cho hai đa thức :P(x)=x^3-2ax+a^2 và Q(x)=x^2 +(3a+1)x+a^2
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Để P(x) = Q(x)
Thì x2 - 2ax + a2 = x2 + (3a + 1)x + a2
=> x2 - 2ax + a2 = x2 + 3ax + x + a2
=> (x2 - 2ax + a2) - (x2 + 3ax + x + a2) = 0
=> x2 - 2ax + a2 - x2 + 3ax - x - a2 = 0
=> (x2 - x2) + (-2ax + 3ax) + (a2 - a2) - x = 0
=> ax - x = 0
=> x(a - 1) = 0
Vậy a = 1
Để \(P\left(x\right)=Q\left(x\right)\)thì \(x^2-2ax+a^2=x^2+\left(3a+1\right).x+a^2\)
\(\Leftrightarrow-2ax=\left(3a+1\right).x\)\(\Leftrightarrow\left(3a+1\right).x+2ax=0\)
\(\Leftrightarrow\left(3a+1+2a\right).x=0\)\(\Leftrightarrow\left(5a+1\right).x=0\)
\(\Leftrightarrow5a+1=0\)\(\Leftrightarrow5a=-1\)\(\Leftrightarrow a=\frac{-1}{5}\)
Vậy \(a=\frac{-1}{5}\)
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\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
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Bài 2:
a) x(x - 3)- y(3 - x)
= x(x - 3) + y(x - 3)
= (x - 3)(x + y) (1)
Thay x = \(\frac{1}{3}\); y = \(\frac{8}{3}\)vào (1)
Ta có: (\(\frac{1}{3}\)- 3)(\(\frac{1}{3}\)+ \(\frac{8}{3}\))
= \(\frac{-8}{3}\). 3
= -8
![](https://rs.olm.vn/images/avt/0.png?1311)
A(x)=x^2-2ax+a^2
Q(x)=x^2+(3a+1)x+a^2
A(1)=Q(3)
=>1-2a+a^2=3^2+3(3a+1)+a^2
=>1-2a=9+9a+3
=>9a+12=-2a+1
=>11a=-11
=>a=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(f\left(x\right)=x^3+2ax+b\)
Vì \(f\left(x\right)⋮\left(x-1\right)\)\(\Rightarrow f\left(1\right)=0\)\(\Leftrightarrow1+2a+b=0\)\(\Leftrightarrow2a+b=-1\)(1)
Vì \(f\left(x\right)\)chia \(x+2\)dư \(3\) \(\Rightarrow f\left(-2\right)=3\)
\(\Leftrightarrow-8-4a+b=3\Leftrightarrow-4a+b=11\Leftrightarrow4a-b=-11\)(2)
Cộng (1) với (2) ta được \(2a+b+4a-b=6a=-1-11=-12\)\(\Rightarrow a=-2\)
\(\Rightarrow b=3\)
Vậy \(a=-2;b=3\)
câu hỏi của đề là gì vậy bn