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a: =>6x^2+2xb-15x-5b=ax^2+x+c
=>6x^2+x(2b-15)-5b=ax^2+x+c
=>a=6; 2b-15=1; -5b=c
=>a=6; b=8; c=-40
b: =>ax^3-ax^2-ax+bx^2-bx-b=ax^3+cx^2-1
=>x^2(-a+b)+x(-a-b)-b=cx^2-1
=>-b=-1; -a+b=c; -a-b=0
=>b=1; c=b-a; a=-b=-1
=>c=b-a=1-(-1)=2; b=1; a=-1
a: \(\left(x^2+cx+2\right)\left(ax+b\right)\)
\(=ax^3+bx^2+ac\cdot x^2+bc\cdot x+2ax+2b\)
\(=ax^3+x^2\left(b+ac\right)+x\left(bc+2a\right)+2b\)
Theo đề, ta có: a=1; 2b=-2; b+ac=1 và bc+2a=0
=>a=1; b=-1; c-1=1; bc+2a=0
=>a=1; b=-1; c=2
b: \(\left(x^2-x+1\right)\left(ax^2+bx+c\right)\)
\(=ax^4+bx^3+cx^2-ax^3-bx^2-cx+ax^2+bx+c\)
\(=ax^4+x^3\left(b-a\right)+x^2\left(c-b+a\right)+x\left(-c+b\right)+c\)
Theo đề, ta có:
a=2; b-a=-1; c-b+a=2; -c+b=0; c=1
=>a=2; b=-1+a=-1+2=1; c=1
â) viết lại biểu thức bên trái = (x2+5x-3)(x2-2x-4)+(14+a)x+b-12
Để là phép chia hết thì số dư =0
Số dư chính là (14+a)x+b-12=0 => a+14=0 và b-12=0 <=>a=-14 và b=12
b) làm tương tự phân tích vế trái thành (x3-2x2+4)(x2+9x+18)+(a+32)x2+(b-36)x
số dư là (a+32)x2+(b-36)x=0 =>a=-32 và b=36
c) Tương tự (x2-1)4x+(a+4)x+b
số dư là (a+4)x+b =2x-3 =>a+4=2 và b=-3 <=>a=-2 và b=-3
\(\left(ax^2+bx+c\right)\left(x+1\right)=ax^3+\left(a+b\right)x^2+\left(b+c\right)x+c\)
đồng nhất đa thức trên với đa thức đã cho ta được
\(\left\{{}\begin{matrix}a=1\\a+b=8\\b+c=19\\c=12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=1\\b=7\\c=12\end{matrix}\right.\)
3 phần kia làm tương tự
b: \(\left(ax^2+bx+c\right)\left(x+3\right)\)
\(=ax^3+3ax^2+bx^2+3bx+cx+3c\)
\(=ax^3+x^2\left(3a+b\right)+x\left(3b+c\right)+3c\)
Theo đề, ta có:
\(\left\{{}\begin{matrix}3c=0\\3b+c=-3\\3a+b=2\\a=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=0\\b=-1\\a=1\end{matrix}\right.\)
c: \(\left(x^2+cx+2\right)\left(ax+b\right)\)
\(=a\cdot x^3+bx^2+ac\cdot x^2+bc\cdot x+2a\cdot x+2b\)
\(=a\cdot x^3+x^2\left(b+ac\right)+x\left(bc+2a\right)+2b\)
Theo đề, ta có: 2b=-2; bc+2a=0; b+ac=1; a=1
=>b=-1; a=1; c=2
d: \(\left(x^2+cx+1\right)\left(ax+b\right)\)
\(=a\cdot x^3+bx^2+ac\cdot x^2+bc\cdot x+a\cdot x+b\)
\(=a\cdot x^3+x^2\left(b+ac\right)+x\left(bc+a\right)+b\)
Theo đề, ta có:
b=2; bc+a=-3; b+ac=0; a=1
=>b=2; a=1; bc=-3-a=-3-1=-4
=>b=2; a=1; 2c=-4
=>b=2; a=1; c=-2
1 ) Ta có :
\(ax+2x+ay+2y+4\)
\(=x\left(a+2\right)+y\left(a+2\right)+4\)
\(=\left(x+y\right)\left(a+2\right)+4\)
\(=\left(a-2\right)\left(a+2\right)+4\) ( do \(x+y=a-2\) )
\(=a^2-4+4\)
\(=a^2\left(đpcm\right)\)
2 ) \(\left(ax+b\right)\left(x^2-x-1\right)=ax^3+cx^2-1\)
\(\Leftrightarrow ax^3+bx^2-ax^2-bx-ax-b=ax^3+cx^2-1\)
\(\Leftrightarrow ax^3+x^2\left(b-a\right)-\left(b+a\right)x-b=ax^3+x^2c-0.x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}b-a=c\\b+a=0\\b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-a=c\\1+a=0\\b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-a=c\\a=-1\\b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}c=2\\a=-1\\b=1\end{matrix}\right.\)
Vậy \(a=-1;b=1;c=2\)
Ta có:
\(ax+2x+ay+2y+4\)
\(=\left(ax+ay\right)+\left(2x+2y\right)+4\)
\(=a\left(x+y\right)+2\left(x+y\right)+4\)
\(=\left(x+y\right)\left(a+2\right)+4\)
Thay \(x+y=a-2\), ta được
\(=\left(a-2\right)\left(a+2\right)+4\)
\(=a^2-4+4\)
\(=a^2\)
a/ \(-4x^3\cdot\left(ax^2+bx+c\right)=-8x^5+12x^4-20x^3\)
\(\Leftrightarrow-4ax^5-4bx^4-4cx^3=-8x^5+12x^4-20x^3\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-\dfrac{8}{-4}=\dfrac{8}{4}=2\\b=-\dfrac{12}{4}=-3\\c=-\dfrac{20}{-4}=5\end{matrix}\right.\)
Vậy......................
b/ \(-2x^3\cdot\left(ax^2-bx-c\right)=-4x^5+6x^4+2x^3\)
\(\Leftrightarrow-2ax^5+2bx^4+2cx^3=-4x^5+6x^4+2x^3\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2\\b=3\\c=1\end{matrix}\right.\)
b, \(ax^3+bx^2+5x-50⋮\left(x^2+3x-10\right)\)
\(\Rightarrow f\left(x\right)=ax^3+bx^2+5x-50⋮\left(x-2\right)\left(x+5\right)\)\(\Leftrightarrow\left\{{}\begin{matrix}f\left(2\right)=0\\f\left(-5\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}f\left(2\right)=8a+4b+10-50=0\\f\left(-5\right)=-125a+25b-25-50=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}f\left(2\right)=4\left(2a+b\right)=40\\f\left(-5\right)=-25\left(5a-b\right)=75\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}f\left(2\right)=2a+b=1\\f\left(-5\right)=5a-b=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{2}{7}\\b=\dfrac{11}{7}\end{matrix}\right.\)
a) \(\left(2x-5\right)\left(3x+b\right)=ax^2+bx+c\)
\(\Rightarrow6x^2+2bx-15x-5b=ax^2+bx+c\)
\(\Rightarrow6x^2+\left(2b-15\right)x-5b=ax^2+bx+c\)
\(\Rightarrow\left\{{}\begin{matrix}a=6\\2b-15=b\\-5b=c\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=6\\b=15\\c=-75\end{matrix}\right.\)
b) \(\left(ax+b\right)\left(x^2-x-1\right)=ax^3+cx^2-1\)
\(\Rightarrow ax^3-ax^2-ax+bx^2-bx-b=ax^3+cx^2-1\)
\(\Rightarrow ax^3-ax^2+bx^2-ax-bx-b=ax^3+cx^2-1\)
\(\Rightarrow ax^3+\left(b-a\right)x^2-\left(a+b\right)x-b=ax^3+cx^2-1\)
\(\Rightarrow\left\{{}\begin{matrix}b-a=c\\-\left(a+b\right)=0\\-b=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b-a=c\\a=-b\\b=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-1\\b=1\\c=2\end{matrix}\right.\)
c) \(ax\left(x-4\right)-b\left(x+6\right)+5=2x^2+5x\left(a-b\right)-6x+c\)
\(\Rightarrow ax^2-4ax-bx-6b+5=2x^2+\left(5a-5b\right)x-6x+c\)
\(\Rightarrow ax^2-\left(4a+b\right)x-\left(5a-5b\right)x-6b+5=2x^2-6x+c\)
\(\Rightarrow ax^2-\left(4a+b+5a-5b\right)x-6b+5=2x^2-6x+c\)
\(\Rightarrow ax^2-\left(9a-4b\right)x-6b+5=2x^2-6x+c\)
\(\Rightarrow\left\{{}\begin{matrix}a=2\\-\left(9a-4b\right)=-6\\-6b+5=c\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=2\\b=\dfrac{9a-6}{4}\\c=-6b+5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=2\\b=3\\c=-13\end{matrix}\right.\)