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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
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) \(8x^2+27=\left(x-1\right)^3+\left(x+4\right)^3\)
\(\Leftrightarrow8x^3+27=x^3-2x^2+x-x^2+2x-1+x^3+8x^2+16x+4x^2+32x+64\)
\(\Leftrightarrow8x^3+27=2x^3+9x^2+51x+63\)
\(\Leftrightarrow8x^3+27-2x^3-9x^2-51x-63=0\)
\(\Leftrightarrow6x^3-36-9x^2-51x=0\)
\(\Leftrightarrow3\left(2x^3-12-3x^2-17x\right)=0\)
\(\Leftrightarrow3\left(2x^2+3x-8x-12\right)\left(x+1\right)=0\)
\(\Leftrightarrow3\left(2x^2+3x-8x-12\right)\left(x+1\right)=0\)
\(\Leftrightarrow3\left[x\left(2x+3\right)-4\left(2x+3\right)\right]\left(x+1\right)=0\)
\(\Leftrightarrow3\left(2x+3\right)\left(x-4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}2x+3=0\\x-4=0\\x+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{3}{2}\\x=4\\x=-1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=-\frac{3}{2}\\x=4\\x=-1\end{cases}}\)
tớ tưởng áp dụng công thức: \(\left(A+B\right)^3=A^3+B^3+3AB\left(A+B\right)\)
và \(\left(A-B\right)^3=A^3-B^3-3AB\left(A-B\right)\)
\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)
\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)
\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=1.\end{matrix}\right.\)
\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)
\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)
\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)
\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)
\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)
\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)
\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)
a) Giả sử phép chia có thương là : q(x)
Khi đó , ta có : ax3 + bx - 24 = ( x + 1)( x + 3)q(x) , với mọi x ( 1)
Chọn các giá trị riêng của x sao cho :
( x + 1)( x + 3) = 0
Suy ra : x = -1 hoặc x = - 3
* Với x = -1 thì :
( 1) <=> -a -b - 24 = 0
<=> -( a + b) = 24
<=>a + b = -24 ( 2)
* Với x = -3 , thì :
( 1) <=> - 27a - 3b - 24 = 0
<=> -( 27a + 3b) = 24
<=> 27a + 3b = - 24 ( 3)
Từ ( 2 ; 3) suy ra a = 2 ; b = - 26
Vậy , ....
b) Do đa thức chia có bậc 4 ,đa thức bị chia có bậc 2 suy ra thương có bậc 2
Giả sử thương là : cx2 + dx + e
Ta có : x4 + ax2 + b = ( x2 + x + 1)( cx2 + dx + e)
x4 + ax2 + b = cx4 + dx3 + ex2 + cx3 + dx2 + ex + cx2 + dx + e
x4 + ax2 + b = cx4 + x3( d + c) + x2(e + d + c) + x( e + d) + e
Đồng nhất hệ số , ta có :
* c = 1
* d + c = 0 --> d + 1 = 0 --> d = -1
* e + d + c = a --> a = 1 - 1 + 1 = 1
* e + d = 0 e - 1 = 0 --> e = 1
* e = b --> b = 1
Vậy , a = 1 ; b = 1 thỏa mãn điều kiện đề bài
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 \(\left(x^3+ax+b\right)⋮\left(x^2+x-2\right)\)