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Èo,phân tích ra tưởng cái hệ 3 ẩn r định bỏ cuộc và cái kết:(
Ta có:
\(f\left(x\right)=\left(x-2\right)\cdot Q\left(x\right)+5\)
\(f\left(x\right)=\left(x+1\right)\cdot K\left(x\right)-4\)
Theo định lý Huy ĐZ ta có:
\(f\left(2\right)=5\Rightarrow8+4a+2b+c=5\left(1\right)\)
\(\Rightarrow f\left(-1\right)=-4\Rightarrow-1+a-b+c=-4\left(2\right)\)
Lấy \(\left(1\right)-\left(2\right)\) ta được:
\(9+3a+3b=9\Leftrightarrow a+b=0\)
Khi đó:
\(\left(a^3+b^3\right)\left(b^5+c^5\right)\left(c^7+d^7\right)\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)\left(b^5+c^5\right)\left(c^7+a^7\right)\)
\(=0\) ( theo Huy ĐZ thì \(a+b=0\) )
Ap dung dinh ly Bozout ta co
\(f\left(2\right)=2^3+a.2^2+b.2+c=5\)
<=> \(4a+2b+c=-3\) (1)
tuong tu \(f\left(-1\right)=\left(-1\right)^3+a-b+c=-4\)
<=> \(a-b+c=-3\) (2)
tu (1) va (2) => \(4a+2b=a-b=-3\)
=> a=b+-3
=> \(4\left(b-3\right)+2b=-3\Rightarrow b=\frac{3}{2}\)
=> \(a=-\frac{3}{2}\)
=> \(\left(a^3+b^3\right)=\left(a+b\right)\left(a^2-ab+b^2\right)=\left(\frac{3}{2}-\frac{3}{2}\right)\left(a^2-ab+b^2\right)=0\)
=> gia tri bieu thuc =0
Bài 1:
\(2x^4+ax^2+bx+c⋮x-2\\ \Leftrightarrow2x^4+ax^2+bx+c=\left(x-2\right)\cdot a\left(x\right)\)
Thay \(x=2\Leftrightarrow32+4a+2b+c=0\Leftrightarrow4a+2b+c=-32\left(1\right)\)
\(2x^4+ax^2+bx+c:\left(x^2-1\right)R2x\\ \Leftrightarrow2x^4+ax^2+bx+c=\left(x-1\right)\left(x+1\right)\cdot b\left(x\right)+2x\)
Thay \(x=1\Leftrightarrow2+a+b+c=2\Leftrightarrow a+b+c=0\left(2\right)\)
Thay \(x=-1\Leftrightarrow2+a-b+c=-2\Leftrightarrow a-b+c=-4\left(3\right)\)
Từ \(\left(1\right)\left(2\right)\left(3\right)\Leftrightarrow\left\{{}\begin{matrix}4a+2b+c=-32\\a+b+c=0\\a-b+c=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{34}{3}\\b=2\\c=\dfrac{28}{3}\end{matrix}\right.\)
Bài 2:
Do \(f\left(x\right):x^2+x-12\) được thương bậc 2 nên dư bậc 1
Gọi đa thức dư là \(ax+b\)
Vì \(f\left(x\right):x^2+x-12\) được thương là \(x^2+3\) và còn dư nên
\(f\left(x\right)=\left(x^2+x-12\right)\left(x^2+3\right)+ax+b\\ \Leftrightarrow f\left(x\right)=\left(x+4\right)\left(x-3\right)\left(x^2+3\right)+ax+b\)
Thay \(x=3\Leftrightarrow f\left(3\right)=3a+b\)
Mà \(f\left(x\right):\left(x-3\right)R2\Leftrightarrow f\left(3\right)=2\Leftrightarrow3a+b=2\left(1\right)\)
Thay \(x=-4\Leftrightarrow f\left(-4\right)=-4a+b\)
Mà \(f\left(x\right):\left(x+4\right)R9\Leftrightarrow f\left(-4\right)=9\Leftrightarrow-4a+b=-9\left(2\right)\)
Từ \(\left(1\right)\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}3a+b=2\\-4a+b=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-1\\b=5\end{matrix}\right.\)
Do đó \(f\left(x\right)=\left(x^2+x-12\right)\left(x^2+3\right)-x+5\)
\(\Leftrightarrow f\left(x\right)=x^4+3x^2+x^3+3x-12x^2-36-x+5\\ \Leftrightarrow f\left(x\right)=x^4+x^3-9x^2+2x-31\)
Vì \(f\left(x\right)⋮x-2;f\left(x\right):x^2-1\) dư 1\(\Rightarrow\left\{{}\begin{matrix}f\left(x\right)=g\left(x\right)\cdot\left(x-2\right)\\f\left(x\right)=q\left(x\right)\left(x^2-1\right)+x=q\left(x\right)\left(x-1\right)\left(x+1\right)+x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}f\left(2\right)=0\\f\left(1\right)=1\\f\left(-1\right)=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}32+4a+2b+c=0\\2+a+b+c=1\\2+a-b+c=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}4a+2b+c=-32\left(1\right)\\a+b+c=-1\left(2\right)\\a-b+c=-3\left(3\right)\end{matrix}\right.\)
Trừ từng vế của (2) cho (3) ta được:
\(\Rightarrow2b=2\Rightarrow b=1\)
Thay b=1 vào lần lượt (1) ,(2),(3) ta được:
\(\Rightarrow\left\{{}\begin{matrix}4a+2+c=-32\\a+1+c=-1\\a-1+c=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4a+c=-34\\a+c=-2\\a+c=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4a+c=-34\left(4\right)\\a+c=-2\left(5\right)\end{matrix}\right.\)
Trừ từng vế của (4) cho (5) ta được:
\(\Rightarrow3a=-32\Rightarrow a=-\dfrac{32}{3}\Rightarrow c=-2+\dfrac{32}{3}=\dfrac{26}{3}\) Vậy...
\(f\left(x\right)=2x^4+ax^2+bx+c\)
\(=2x^4-4x^3+4x^3-8x^2+\left(a+8\right)x^2-x\left(2a+16\right)+\left(2a+16+b\right)x-2\left(2a+16+b\right)+4a+32+2b+c\)
\(=\left(x-2\right)\left(2x^3+4x^2+x\left(a+8\right)+2a+16+b\right)+4a+2b+32+c\)
=>\(\dfrac{f\left(x\right)}{x-2}=2x^3+4x^2+x\left(a+8\right)+2a+16+b+\dfrac{4a+2b+32+c}{x-2}\)
f(x) chia hết cho x-2 nên \(4a+2b+32+c=0\)(1)
\(f\left(x\right)=2x^4+ax^2+bx+c\)
\(=2x^4-4x^3+6x^2+4x^3-16x^2+12x+\left(a+10\right)x^2-4x\left(a+10\right)+3a+30+x\left(4a+28+b\right)+c-3a-30\)
\(=\left(x^2-4x+3\right)\left(2x^2+4x+a+10\right)\)+x(4a+28+b)+c-3a-30
f(x) chia cho x2-4x+3 dư -x+2 nên ta có:
\(\left\{{}\begin{matrix}4a+28+b=-1\\c-3a-30=2\end{matrix}\right.\)(2)
Từ (1),(2) ta có hệ phương trình:
\(\left\{{}\begin{matrix}4a+2b+32+c=0\\4a+b+28=-1\\c-3a=32\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4a+2b+c=-32\\4a+b=-29\\-3a+c=32\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b+c=-3\\-3a+c=32\\4a+b=-29\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b+3a=-35\\4a+b=-29\\b+c=-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-a=-6\\4a+b=-29\\b+c=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=6\\b=-29-4a=-29-4\cdot6=-53\\c=-3-b=-3-\left(-53\right)=50\end{matrix}\right.\)