\(B=\dfrac{2^{24}\cdot3^5-2^{24}\cdot3^4}{2^{24}\cdot3^5}+1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{301}-\dfrac{1}{303}\)

\(=\dfrac{2^{24}\cdot3^4\left(3-1\right)}{2^{24}\cdot3^5}+\dfrac{302}{303}\)

\(=\dfrac{2}{3}+\dfrac{302}{303}=\dfrac{202+302}{303}=\dfrac{504}{303}\)

=168/101

giúp mk ik, năn nỉ đó.

\(=\dfrac{2^{19}\cdot3^9-3\cdot3^8\cdot2^{18}\cdot5}{2^{19}\cdot3^9+2^{20}\cdot3^{10}}=\dfrac{-3^{10}\cdot2^{18}}{2^{19}\cdot3^9\cdot7}=-\dfrac{3}{14}\)

`B=(1/(3-sqrtx)-1/(3+sqrtx))*(3+sqrtx)/sqrtx(x>=0,x ne 9)`

`B=((3+sqrtx)/(9-x)-(3-sqrtx)/(9-x))*(3+sqrtx)/sqrtx`

`B=((3+sqrtx-3+sqrtx)/(9-x))*(3+sqrtx)/sqrtx`

`B=(2sqrtx)/((3-sqrtx)(3+sqrtx))*(3+sqrtx)/sqrtx`

`B=2/(3-sqrtx)`

`B>1/2`

`<=>2/(3-sqrtx)-1/2>0`

`<=>(4-3+sqrtx)/[2(3-sqrtx)]>0`

`<=>(sqrtx+1)/(2(3-sqrtx))>0`

Mà `sqrtx+1>=1>0`

`<=>2(3-sqrtx)>0`

`<=>3-sqrtx>0`

`<=>sqrtx<3`

`<=>x<9`

\(=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}.\dfrac{12}{7}\)

\(=\dfrac{2}{3}+\dfrac{7.3.2.2}{3.7.3.2.3}\)

\(=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)

TICK CHO MÌNH NHÉ

Giải:

\(\dfrac{2}{3}\) + \(\dfrac{1}{3}\) . (\(-\dfrac{4}{9}\) + \(\dfrac{5}{6}\) ) : \(\dfrac{7}{12}\)

 = \(\dfrac{2}{3}\)  + \(^{\dfrac{1}{3}}\) . \(\dfrac{7}{18}\) : \(\dfrac{7}{12}\)

 = \(\dfrac{2}{3}\) + \(\dfrac{7}{54}\) : \(\dfrac{7}{12}\)

 = \(\dfrac{2}{3}\) + \(\dfrac{2}{9}\)

 = \(\dfrac{8}{9}\)

2 /3 + 1/ 3 . ( − 4 9 + 5 6 ) : 7 /12

= 2/ 3 + 1 /3 . 7 /18 . 12/ 7

= 2/ 3 + 7 /48 . 12 /7

= 2/ 3 + 1/ 4

= 11/ 12

\(=\dfrac{2}{3}+\dfrac{1}{3}.\left(\dfrac{7}{18}\right):\dfrac{7}{12}\)

\(=\dfrac{2}{3}+\dfrac{7}{54}:\dfrac{7}{12}\)

\(=\dfrac{2}{3}+\dfrac{2}{9}\)

\(=\dfrac{8}{9}\)

\(B=\dfrac{2x^2-12x+25}{x^2-6x+12}=\dfrac{2\left(x^2-6x+12\right)+1}{x^2-6x+12}=2+\dfrac{1}{x^2-6x+9+4}=2+\dfrac{1}{\left(x-3\right)^2+4}\le2+\dfrac{1}{4}=\dfrac{9}{4}\)

Không có min nha bạn . Chỉ có max thôi 

Dấu = xảy ra khi x=3

\(B=\frac{1}{3}.b+\frac{2}{9}.b– b: \frac{9}{4}= \frac{1}{3}.b+\frac{2}{9}.b – b. \frac{4}{9}\)

\(=b(\frac{1}{3}+\frac{2}{9}-\frac{4}{9})=b. (\frac{3}{9}+\frac{2}{9}-\frac{4}{9})= b. \frac{1}{9} \)

Thay \(b=\frac{9}{10}\) vào B, ta được

B= \(b=\frac{9}{10}. \frac{1}{9}= \frac{1}{10}\)

\(log_2\dfrac{9}{10}+log_330=\) ? bạn chắc đề đúng chứ, 2 cơ số ko giống nhau, rút gọn cũng được nhưng nó sẽ không gọn trên thực tế.

\(log_3\dfrac{5}{9}-2log_3\sqrt{5}=log_3\dfrac{5}{9}-log_35=log_3\left(\dfrac{1}{9}\right)=log_33^{-2}=-2\)

\(log_2\dfrac{16}{3}+2log_2\sqrt{6}=log_2\dfrac{16}{3}+log_26=log_2\left(\dfrac{16}{3}.6\right)=log_232=log_22^5=5\)

\(\frac{2^{19}.27^3+15^4.4^9.9^4}{6^9.2^{10}+12^{10}}\)

\(=\frac{2^{19}.\left(3^3\right)^3+\left(3.5\right)^4.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)

\(=\frac{2^{19}.3^9+3^{12}.5^4.2^{18}}{2^{19}.3^9+2^{20}.3^{10}}\)

\(=\frac{2^{18}.3^9.\left(2+3^3+5^4\right)}{2^{19}.3^9.\left(1+2+3\right)}\)

\(=\frac{654}{2.6}\)

\(=\frac{109}{2}\)

Chúc bn học tốt !!!!