Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)\)
\(=\dfrac{1}{2}x^3-x^2+\dfrac{3}{2}x-5x^2+10x-15\)
\(=\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)
\(\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)\)
\(=\dfrac{1}{2}x^3-5x^2-x^2+10x+\dfrac{3}{2}x-15\)
\(=\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)
\(\left(x^2-2x+3\right)\left(1212x-5\right)\)
\(=1212x^3-5x^2-2424x^2+10x+3636x-15\)
\(=1212x^3-2429x^2+3646x-15\)
\(=1212x^3-5x^2-2424x^2+10x+3636x-15\\ =1212x^3-2429x^2+3646x-15\)
a) Ta có : P = (x + 5)(ax2 + bx + 25)
= ax3 + bx2 + 25x + 5ax2 + 5bx + 125
= ax3 + (bx2 + 5ax2) + (25x + 5bx) + 125
= ax3 + x2(b + 5a) + x(25 + 5b) + 125
a) Ta có : P = (x + 5)(ax2 + bx + 25)
= ax3 + bx2 + 25x + 5ax2 + 5bx + 125
= ax3 + (bx2 + 5ax2) + (25x + 5bx) + 125
= ax3 + x2(b + 5a) + x(25 + 5b) + 125
b)\(P=ax^3+x^2\left(b+5a\right)+x\left(5b+25\right)+125\)
\(Q=x^3+125\). ĐỒng nhất 2 đa thức ta có:
\(\hept{\begin{cases}ax^3=x^3\\x^2\left(b+5a\right)+x\left(5b+25\right)=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a=1\\x^2\left(b+5a\right)+x\left(5b+25\right)=0\end{cases}}\)
\(\Rightarrow x^2\left(b+5\right)+5x\left(b+5\right)=0\)
\(\Rightarrow\left(x^2+5x\right)\left(b+5\right)=0\)
\(\Rightarrow b=-5\). Vậy...
1. Ta có:
\(P=ax^3+bx^2+25x+5ax^2+5bx+125=ax^3+\left(b+5a\right)x^2+\left(25+5b\right)x+125\)
Vậy để P = Q thì \(\hept{\begin{cases}a=1\\b+5a=0\\25+5b=0\end{cases}\Rightarrow\hept{\begin{cases}a=1\\b=-5\end{cases}}}\)
2. Hoàn toàn tương tự.
\(a,P=\left(x-a\right)\left(x-b\right)\left(x-c\right)\)
\(=(x^2-ax-bx+ac)\left(x-c\right)\)
\(=x^3-cx^2-ax^2+cax-bx^2+bcx+abx-abc\)
\(=x^3-x^2\left(a+b+c\right)+x\left(ab+bc+ca\right)-abc\)
\(=x^3-12x^2+47x-60\)
\(b,\) Ta có \(\left(x-4\right)^3=x^3-12x^2+48x-64\)
\(\Rightarrow P=\left(x-4\right)^3-\left(x+4\right)\)
Đặt \(t=x-4\)
\(\Rightarrow P=t^3-t\)
\(\Rightarrow P=t\left(t-1\right)\left(t+1\right)\)
\(\Rightarrow P=\left(x-4\right)\left(x-3\right)\left(x-5\right)\)
\(\left|x\right|=3\Rightarrow x=\orbr{\begin{cases}3\\-3\end{cases}}\)
Với \(x=3\Rightarrow P=0\)
Với \(x=-3\Rightarrow P=-336\)
\(=\dfrac{1}{2}x^3-5x^2-x^2+10x+\dfrac{3}{2}x-15=\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)
(x^2−2x+3)(12x−5)
=12x^3−x^2+32x−5x^2+10x−15
=12x^3−6x^2+23/2x−15