\(\text{a) x}^4+2x^2+1=\left(x^2+1\right)^2\)

\(\text{b) 3}ax^2+3bx^2+ãx+bx+5a+5b=\left(3ax^2+3bx^2\right) +\left(ax+bx\right)+\left(5a+5b\right)=3x^2+x\left(a+b\right)+5\left(a+b\right)=\left(a+b\right)\left(3x^2+x+5\right)\)

\(\text{c) a}x^2-bx^2-2ax+2bx-3a+3b=\left(\text{a}x^2-bx^2\right)-\left(2ax-2bx\right)-\left(3a-3b\right)=x^2\left(a-b\right)-2x\left(a-b\right)-3\left(a-b\right)=\left(x^2-2x-3\right)\left(a-b\right)\)

a) \(x^4+2x^2+1\)

\(=x^2\left(x^2+1\right)+\left(x^2+1\right)\)

\(=\left(x^2+1\right)^2\)

b) \(3ax^2+3bx^2+ax+bx+5a+5b\)

\(=3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\)

\(=\left(a+b\right)\left(3x^2+x+5\right)\)

c) \(ax^2-bx^2-2ax+2bx-3a+3b\)

\(=x^2\left(a-b\right)-2x\left(a-b\right)-3\left(a-b\right)\)

\(=\left(a-b\right)\left(x^2-2x-3\right)\)

\(=\left(a-b\right)\left(x-3\right)\left(x+1\right)\)

Câu d mình chịu

ax²+cx²-ay+ay²-cy+cy² 

= x²(a+c)-y(a+c)+y²(a+c)

= (a+c)(x²-y+y²)

ax²+cx²-ay+ay²-cy+cy² 

x²(a+c)-y(a+c)

a+c    x²-y+y²

\(ax^2+cx^2-ay+ay^2-cy+cy^2\)

\(=x^2\left(a+c\right)-y\left(a+c\right)+y^2\left(a+c\right)\)

\(=\left(a+c\right)\left(x^2-y+y^2\right)\)

= (ax+a-ay) + (by-bx-b)

= a nhân ( x+1-y) + b nhân ( y-x-1 )

= a nhân ( x+1-y) - b nhân ( x+1-y )          

= (x+1-y) nhân (a-b) 

\(a,=x^2-mx-nx+mn=x\left(x-m\right)-n\left(x-m\right)=\left(x-n\right)\left(x-m\right)\\ b,=a\left(x-y\right)-b\left(x-y\right)+\left(a-b\right)\\ =\left(x-y\right)\left(a-b\right)+\left(a-b\right)=\left(a-b\right)\left(x-y+1\right)\)

b: \(=a\left(x-y\right)-b\left(x-y\right)+a-b\)

\(=\left(x-y+1\right)\left(a-b\right)\)

a) \(x^2-\left(m+n\right)x+mn=\left(x^2-n\cdot x\right)-\left(m\cdot x-m\cdot n\right)=x\left(x-n\right)-m\left(x-n\right)=\left(x-m\right)\left(x-n\right)\)

b) \(ax+by+a-bx-ay-b\)

   \(=\left(ax-ay+a\right)-\left(bx-by+b\right)\)

   \(=a\left(x-y+1\right)-b\left(x-y+1\right)\)

   \(\left(a-b\right)\left(x-y+1\right)\)

\(ax^2+bx^2-bx-ax+cx^2-cx\)

\(=\left(ax^2-ax\right)+\left(bx^2-bx\right)+\left(cx^2-cx\right)\)

\(=ax\left(x-1\right)+bx\left(x-1\right)+cx\left(x-1\right)\)

\(=\left(x-1\right)\left(ax+bx+cx\right)\)

\(=x\left(x-1\right)\left(a+b+c\right)\)

ax²+bx²-bx-ax+cx²-cx

= (ax² - ax ) + (bx² -bx ) + ( cx² - cx )

= a(x) + b(x) + c(x)

= (x)(a+b+c)

a) x²y+xy+x+1= (x²y+xy)+(x+1)=xy(x+10+(x+1)=(x+1)(xy+1)

b) x²-(a+b)x+ab=x²-ax-bx+ab=(x²-ax)-(bx-ab)=x(x-a)-b(x-a)=(x-a)(x-b)

c) ax²+ay-bx²-by=(ax²+ay)-(bx²+by)=a(x²+y)-b(x²+y)=(a-b)(x²+y)

d) ax-2x-a²+2a=(ax-2x)-(a²-2a)=x(a-2)-a(a-2)=(a-2)(x-a)

e) 2x²+4ax+x+2a=(2x²+4ax)+(x+2a)=2x(x+2a)+(x+2a)=(x+2a)(2x+1)

f) x³+ax²+x+a=(x³+ax²)+(x+a)=x²(x+a)+(x+a)=(x²+1)(x+a)

g: Ta có: \(x^4+2x^3-4x-4\)

\(=\left(x^2-2\right)\left(x^2+2\right)-2x\left(x^2-2\right)\)

\(=\left(x^2-2\right)\cdot\left(x^2+2x+2\right)\)

a ) xy + 1 - x - y

= x ( y - 1 ) + 1 - y

= x ( y - 1 ) - ( y - 1 )

= ( x - 1 ) ( y - 1 )

b ) x² + ab + ac + bx

= ( b + x )x + a( b + c )

= ( b + x )1x + 1a( b + c )

= ( b + x ) ( x + a ) ( b + c )

c ) ax + bx - cx + a + b - c

= ( a + b - c )x + a + b - c

= ( a + b - c )x + ( a + b - c )1

= ( a + b - c ) ( x + 1 )

\(=x\left(a-b\right)-\left(a^2-2ab+b^2\right)=\)

\(=\left(a-b\right)x-\left(a-b\right)^2=\)

\(=\left(a-b\right)\left[x-\left(a-b\right)\right]=\left(a-b\right)\left(x-a+b\right)\)

\(\left(x+a\right)\left(x+b\right)\left(x+c\right)=\left(x^2+bx+ax+ab\right)\left(x+c\right)\)

\(=x^3+cx^2+bx^2+bcx+ax^2+acx+abx+abc\)

\(=x^3+\left(a+b+c\right)x^2+\left(ab+ac+bc\right)x+abc\)

Đồnh nhất đa thức trên với đa thức \(x^3+ax^2+bx+c\),ta đc hệ điều kiện:

\(\hept{\begin{cases}a+b+c=a\left(1\right)\\ab+ac+bc=b\left(2\right)\\abc=c\left(3\right)\end{cases}}\)

Từ \(\left(1\right)a+b+c=a=>b+c=0=>c=-b\)

Thay vào (2),ta đc: \(ab+a.\left(-b\right)+b.\left(-b\right)=b=>ab-ab-b^2=b=>-b^2=b\)

\(=>b^2+b=0=>b\left(b+1\right)=0=>\orbr{\begin{cases}b=0\\b=-1\end{cases}}\)

+b=0 thì từ (1) suy ra c=0 ; a tùy ý

+b=-1 thì từ (1) suy ra c=1

Mà theo (3)\(abc=c=>a=\frac{c}{bc}=\frac{1}{-1}=-1\)

Vậy a=-1 hoặc a tùy ý ;b=0 hoặc b=-1;c=0 hoặc c=1

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