-z\right)\)
a) a3+a2c-abc+b2c+b3 =(a3+b3)+(a2c-abc+b2c)=(a+b)(a2-ab+b2)+c(a2-ab+b2)=(a2-ab+b2)(a+b-c)
b) x3-7x-6 = x3+x2-x2-x-6x-6=x2(x+1)-x(x+1)-6(x+1)=(x+1)(x2-x-6)=(x+1)(x-3)(x+2)
c) x3-x2-14x+24=x3-2x2+x2-2x-12x+24=x2(x-2)+x(x-2)-12(x-2)=(x-2)(x2+x-12)=(x-2)(x+4)(x-3)
Thank bn.