Áp dụng hằng đẳng thức, khai triển các biểu thức sau:
a, \(\left(2x+y+3\right)^2\)
b, \(\left(x-2y+1\right)^2\)
c, \(\left(x^2-2xy^2-3\right)^2\)
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\(a.\left(2xy-3\right)^2=4x^2y^2-12xy+9\)
\(b.\left(\dfrac{1}{2}x+\dfrac{1}{3}\right)^2=\dfrac{1}{4}x^2+\dfrac{1}{3}x+\dfrac{1}{9}\)
Giải:
a) \(\left(2x+y+3\right)^2\)
\(=\left(2x+y\right)^2+2.3\left(2x+y\right)+3^2\)
\(=\left(2x\right)^2+2.2x.y+y^2+2.3\left(2x+y\right)+3^2\)
\(=4x^2+4xy+y^2+12x+6y+9\)
Vậy ...
b) \(\left(x-2y+1\right)^2\)
\(=\left(x-2y\right)^2+2\left(x-2y\right)+1^2\)
\(=x^2-2.x.2y+\left(2y\right)^2+2x-4y+1^2\)
\(=x^2-4xy+4y^2+2x-4y+1\)
Vậy ...
c) \(\left(x^2-2xy^2-3\right)^2\)
\(=\left(x^2-2xy^2\right)^2+2.3.\left(x^2-2xy^2\right)-3^2\)
\(=\left(x^2\right)^2-2.x^2.2xy^2+\left(2xy^2\right)^2+2.3.\left(x^2-2xy^2\right)-3^2\)
\(=x^4-4x^3y^2+4x^2y^4+6x^2-12xy^2-9\)
Vậy ...
a,\(\left(x^2+2xy\right)^3=\left(x^2\right)^3+3.\left(x^2\right)^2.2xy+3.\left(2xy\right)^2.x^2+\left(2xy\right)^3\)
\(=x^6+6x^5y+12x^4y^2+8x^3y^3\)
b,\(\left(3x^2-2y\right)^3=\left(3x^2\right)^3-3.\left(3x^2\right)^2.2y+3.\left(2y\right)^2.3x^2-\left(2y\right)^3\)
\(=27x^6-54x^4y+36y^2x^2-8y^3\)
c,\(\left(2x^3-y^2\right)^3=8x^9-12x^6y^2+6x^3y^4-y^6\)
a) \(\left(3x^2-2y^3\right)^2\)
\(=\left(3x^2\right)^2-2\cdot3x^2\cdot2y^3+\left(2y^3\right)^2\)
\(=9x^4-12x^2y^3+4y^6\)
b) \(\left(-2x^2-3\right)^2\)
\(=\left(-2x^2\right)^2-2\cdot\left(-2x^2\right)\cdot3+3^2\)
\(=4x^4+12x^2+9\)
a) \(\left(x^2+2xy\right)^3\)
\(=\left(x^2\right)^3+3\left(x^2\right)^22xy+3x^2\left(2xy\right)^2+\left(2xy\right)^3\)
\(=x^6+6x^5y+12x^4y^2+8x^3y^3\)
b) \(\left(3x^2-2y\right)^3\)
\(=\left(3x^2\right)^3-3\left(3x^2\right)^22y+3.3x^2\left(2y\right)^2-\left(2y\right)^3\)
\(=27x^6-54x^4y+36x^2y^2-8y^3\)
c) \(\left(2x^3-y^2\right)^3\)
\(=\left(2x^3\right)^3-3\left(2x^3\right)^2y^2+3.2x^3\left(y^2\right)^2-\left(y^2\right)^3\)
\(=8x^9-12x^6y^2+6x^3y^4-y^6.\)
a) \(\left(2x^2-1\right)^2=\left(2x^2\right)^2-2.2x^2.1+1^2\)
\(=4x^4-4x^2+1\).
b) \(\left(\frac{1}{2}x+3y^2\right)^2=\left(\frac{1}{2}x\right)^2+2.\frac{1}{2}x.3y^2+\left(3y^2\right)^2\)
\(=\frac{1}{4}x^2+3y^2x+9y^4\)
Chúc bn hc tốt!
\(a,\left(2x+y+3\right)^2=4x^2+y^2+9+4xy+12x+6y\)
\(b,\left(x-2y+1\right)^2=x^2+4y^2+1-4xy+2x-4y\)
\(c,\left(x^2-2xy^2-3\right)^2=x^4+2x^2y^4+9-4x^3y^2-6x^2+12xy^2\)