Phân tích đa thức thành nhân tử ( phối hợp các phương pháp )
1) xy( a2 + 2b2 ) - ab( 2x2 + y2 )
2) ( xy - ab )2 + ( bx - ay )2
3) ( 2xy + ab )2 + ( 2ay - bx )2
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1) \(xy\left(a^2+2b^2\right)-ab\left(2x^2+y^2\right)\)
\(=a^2xy+2b^2xy-2x^2ab-y^2ab\)
\(=\left(a^2xy-y^2ab\right)+\left(2b^2xy-2x^2ab\right)\)
\(=ay\left(ax-by\right)+2bx\left(by-ax\right)\)
\(=ay\left(ax-by\right)-2bx\left(ax-by\right)\)
\(=\left(ax-by\right)\left(ay-2bx\right)\)
2) Sửa đề \(\left(xy+ab\right)^2+\left(bx-ay\right)^2\)
\(=\left(xy\right)^2+2xyab+\left(ab\right)^2+\left(bx\right)^2-2xyab+\left(ay\right)^2\)
\(=x^2y^2+a^2b^2+b^2x^2+a^2y^2\)
\(=\left(x^2y^2+b^2x^2\right)+\left(a^2b^2+a^2y^2\right)\)
\(=x^2\left(b^2+y^2\right)+a^2\left(b^2+y^2\right)\)
\(=\left(b^2+y^2\right)\left(a^2+x^2\right)\)
3) \(\left(2xy+ab\right)^2+\left(2ay-bx\right)^2\)
\(=\left(2xy\right)^2+2.2xyab+\left(ab\right)^2+\left(2ay\right)^2-2.2xyab+\left(bx\right)^2\)
\(=4x^2y^2+a^2b^2+4a^2y^2+b^2x^2\)
\(=\left(4x^2y^2+b^2x^2\right)+\left(4a^2y^2+a^2b^2\right)\)
\(=x^2\left(4y^2+b^2\right)+a^2\left(4y^2+b^2\right)\)
\(=\left(4y^2+b^2\right)\left(a^2+x^2\right)\)
Bài 4:
Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Phân tích đa thức thành nhân tử ( phối hợp các phương pháp )
1) x2 - ( a + b )xy + aby2
\(=x^2-axy-bxy+aby^2\)
\(=(x^2-axy)-(bxy+aby^2)\)
\(=x(x-ay)-by(x+ay)\)
\(=(x-ay)(x-by)\)
2) x2 + ( 2a + b )xy + 2aby2
=x2 + 2axy + bxy + 2aby2
=(x2+ bxy) +(2axy+ 2aby2 )
=x(x+ by) +2ay(x+ by)
=(x+ by)(x+2ay)
a) \(2x-72x^3=2x\left(1-36x^2\right)=2x\left(1-6x\right)\left(1+6x\right)\)
f) \(4x^4+1=4x^4+4x^2+1-4x^2=\left(2x^2+1\right)^2-\left(2x\right)^2=\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)\)
\(ax+bx+ay+by\)
\(=x\left(a+b\right)+y\left(a+b\right)\)
\(=\left(x+y\right)\left(a+b\right)\)
\(xy+1-x-y\)
\(=x\left(y-1\right)-\left(y-1\right)\)
\(=\left(x-1\right)\left(y-1\right)\)
a: \(\left(xy+ab\right)^2+\left(bx-ay\right)^2\)
\(=x^2y^2+a^2b^2+x^2b^2+a^2y^2\)
\(=x^2\left(b^2+y^2\right)+a^2\left(b^2+y^2\right)\)
\(=\left(b^2+y^2\right)\left(x^2+a^2\right)\)
\(1,\\ a,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ b,=a^2\left(a-x\right)-y\left(a-x\right)=\left(a^2-y\right)\left(a-x\right)\\ c,=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\\ d,=x\left(x-2y\right)+t\left(x-2y\right)=\left(x+t\right)\left(x-2y\right)\\ 2,\\ \Rightarrow x^2-4x+4-x^2+9=6\\ \Rightarrow-4x=-7\Rightarrow x=\dfrac{7}{4}\\ 3,\\ a,x^2+2x+2=\left(x+1\right)^2+1\ge1>0\\ b,-x^2+4x-5=-\left(x-2\right)^2-1\le-1< 0\)
1. \(xy\left(a^2+2b^2\right)-ab\left(2x^2+y^2\right)\)
\(=xya^2+2xyb^2-2abx^2-aby^2\)
\(=xya^2-aby^2-2abx^2+2xyb^2\)
\(=ay\left(ax-by\right)-2bx\left(ax-by\right)\)
\(=\left(ay-2bx\right)\left(ax-by\right)\)
2. \(xy\left(a^2+2b^2\right)+ab\left(2x^2+y^2\right)\)
\(=xya^2+2xyb^2+2abx^2+aby^2\)
\(=xya^2+aby^2+2abx^2+2xyb^2\)
\(=ay\left(ax+by\right)+2bx\left(ax+by\right)\)
\(=\left(ay+2bx\right)\left(ax+by\right)\)
1) \(xy\left(a^2+2b^2\right)-ab\left(2x^2+y^2\right)\)
\(=a^2xy+2b^2xy-2abx^2-aby^2\)
\(=\left(a^2xy-aby^2\right)+\left(2b^2xy-2abx^2\right)\)
\(=ay\left(ax-by\right)+2bx\left(by-ax\right)\)
\(=ay\left(ax-by\right)-2bx\left(ax-by\right)\)
\(=\left(ax-by\right)\left(ay-2bx\right)\)
2) Sửa đề \(\left(xy+ab\right)^2+\left(bx-ay\right)^2\)
\(=\left(xy\right)^2+2xyab+\left(ab\right)^2+\left(bx\right)^2-2xyab+\left(ay\right)^2\)
\(=x^2y^2+a^2b^2+b^2x^2+a^2y^2\)
\(=\left(x^2y^2+b^2x^2\right)+\left(a^2b^2+a^2y^2\right)\)
\(=x^2\left(b^2+y^2\right)+a^2\left(b^2+y^2\right)\)
\(=\left(b^2+y^2\right)\left(x^2+a^2\right)\)
3) \(\left(2xy+ab\right)^2+\left(2ay-bx\right)^2\)
\(=\left(2xy\right)^2+2.2xyab+\left(ab\right)^2+\left(2ay\right)^2-2.2xyab+\left(bx\right)^2\)
\(=4x^2y^2+4xyab+a^2b^2+4a^2y^2-4xyab+b^2x^2\)
\(=4x^2y^2+4a^2y^2+a^2b^2+b^2x^2\)
\(=4y^2\left(x^2+a^2\right)+b^2\left(a^2+x^2\right)\)
\(=\left(a^2+x^2\right)\left(4y^2+b^2\right)\)