Question

Sa se calculeze:
a) (2x-y)^2
b) (4+2x)^2
c) (x supra 3 + 2y )^3
d) (y supra 2 - 4z)^3
e) (3x-2y+radical din 5 )^2

Answer 1

[tex]a) (2x-y)^2=(2x)^2-2*2x*y+y^2=4 x^{2} -4xy+ y^{2} \\ b) (4+2x)^2= 4^{2}+2*4*2x+(2x)^2=16+16x+4 x^{2} \\ c) ( \frac{x}{3} +2y)^3=( \frac{x}{3} )^3+3*( \frac{x}{3} )^2*2y+3*\frac{x}{3}*(2y)^2+(2y)^3= \\ =\frac{ x^{3} }{27}+6y* \frac{ x^{2}}{9}}+x*4y^{2} +8y^3=\frac{ x^{3} }{27}+ \frac{2}{3} x^{2} y+4x y^{2} +8y^3 \\ d)( \frac{y}{2}-4z )^3=(\frac{y}{2})^3-3*(\frac{y}{2})^2*4z+3*\frac{y}{2}*(4z)^2-(4z)^3= \\ =\frac{y^{3}}{8}-3* \frac{ y^{2} }{4}*4z+3* \frac{y}{2}*16 z^{2} -64 z^{3}= \frac{y^{3}}{8}-3y ^{2}z+24yz^{2}-64z^{3} \\ e) (3x-2y+ \sqrt{5} )^2=(3x)^2-(2y)^2+( \sqrt{5} )^2-2*3x*2y+2*3x* \sqrt{5} \\ -2* 2y*\sqrt{5}=9 x^{2} -4 y^{2} +5-12xy+6 \sqrt{5}x-4 \sqrt{5}y [/tex]