tim a de:
\(\left(x^3-ax^2+1\right):\left(x-1\right)\)
Những câu hỏi liên quan
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
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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
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Tim cap(x;y) \(\varepsilon Z\) de B=-3
\(B=\frac{x^2}{\left(x+y\right)\left(x-y\right)}-\frac{y^2}{\left(x+y\right)\left(1+x\right)}-\frac{x^2y^2}{\left(1+x\right)\left(1-y\right)}\)
cho biểu thức A left(frac{3}{4}+frac{3}{x^2+3x}right)chialeft(frac{x^2}{27-3x}+frac{1}{x+3}right)a, rut gon A b, tim x de a-1c, tim cac gia tri nguyen cua x de A co gia tri nguyencho bthuc B left(frac{x^2}{x^3-4x}+frac{6}{6-3x}+frac{1}{x-2}right)chialeft(x-2+frac{16-x^2}{x+2}right)rut gon B tính b khi /x/ 1/2tim x de b2tim x in z de b in z
Đọc tiếp
cho biểu thức A= \(\left(\frac{3}{4}+\frac{3}{x^2+3x}\right)chia\left(\frac{x^2}{27-3x}+\frac{1}{x+3}\right)\)
a, rut gon A
b, tim x de a<-1
c, tim cac gia tri nguyen cua x de A co gia tri nguyen
cho bthuc B = \(\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x-2}\right)chia\left(x-2+\frac{16-x^2}{x+2}\right)\)rut gon B tính b khi /x/ = 1/2tim x de b=2tim x \(\in\) z de b \(\in\) z
Bài 2:
a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x-2}\right):\left(\dfrac{x^2-4+16-x^2}{x+2}\right)\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\dfrac{x-x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{12}=\dfrac{-1}{6\left(x-2\right)}\)
b: Thay x=1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(\dfrac{1}{2}-2\right)}=\dfrac{-1}{6\cdot\dfrac{-3}{2}}=\dfrac{1}{9}\)
Thay x=-1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(-\dfrac{1}{2}-2\right)}=-\dfrac{1}{15}\)
c: Để B=2 thì \(\dfrac{-1}{6\left(x-2\right)}=2\)
=>6(x-2)=-1/2
=>x-2=-1/12
hay x=23/12
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a)x^3+ax+bx+6⋮left(x-1right)
b)x^4+ax^3+bx^2+5x+1⋮left(x+1right)^2
c)^{x^4+3x^3+ax^2+bx+5⋮left(x-2right)^2}
d)x^4+10x^3+ax^2+bx+7⋮left(x+2right)^2
e)x^4+ax^3+5x^2+bx+1⋮x-1
Cho a+b+c0.tínhleft(a+b+cright)^3+left(b+a-cright)^3+left(c+a-bright)^3
Đọc tiếp
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\(\left(a+b+c\right)^3+\left(b+a-c\right)^3+\left(c+a-b\right)^3\)
Xác định a,b để :
a/ \(\left(x^3+ax+b\right)⋮\left(x^2+x-2\right)\)
b/ \(\left(x^3+ax^2-4\right)⋮x^2+ax+4\)
c/ \(\left(x^4+ax^2+b\right)⋮\left(x^2-x+1\right)\)
d/ \(\left(x^4+4\right)⋮\left(x^2+ax+b\right)\)
a) Đặt \(f\left(x\right)=x^3+ax+b\)
Vì \(f\left(x\right)⋮x^2+x-2\)
\(\Rightarrow f\left(x\right)=\left(x^2+x-2\right)q\left(x\right)\)
\(=\left(x^2-x+2x-2\right)q\left(x\right)\)
\(=\left[x\left(x-1\right)+2\left(x-1\right)\right]q\left(x\right)\)
\(=\left(x-1\right)\left(x+2\right)q\left(x\right)\)
\(\Rightarrow f\left(1\right)=\left(1-1\right)\left(1+2\right)q\left(1\right)\)
\(\Rightarrow f\left(1\right)=0\left(1\right)\)
\(f\left(-2\right)=\left(-2-1\right)\left(-2+2\right)q\left(-2\right)\)
\(\Rightarrow f\left(-2\right)=0\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}f\left(1\right)=0\\f\left(-2\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1+a+b=0\\-8-2a+b=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a+b=-1\\-2a+b=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=-3\\b=2\end{matrix}\right.\)
Vậy a=-3 và b=2 thì \(\left(x^3+ax+b\right)⋮\left(x^2+x-2\right)\)
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Cho
\(f\left(x\right)=ax^3+4x\left(x^2-x\right)+8\)
\(g\left(x\right)=x^3-4x\left(bx+1\right)+c-3\)
Trong do a,b,c la hang.Xac dinh a,b,c de \(f\left(x\right)=g\left(x\right)\)
Cho da thuc \(f\left(x\right)=x^3+2x^2+ax+1\)
TIm a biet no \(f\left(x\right)=-2\)
Khi \(f\left(x\right)=-2\)
Ta có : \(f\left(-2\right)=\left(-2\right)^3+2.\left(-2\right)^2+a.\left(-2\right)+1\)
\(=-8+8+a.\left(-2\right)+1\)
\(=a.\left(-2\right)+1\)
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kun ơi, nó là;
x3 + 2x2 +ax +1 = -2
bài này bn ấy cho thiếu x=? thì làm sao tính dc a? pk kun
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Cho \(f\left(x\right)=ax^3+4x\left(x^2-1\right)+8\) và \(g\left(x\right)=x^3+4x\left(bx+1\right)+c-3\) xác định a, b, c để \(f\left(x\right)=g\left(x\right)\)
