Question

va rog help nu am habar cum sa fac

Answer 1

$\displaystyle 1a).2(x-1)-3(x-2)=2x-2-3x+6=-x+4 \\ \\ b).2(x+3)-(x+1)=2x+6-x-1=x+5 \\ \\c).2(x+5)-(x-1)=2x+10-x+1=x+11 \\ \\ d).x(2x-1)-2x^2=2x^2-x-2x^2=-x\\ \\ e).x(4x-1)-4x^2=4x^2-x-4x^2=-x \\ \\ f).5(x-1)-10x=5x-5-10x=-5x-5$

$\displaystyle 2a).(x-1)(x+1)=x^2-1^2=x^2-1 \\ \\b).(x-2)(x+1)=x^2+x-2x-2=x^2-x-2 \\ \\ c).(x+5)x=x^2+5x\\ \\ d).x^2(x+1)+x^2=x^3+x^2+x^2=x^3+2x^2\\ \\ e).x(x+y)+x^2=x^2+xy+x^2=2x^2+xy \\ \\ f).(5x)^2(x+1)=25x^2(x+1)=25x^3+25x^2$

$\displaystyle 3a).(x+1)(x-1)-(x+2)(x-2)=x^2-1^2-(x^2-2^2)=\\ \\ = x^2-1-(x^2-4)=x^2-1-x^2+4=3 \\ \\ b).(x+1)(x+2)-x(x-1)=x^2+2x+x+2-x^2+x=4x+2 \\ \\ c).(x+7)(2x-1)-( \sqrt{2} x)^2=2x^2-x+14x-7-2x^2= 13x-7 \\ \\ d).2(3x^2-1)-x(x+1)=6x^2-2-x^2-x=5x^2-x-2 \\ \\ e).(x-2)(x+1)-x(x-3)=x^2+x-2x-2-x^2+3x=2x-2\\ \\ f).(3x+1)(4x-5)-12x^2=12x^2-15x+4x-5-12x^2=-11x-5$

$\displaystyle 4a).x^1 \cdot x^2 \cdot x^3-x^6=x^{1+2+3}-x^6=x^6-x^6=0 \\ \\ b).(x-1)^1(x-1)^2(x-1)^3-(x-1)^6=(x-1)^{1+2+3}-(x-1)^6=\\ \\ =(x-1)^6-(x-1)^6=0 \\ \\ c).(x+ \sqrt{2} )^1(x+ \sqrt{2} )^2-(x+ \sqrt{2} )^3=(x+ \sqrt{2} )^{1+2}-(x+ \sqrt{2} )^3= \\ \\=(x+ \sqrt{2} )^3-(x+ \sqrt{2} )^3=0\\ \\ d).(x \sqrt{2} +1)^2(x \sqrt{2} +1)^1-(x \sqrt{2} +1)^3=(x \sqrt{2} +1)^{2+1}-(x \sqrt{2} +1)^3= \\ \\ =(x \sqrt{2} +1)^3-(x \sqrt{2} +1)^3=0$

$\displaystyle 5a).S=x+2x+3x+...+2014x=x(1+2+3+...+2014)=\\ \\ =x \cdot \frac{2014(2014 +1)}{2} =x \cdot \frac{2014 \cdot 2015}{2} =x \cdot \frac{4058210}{2} = 2029105x \\ \\ b).(S:1007-x):2014=(2029105x:1007-x):2014= \\ \\ =(2015x-x):2014=2014x:2014=x$