25 - ( 15 - x + 303) + 303 = 25 - 15 + (303 - 303) + x = x + 10

Tách 134134 =7.11.13.67.2

547547=7.11.13.547

1001 =7.11.13

3003=3x7x11x13

Có mỗi liên kết ở 7.11.13 , 

Bằng 133/550 nka

Tách 134134 =7.11.13.67.2

547547=7.11.13.547

1001 =7.11.13

3003=3x7x11x13

Có mỗi liên kết ở 7.11.13 , tự làm phần tiếp nhé .

\(\frac{202}{303}=\frac{202:101}{303:101}=\frac{2}{3}\)

202/303=2/3

202/303=2/3

Chúc bạn học giỏi nha!!!

K cho mik vs nhé

Không thể rút gọn được nha I can rest your love on me

a) 2x + (-61) - (21 - 61) = 2x - 21 + (61 - 61) = 2x - 21.

b) (- 3 - x + 5) + 3 = (- 3 + 3) + 5 - x = 5 - x.

c) 11- (13 - x) + (13 - 11) = (11- 11) + (13- 13) + x = x

d) 25 - ( 15 - x + 303) + 303 = 25 - 15 + (303 - 303) + x = x + 10

808/909=8/9

111/333=1/3=3/9

202/303=2/3

8/9+3/9-2/9=11/9-2/9

                  =9/9=1

            Học tốt               

8/9 + 37/111 + -2/3

Tự làm nha

808/909+111/333-202/303=8/9+1/3+2/3=8/9+1=17/9

tổng bằng \(\frac{5}{9}\)

\(\frac{808}{909}+\frac{111}{333}+\left(-\frac{202}{303}\right)\)

\(=\frac{8}{9}+\frac{1}{3}-\frac{2}{3}\)

\(=\frac{8}{9}+\frac{3}{9}-\frac{6}{9}\)

\(=\frac{5}{9}\)

\(\frac{808}{909}+\frac{111}{333}+\frac{-202}{303}\)

\(=\frac{8}{9}+\frac{1}{3}+\frac{-2}{3}=\frac{8}{9}+\frac{1+(-2)}{3}=\frac{8}{9}+\frac{-1}{3}=\frac{24+(-9)}{27}=\frac{15}{27}=\frac{5}{9}\)

\(log_{a^3}b.log_ba=\dfrac{1}{3}.log_ab.log_ba=\dfrac{1}{3}\)

\(log_{a^{10}}b^5.log_{b^3}a^9=\dfrac{1}{10}.5.log_ab.\dfrac{1}{3}.9.log_ba=\dfrac{3}{2}\)

\(log_{a^{107}}b^{101}.log_{b^{303}}a^{428}=\dfrac{1}{107}.101.log_ab.\dfrac{1}{303}.428.log_ba=\dfrac{4}{3}.log_ab.log_ba=\dfrac{4}{3}\)

a: \(log_{a^3}b\cdot log_ba=\dfrac{1}{3}\cdot log_ab\cdot log_ba=\dfrac{1}{3}\)

b: \(log_{a^{10}}b^5\cdot log_{b^3}a^9\)

\(=\dfrac{1}{10}\cdot log_ab^5\cdot\dfrac{1}{3}\cdot log_ba^9\)

\(=\dfrac{1}{30}\cdot5\cdot log_ab\cdot9\cdot log_ba=\dfrac{45}{30}=\dfrac{3}{2}\)

c: \(log_{a^{107}}b^{101}\cdot log_{b^{303}}a^{428}\)

\(=\dfrac{1}{107}\cdot log_ab^{101}\cdot\dfrac{1}{303}\cdot log_ba^{428}\)

\(=\dfrac{1}{107}\cdot101\cdot log_ab\cdot\dfrac{1}{303}\cdot428\cdot log_ba\)

\(=4\cdot\dfrac{1}{3}=\dfrac{4}{3}\)