\(=\frac{50-\frac{4}{13}+\frac{2}{15}-\frac{2}{17}}{2\left(50-\frac{4}{13}+\frac{2}{15}-\frac{2}{17}\right)}=\frac{1}{2}\)

\(A=\frac{50-\frac{4}{13}+\frac{2}{15}-\frac{2}{17}}{100-\frac{8}{13}+\frac{4}{15}-\frac{4}{17}}\)

\(=\frac{50-\frac{4}{13}+\frac{2}{15}-\frac{2}{17}}{2\left(50-\frac{4}{13}+\frac{2}{15}-\frac{2}{17}\right)}\)

\(=\frac{1}{2}\)

\(B=\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+......+\frac{9}{1999.2009}\)

\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{19}-\frac{1}{39}+....+\frac{1}{1999}-\frac{1}{2009}\right)\)

\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{2009}\right)\)

\(=\frac{1}{19}+\frac{9}{10}\cdot\frac{1990}{38171}\)

\(=\frac{200}{2009}\)

cái này......hình như ở trên học 24 mà

NĐT_2004_asd nằm mơ đi con nhé nếu giải mày đc 20 điểm hả

\(\frac{50-\frac{4}{13}+\frac{2}{15}-\frac{2}{17}}{100-\frac{8}{13}+\frac{4}{15}-\frac{4}{17}}\)= \(\frac{50-\frac{4}{13}+\frac{2}{15}-\frac{2}{17}}{2.\left(50-\frac{4}{13}+\frac{2}{15}-\frac{2}{17}\right)}\)= \(\frac{1}{2.1}\)=\(\frac{1}{2}\)

\(=\left(\frac{4}{17}+\frac{13}{17}\right)-\left(\frac{7}{15}+\frac{8}{15}\right)+\left(\frac{10}{16}-\frac{7}{16}\right)\)

\(=1-1+\frac{3}{16}\)

\(\frac{3}{16}\)

gộp 1,3 

2,4

5,6

thì ra 17/17-15/15+3/16=3/16

hk tốt!

I

Dễ mà kb vs mk đi

giải giúp mình bài 1 rồi mình addfriend cho

1.a, vì (x+2/3)^2 \(\ge0\Rightarrow\left(x+\frac{2}{3}\right)^2+\frac{1}{2}\ge\frac{1}{2}\)

vậy GTNN của biểu thức A=1/2 đạt được khi x=-2/3

b. ta có: \(\left(x-\frac{1}{2}\right)^2+2\ge2\Rightarrow\frac{1}{\left(x-\frac{1}{2}\right)^2+2}\le\frac{1}{2}\Rightarrow\frac{2}{\left(x-\frac{1}{2}\right)^2+2}\le1\)

Vậy Max của B=1 đạt được khi x=1/2

2.a, tự làm đi. gợi ý: dế thấy dược là mấu sẽ =tử.2 nên bạn đặt 2 làm chung ở mẫu sẽ ra giống ý như tử rồi rút gọn còn 1/2

b. \(\frac{1}{19}+\frac{9}{19.29}...\frac{9}{1999.2009}=\frac{1}{19}+9\left(\frac{1}{19.29}+\frac{1}{29.39}...\frac{1}{1999.2009}\right)\)

\(=\frac{1}{19}+9.\frac{1}{10}\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}...+\frac{1}{1999}-\frac{1}{2009}\right)\)

\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{2009}\right)=\frac{200}{2009}\)

\(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)

\(A=49\frac{8}{23}-5\frac{7}{32}+14\frac{8}{23}\)

\(A= \left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)

\(A=\left[\left(49-14\right)-\left(\frac{8}{23}-\frac{8}{23}\right)\right]-5\frac{7}{32}\)

\(A=\left[35-0\right]-5\frac{7}{32}\)

\(A=35-5\frac{7}{32}\)

\(A=\frac{953}{32}\)

\(B=71\frac{38}{45}-\left(43\frac{38}{45}-1\frac{17}{57}\right)\)

\(B=71\frac{38}{45}-\frac{36377}{855}\)

\(B=\frac{1670}{57}\)

\(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right):\frac{4}{5}\)

\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]:\frac{4}{5}\)

\(C=\left[\frac{51}{8}:\frac{7}{12}\right]:\frac{4}{5}\)

\(C=\frac{153}{14}:\frac{4}{5}\)

\(C=\frac{765}{56}\)

\(D=\left[\left(\frac{10}{15}-\frac{2}{3}\right):\frac{1}{7}\right]\cdot0,15-\frac{1}{4}\)

\(D=\left[0:\frac{1}{7}\right]\cdot\frac{3}{20}-\frac{1}{4}\)

\(D=0\cdot\frac{3}{20}-\frac{1}{4}\)

\(D=0-\frac{1}{4}\)

\(D=-\frac{1}{4}\)

\(E=\frac{13}{30}+\frac{28}{45}\cdot2\frac{1}{2}-\left[\left(\frac{1}{2}+\frac{1}{3}\right):\frac{53}{90}\right]:\frac{50}{53}\)

\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\left[\frac{5}{6}:\frac{53}{90}\right]:\frac{50}{53}\)

\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{75}{53}:\frac{50}{53}\)

\(E=\frac{13}{30}+\frac{14}{9}-\frac{3}{2}\)

\(\)\(E=\frac{22}{45}\)

CHUC BAN HOC TOT >.<

a) \(\frac{31}{17}+\frac{-5}{13}+\frac{-8}{13}-\frac{14}{17}\)

\(=\left(\frac{31}{17}-\frac{14}{17}\right)+\left(\frac{-5}{13}+\frac{-8}{13}\right)\)

\(=1+\left(-1\right)\)

\(=0\)

b) \(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+\frac{5}{7}\)

\(=\frac{-5}{7}.\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{5}{7}\)

\(=\frac{-5}{7}.1+\frac{5}{7}\)

\(=\frac{-5}{7}+\frac{5}{7}\)

\(=0\)

Ta có:

\(VT:\frac{4^{15}}{7^{30}}=\frac{\left(2^2\right)^{15}}{7^{30}}=\frac{2^{30}}{7^{30}}\)

\(VP:\frac{8^{10}\cdot3^{30}}{7^{30}.4^{15}}=\frac{\left(2^3\right)^{10}.3^{30}}{7^{30}.\left(2^2\right)^{15}}=\frac{2^{30}.3^{30}}{7^{30}.2^{30}}=\frac{3^{30}}{7^{30}}\)

Ta thấy :\(\frac{2^{30}}{7^{30}}vs\frac{3^{30}}{7^{30}}\)có:

\(\orbr{\begin{cases}2^{30}< 3^{30}\\7^{30}=7^{30}\end{cases}\Rightarrow\frac{2^{30}}{7^{30}}< \frac{3^{30}}{7^{30}}\Leftrightarrow\frac{4^{15}}{7^{30}}< \frac{8^{10}.3^{30}}{7^{30}.4^{15}}}\)

Chúc bn hok tốt

< minh khong viet cach giai