Tìm a, b biết :
\(x^4-3x+2=\left(x-1\right)\left(x^3+ax^2+bx-2\right)\)
1.tìm a,b để:
a)\(x^3+ax+bx+6⋮\left(x-1\right)\)
b)\(x^4+ax^3+bx^2+5x+1⋮\left(x+1\right)^2\)
c)\(^{x^4+3x^3+ax^2+bx+5⋮\left(x-2\right)^2}\)
d)\(x^4+10x^3+ax^2+bx+7⋮\left(x+2\right)^2\)
e)\(x^4+ax^3+5x^2+bx+1⋮x-1\)
2.Cho a+b+c=0.tính\(\left(a+b+c\right)^3+\left(b+a-c\right)^3+\left(c+a-b\right)^3\)
bài 2:
\(A=\left(a+b+c\right)^3+\left(b+a-c\right)^3+\left(c+a-b\right)^3\)
\(=\left(c+b+a-2c\right)^3+\left(c+a+b-2b\right)^3\)
\(=\left(-2c\right)^3+\left(-2b\right)^3=-8\left(b+c\right)\)
sao nữa nhỉ :v
Tìm a, b biết:
1./ \(x^4-3x+2=\left(x-1\right)\left(x^3+bx^2+ax-2\right)\)
2./ \(x^4+x^3-x-1=\left(x^2-1\right)\left(x^2+ax+b\right)\)
Giải hộ mình nha, mình cần gấp !
a)\(x^3+ax+bx+6⋮\left(x-1\right)\)
b)\(x^4+ax^3+bx^2+5x+1⋮\left(x+1\right)^2\)
c)\(^{x^4+3x^3+ax^2+bx+5⋮\left(x-2\right)^2}\)
d)\(x^4+10x^3+ax^2+bx+7⋮\left(x+2\right)^2\)
e)\(x^4+ax^3+5x^2+bx+1⋮x-1\)
Cho a+b+c=0.tính\(\left(a+b+c\right)^3+\left(b+a-c\right)^3+\left(c+a-b\right)^3\)
Tìm a, b biết :
a, \(x^4+ax^2+b⋮x^2-x+1\)
b, \(ax^3+bx^2+5x-50⋮\left(x^2+3x-10\right)\)
c, \(ax^4+bx^3+1⋮\left(x-1\right)^2\)
d, \(x^4+4⋮\left(x^2+ax+b\right)\)
b, \(ax^3+bx^2+5x-50⋮\left(x^2+3x-10\right)\)
\(\Rightarrow f\left(x\right)=ax^3+bx^2+5x-50⋮\left(x-2\right)\left(x+5\right)\)\(\Leftrightarrow\left\{{}\begin{matrix}f\left(2\right)=0\\f\left(-5\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}f\left(2\right)=8a+4b+10-50=0\\f\left(-5\right)=-125a+25b-25-50=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}f\left(2\right)=4\left(2a+b\right)=40\\f\left(-5\right)=-25\left(5a-b\right)=75\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}f\left(2\right)=2a+b=1\\f\left(-5\right)=5a-b=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{2}{7}\\b=\dfrac{11}{7}\end{matrix}\right.\)
Tìm hệ số a,b,c biết
a, \(-3x^2\left(2ax^2-bx+c\right)=6x^5+9x^4-3c^2\forall x\)
b,\(\left(x^2+cx+2\right)\left(a+b\right)=x^3+x^2-2\forall x\)
c,\(\left(ax^2+bx+c\right)+\left(x+3\right)=x^2+2x-3x\forall x\)
Help me!!
Xác định a, b để \(f\left(x\right)⋮g\left(x\right)\)
a) f(x)= \(2x^3-3x^2+ax+b\) ; \(g\left(x\right)=x^2+x+2\)
b) \(f\left(x\right)=2x^4+ax^2+b\) ; \(g\left(x\right)=x^2-x-3\)
c) \(f\left(x\right)=3x^4-8x^3-10x^2+ax-b\) ; \(g\left(x\right)=3x^2-2x+1\)
d) \(f\left(x\right)=ax^3+bx^2-11x+30\) ; \(g\left(x\right)=x^2-3x-10\)
tìm a ; b sao cho :
a, \(\left(2x^3-x^2+ax+b\right)⋮\left(x^2-1\right)\)
b, \(\left(x^4+ax^2+bx-1\right)⋮\left(x^2-1\right)\)
c, \(\left[x^4+x^3 +ax^2+\left(a+b\right)x+2b+1\right]⋮\left(x^3+ax+b\right)\)
a: \(\dfrac{2x^3-x^2+ax+b}{x^2-1}\)
\(=\dfrac{2x^3-2x-x^2+1+\left(a+2\right)x+b-1}{x^2-1}\)
\(=2x-1+\dfrac{\left(a+2\right)x+b-1}{x^2-1}\)
Để đây là phép chia hết thì a+2=0 và b-1=0
=>a=-2; b=1
b: \(\Leftrightarrow x^4-1+ax^2-a+bx+a⋮x^2-1\)
=>bx+a=0
=>a=b=0
tìm a, b, c để hso \(f\left(x\right)=ax^2+bx+c\) có đạo hàm \(f'\left(x\right)\) thỏa mãn \(f\left(x\right)+\left(x-1\right)f'\left(x\right)=3x^2\) voi mọi x thuoc R
\(f'\left(x\right)=2ax+b\)
\(f\left(x\right)+\left(x-1\right)f'\left(x\right)=ax^2+bx+c+\left(x-1\right)\left(2ax+b\right)\)
\(=3ax^2+\left(2b-2a\right)x+c-b\)
Yêu cầu bài toán thỏa mãn khi: \(\left\{{}\begin{matrix}3a=3\\2b-2a=0\\c-b=0\end{matrix}\right.\) \(\Leftrightarrow a=b=c=1\)
Xác định a, b để f(x) \(⋮\) g(x)
a) \(f\left(x\right)=2x^3-3x^2+ax+b\) ; \(g\left(x\right)=x^2+x+2\)
b) \(f\left(x\right)=2x^4+2x^2+b\) ; \(g\left(x\right)=x^2-x-3\)
c) \(f\left(x\right)=3x^4-8x^3-10x^2+ax-b\) ; \(g\left(x\right)=3x^2-2x+1\)
d) \(f\left(x\right)=ax^3+bx^2-11x+30\) ; \(g\left(x\right)=x^2-3x-10\)