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a ) \(\left(x+y\right)^5=x^5+5x^4y+10x^3y^2+5xy^4+y^5\)
b ) \(\left(x-3y\right)^6=\left[x+\left(-3y\right)\right]^6\)
\(=x^6+6x^5\left(-3y\right)+15x^4\left(-3y\right)^2+20x^3\left(-3y\right)^3+15x^2\left(-3y\right)^4+6x\left(-3y\right)^5+\left(-3y\right)^6\)
\(=x^6-18x^5y+135x^4y^2-540x^3y^3+1215x^2y^4-1458xy^5+729y^6\)
Chúc bạn học tốt
\(\left(x^3-1\right)\left(x^3+1\right)=\left(x^3\right)^2-1^2=x^6-1\)
\(\left(x^3-1\right)\left(x^3+1\right)\)
\(=\left(x^3\right)^2-1^2\)
\(=x^3-1\)
Z thôi T nha
1.
a) \({\left( {x + 3} \right)^3} = {x^3} + 3.{x^2}.3 + 3.x{.3^2} + {3^3} = {x^3} + 9{x^2} + 27x + 27\)
b) \({\left( {x + 2y} \right)^3} = {x^3} + 3.{x^2}.2y + 3.x.{\left( {2y} \right)^2} + {\left( {3y} \right)^3} = {x^3} + 6{x^2}y + 12x{y^2} + 27{y^3}\)
2.
\(\begin{array}{l}{\left( {2x + y} \right)^3} - 8{x^3} - {y^3} = {\left( {2x} \right)^3} + 3.{\left( {2x} \right)^2}.y + 3.2x.{y^2} + {y^3} - 8{x^3} - {y^3}\\ = 8{x^3} + 12{x^2}y + 6x{y^2} + {y^3} - 8{x^3} - {y^3}\\ = \left( {8{x^3} - 8{x^3}} \right) + 12{x^2}y + 6x{y^2} + \left( {{y^3} - {y^3}} \right)\\ = 12{x^2}y + 6x{y^2}\end{array}\)
khai triển :
\(\left(2x^2+3y\right)^3=\left(2x^2\right)^3+3.\left(2x^2\right)^2.3y+3.2x^2.\left(3y\right)^2+\left(3y\right)^3\)
\(=8x^6+3.4x^4.3y+3.2x^2.9y+27y^3=8x^6+36x^4y+54x^2y+27y^3\)
Vậy hệ số của x4y trong khai triển.... là 36
Ta có : A = (x - 1 + y)2
= [(x - 1) + y]2
= (x - 1)2 + 2(x - 1)y + y2
= x2 - 2x + 1 + 2xy - 2y + y2
= x2 + y2 + 1 - 2x - 2y + 2xy
( x - y )4
= x4 - y4
= (x2)2 - (y2)2
=\(\orbr{ }\left(x^2\right)-\left(y^2\right)]^2\)
\(\left(4x-1\right)^2+\left(x+3\right)^2=16x^2-8x+1+x^2+6x+9\)
\(=17x^2-2x+10\)
\(\left(x-y+1\right)^3=x^3-y^3+1-3x^2y+3xy^2+3x^2+3x+3y^2-3y-6xy\)
\(\left(4x-1\right)^2+\left(x+3\right)^2=16x^2-8x+1+x^2+6x+9\) \(=17x^2-2x+10\)
\(\left(x-y+1\right)^3=\left(x-y\right)^3+3\left(x-y\right)^2+3\left(x-y\right)+1\)
\(\left(x-\dfrac{y}{x}\right)^3\\ =x^3-3\cdot x^2\cdot\dfrac{y}{x}+3\cdot x\cdot\left(\dfrac{y}{x}\right)^2-\left(\dfrac{y}{x}\right)^3\\ =x^3-3xy+3x\cdot\dfrac{y^2}{x^2}-\dfrac{y^3}{x^3}\\=x^3-3xy+\dfrac{3y^2}{x}-\dfrac{y^3}{x^3}\)
\(\left(x-\dfrac{y}{x}\right)^3=x^3-\dfrac{3x^2.y}{x}+3x.\left(\dfrac{y}{x}\right)^2-\left(\dfrac{y}{x}\right)^3=x^3-3xy+\dfrac{3y^2}{x}-\dfrac{y^3}{x^3}\)