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\(\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)\)
\(=\dfrac{1}{2}x^3-x^2+\dfrac{3}{2}x-5x^2+10x-15\)
\(=\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)
\(\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)\)
\(=\dfrac{1}{2}x^3-5x^2-x^2+10x+\dfrac{3}{2}x-15\)
\(=\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)
\(\left(x^2-2x+3\right)\left(1212x-5\right)\)
\(=1212x^3-5x^2-2424x^2+10x+3636x-15\)
\(=1212x^3-2429x^2+3646x-15\)
\(=1212x^3-5x^2-2424x^2+10x+3636x-15\\ =1212x^3-2429x^2+3646x-15\)
Tham khảo thôi nha .
a) \(P=\left(x+5\right)\left(ax^2+bx+25\right)\)
\(=ax^3+bx^2+25x+5ax^2+5bx+125\)
\(=ax^3+\left(5a+b\right)x^2+\left(5b+25\right)x+125\)
b) Nếu theo đề bài \(\forall x\)thì \(P=Q\)
\(\Leftrightarrow ax^3+\left(5ab\right)x^2+\left(5b+25\right)x+125\)( P)
\(=x^3+125\forall x\)
\(\Leftrightarrow\hept{\begin{cases}a=1\\5a+b=0\\5b+25=0\end{cases}}\)'
\(\Leftrightarrow\hept{\begin{cases}a=1\\b=-5\end{cases}}\)
Vậy ..........
a ) \(\left(5x+2y\right)^2=25x^2+20xy+4y^2\)
b ) \(\left(-3x+2\right)^2=9x^2-12x+4\)
c ) \(\left(\dfrac{2}{3}x+\dfrac{1}{3}y\right)^2=\dfrac{4}{9}x^2+\dfrac{4}{9}xy+\dfrac{1}{9}y^2\)
d ) \(\left(2x-\dfrac{5}{2}y\right)^2=4x^2-10xy+\dfrac{25}{4}y^2\)
e ) \(\left(x+\dfrac{4}{3}y^2\right)^2=x^2+\dfrac{8}{3}xy^2+\dfrac{16}{9}y^4\)
f ) \(\left(2x^2+\dfrac{5}{3}y\right)^2=4x^4+\dfrac{20}{3}x^2y+\dfrac{25}{9}y^2\)
Bài 2:
\(=\dfrac{3x^4+3x^2+x^3+x-3x^2-3+5x-5}{x^2+1}\)
\(=3x^2+x-3+\dfrac{5x-5}{x^2+1}\)
Bài 3:
\(\dfrac{A}{B}=\dfrac{2x^3-x^2-x+1}{x^2-2x}\)
\(=\dfrac{2x^3-4x^2+3x^2-6x+5x+1}{x^2-2x}\)
\(=2x^2+3+\dfrac{5x+1}{x^2-2x}\)
=>\(2x^3-x^2-x+1=\left(x^2-2x\right)\left(2x^2+3\right)+5x+1\)
\(=\dfrac{1}{2}x^3-5x^2-x^2+10x+\dfrac{3}{2}x-15=\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)
(x^2−2x+3)(12x−5)
=12x^3−x^2+32x−5x^2+10x−15
=12x^3−6x^2+23/2x−15