\(a,=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}\\ =\left(5-12+6-4\right)\sqrt{5}\\ =-5\sqrt{5}\)

\(b,=2.\dfrac{3\sqrt{3}}{2}-\dfrac{4\sqrt{3}}{3}-\dfrac{2}{5}.\dfrac{5\sqrt{3}}{4}\\ =3\sqrt{3}-\dfrac{4\sqrt{3}}{3}-\dfrac{\sqrt{3}}{2}\\ =\dfrac{7\sqrt{3}}{6}\)

ab , bc có dấu vecto ko 

vecto hả bạn 

\(1,\\ =2x^2+9x-2x-9\\ =x\left(2x+9\right)-\left(2x+9\right)\\ =\left(x-1\right)\left(2x+9\right)\\ 2,\\ =x^3\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^3-1\right)\left(x^2-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)\\ =\left(x-1\right)\left(x^2+x+1+x+1\right)\\ =\left(x-1\right)\left(x^2+2x+2\right)\)

\(1,45-3\left(x-2\right)=12\\ =>3\left(x-2\right)=45-12\\ =>3\left(x-2\right)=33\\ =>x-2=33:3\\ =>x-2=11\\ =>x=11+2=13\\ 2,120:\left[3\left(x-25\right)\right]=10\\ =>3\left(x-25\right)=120:10\\ =>3\left(x-25\right)=12\\ =>x-25=12:3\\ =>x-25=4\\ =>x=4+25=29\)

\(a,96=3.2^5\\ 120=2^3.3.5\\ =>UCLN\left(96,120\right)=2^3.3=24\)

\(b,17=1.17\\ 20=4.5\)

=> ko có UCLN 

\(c,36=2^2.3^2\\ 60=3.5.2^2\\ 72=2^3.3^2\\ =>UCLN\left(36;60;72\right)=2^2.3=12\)

\(d,240=2^4.3.5\\ 360=2^3.3^2.5\\ =>UCLN\left(240;360\right)=2^3.3.5=120\)

do nguyên tử Z  có 2 lớp e,lớp thứ 2 có 2 e.

=>  \(1s^22s^2\)

=> \(Z=2\)

\(1,\left(x-2\right)^3=x^3-3.x^2.2+3.x.2^2-2^3\\ =x^3-6x^2+12x-8\\ 2,\left(y-3\right)^3=y^3-3.y^2.3+3y.9-27\\ =y^3-9y^2+27y-27\\ 3,\left(2x-5\right)^3=\left(2x\right)^3-3.4x^2.5+3.2x.25-125\\ =8x^3-60x^2+375x-125\)

\(đkx\ge0\\ P=\dfrac{\sqrt{x}+7}{\sqrt{x}+2}\\ =1+\dfrac{5}{\sqrt{x}+2}\)

=>  \(\sqrt{x}+2\inƯ\left(5\right)\)

mà \(Ư\left(5\right)=\left\{1;-1;5;-5\right\}\)

=> \(x=9\left(thoamanđk\right)\)

Vậy \(x=9\) thì P nguyên