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a) \(x^3\left(3x^2-x-\dfrac{1}{2}\right)=3x^5-x^4-\dfrac{1}{2}x^3\)
b) \(\left(5xy-x^2+y\right)\dfrac{2}{5}xy^2=2x^2y^3-\dfrac{2}{5}x^3y^2+\dfrac{2}{5}xy^3\)
c) \(\left(4x^3-3xy^2+2xy\right)\left(-\dfrac{1}{3}x^2y\right)=\dfrac{-4}{3}x^5y+x^3y^3-\dfrac{2}{3}x^3y^2\)
Bài 1 :
a) \(x^4-4x^2-4x-1\)
\(=x^4-\left(4x^2+4x+1\right)\)
\(=x^4-\left(2x+1\right)^2\)
\(=\left(x^2-2x-1\right)\left(x^2+2x+1\right)\)
b) \(x^2+2x-15\)
\(=x^2+2x+1-16\)
\(=\left(x+1\right)^2-4^2\)
\(=\left(x+1+4\right)\left(x+1-4\right)=\left(x+5\right)\left(x-3\right)\)
c) \(x^3y-2x^2y^2+5xy\)
\(=xy\left(x^2-2xy+5\right)\)
B2:
a) \(2\left(x-1\right)^2-\left(2x+3\right)\left(2x-3\right)\)
\(=2\left(x^2-2x+1\right)-\left(4x^2-9\right)\)
\(=2x^2-4x+2-4x^2+9\)
\(=-2x^2-4x+11\)
b) \(\left(x+3\right)^2-2\left(x+3\right)\left(x-3\right)+\left(x-3\right)^2\)
\(=\left(x+3-x+3\right)^2=6^2=36\)
c) \(4\left(x-1\right)\left(x+3\right)+5\left(2x+1\right)^2-2\left(5-3x\right)^2\)
\(=4\left(x^2+2x-3\right)+5\left(4x^2+4x+1\right)-2\left(9x^2-30x+25\right)\)
\(=4x^2+8x-12+20x^2+20x+5-18x^2+60x-50\)
\(=6x^2+88x-57\)
b)x^2+y^2=x+y+8
=>4x^2+4y^2-4x-4y=32
=>4x^2-4x+1+4y^2-4y+1=34
=>(2x-1)^2+(2y-1)^2=9+25=25+9
đến đây thì dễ rồi
y^2+2xy-3x-2=0
=>y^2+2xy+x^2=x^2+3x+2
=>(x+y)^2=(x+2)(x+1)
đến đây thì bn tự lm nha
a) -5xy(3x2y – 5xy +y2)
=-15x3y2+25x2y2-5xy3
b) (x + 8)2 -2(x + 8) (x – 2) + (x – 2)2
=[(x+8)-(x-2)]2
=(x+8-x+2)2
=102
=100
\(a,-5xy\left(3x^2y-5xy+y^2\right)=-15x^3y^2+25x^2y^2-5xy^3\)
\(b,\left(x+8\right)^2-2\left(x+8\right)\left(x-2\right)+\left(x-2\right)^2=\left[x+8-\left(x-2\right)\right]^2=\left[x+8-x+2\right]^2=10^2=100\)