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x^4+ax^3+bx^2+ c x^3+3x+2 x+a x^4+ 3x^2+2x ax^3+x^2(b-3)- 2x+ c ax^3 +3xa+2a x^2(b-3)-x(2-3a)+(c-2a)
Để \(x^4+ax^3+bx^2+c⋮x^3+3x+2\) thì \(x^2\left(b-3\right)-x\left(2-3a\right)+\left(c-2a\right)=0\)
\(\Leftrightarrow a=\frac{2}{3};b=3;c=\frac{4}{3}\)
\(\left(ax^2+bx+c\right)\left(x+1\right)=ax^3+\left(a+b\right)x^2+\left(b+c\right)x+c\)
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
b, \(ax^3+bx^2+5x-50⋮\left(x^2+3x-10\right)\)
\(\Rightarrow f\left(x\right)=ax^3+bx^2+5x-50⋮\left(x-2\right)\left(x+5\right)\)\(\Leftrightarrow\left\{{}\begin{matrix}f\left(2\right)=0\\f\left(-5\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}f\left(2\right)=8a+4b+10-50=0\\f\left(-5\right)=-125a+25b-25-50=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}f\left(2\right)=4\left(2a+b\right)=40\\f\left(-5\right)=-25\left(5a-b\right)=75\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}f\left(2\right)=2a+b=1\\f\left(-5\right)=5a-b=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{2}{7}\\b=\dfrac{11}{7}\end{matrix}\right.\)
2, \(\frac{x^2}{2}+\frac{y^2}{3}+\frac{z^2}{4}=\frac{x^2+y^2+z^2}{5}\)
<=>\(\left(\frac{x^2}{2}-\frac{x^2}{5}\right)+\left(\frac{y^2}{3}-\frac{y^2}{5}\right)+\left(\frac{z^2}{4}-\frac{z^2}{5}\right)=0\)
<=>\(\frac{3}{10}x^2+\frac{2}{15}y^2+\frac{1}{20}z^2=0\)
<=>x=y=z=0
4,
a, \(\frac{1}{x\left(x^2+1\right)}=\frac{a}{x}+\frac{bx+c}{x^2+1}\)
=>\(\frac{1}{x\left(x^2+1\right)}=\frac{ax^2+a+bx^2+cx}{x\left(x^2+1\right)}=\frac{\left(a+b\right)x^2+cx+a}{x\left(x^2+1\right)}\)
Đồng nhất 2 phân thức ta được:
\(\hept{\begin{cases}a+b=0\\c=0\\a=1\end{cases}\Leftrightarrow\hept{\begin{cases}b=-1\\c=0\\a=1\end{cases}}}\)
b,a=1/4,b=-1/4
c, a=-1,b=1,c=1
\(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(A=\left[\frac{2\left(x+2\right)}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right]:\left(\frac{2}{x^2-4}-\frac{x+2}{x^2-4}\right)\)
\(A=\frac{2x+4-4}{\left(x+2\right)^2}:\frac{2-x-2}{x^2-4}\)
\(A=\frac{2x}{\left(x+2\right)^2}.\frac{x^2-4}{-x}=\frac{2\left(x-2\right)}{-\left(x+2\right)}=\frac{-2\left(x-2\right)}{x+2}\)