Cho \(\frac{a}{4}=\frac{b}{7}\) . CMR:
a) \(\frac{a+4}{4}=\frac{b+7}{7}\)
b) \(\frac{a+8}{b+14}=\frac{4}{7}\)
Tính hợp lí:
A = \(\left(\frac{-4}{5}+\frac{4}{3}\right)+\left(\frac{-5}{4}+\frac{14}{5}\right)-\frac{7}{3}\)
B= \(\frac{8}{3}.\frac{2}{5}.\frac{3}{8}.10.\frac{19}{92}\)
C= \(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{14}+1\frac{5}{7}\)
A= (-4/5+4/3)+(-5/4+14/5)-7/3
= 8/15+31/20-7/3
= 25/12-7/3
= -1/4
B= 8/3.2/5.3/8.10.19/92
= 16/15.3/8.10.19/92
= 2/5.10.19/92
= 4.19/92
= 19/23
C= \(\frac{-5}{7}\).\(\frac{2}{11}\)+\(\frac{-5}{7}\).\(\frac{9}{14}\)+1\(\frac{5}{7}\)
=\(\frac{-5}{7}\).\(\frac{2}{11}\)+\(\frac{-5}{7}\).\(\frac{9}{14}\)+\(\frac{12}{7}\)
= \(\frac{-10}{77}\)+\(\frac{-5}{7}\).\(\frac{9}{14}\)+\(\frac{12}{7}\)
= \(\frac{-10}{77}\)+\(\frac{-45}{98}\)+\(\frac{12}{7}\)
= \(\frac{-635}{1078}\)+\(\frac{12}{7}\)
= \(\frac{1213}{1078}\)
Tính:
a) \(A=\frac{\frac{7}{8}+\frac{7}{27}-\frac{7}{49}}{\frac{11}{8}+\frac{11}{27}-\frac{11}{49}}\)
b)\(B=\frac{\frac{8}{9}-\frac{8}{27}-\frac{8}{81}+\frac{8}{243}}{4-\frac{4}{3}-\frac{4}{9}+\frac{4}{27}}\)
c)\(C=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}-\frac{3}{293}}\)
\(c)\) \(C=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}-\frac{3}{293}}\)
\(C=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{193}\right)}\)
\(C=\frac{2}{3}\)
Bạn Cô nàng Thiên Bình làm đúng hết òi =.=
a=7.[1/8+1/27-1/49]
------------------------
11.[1/8+1/27-1/49]
=7/11
cau b,c tuong tu nha h mk
a)\(A=\frac{\frac{7}{8}+\frac{7}{27}-\frac{7}{49}}{\frac{11}{8}+\frac{11}{27}-\frac{11}{49}}\)
\(A=\frac{7.\left(\frac{1}{8}+\frac{1}{27}-\frac{1}{49}\right)}{11.\left(\frac{1}{8}+\frac{1}{27}-\frac{1}{49}\right)}\).
\(A=\frac{7}{11}\)
b)\(B=\frac{\frac{8}{9}-\frac{8}{27}-\frac{8}{81}+\frac{8}{243}}{4-\frac{4}{3}-\frac{4}{9}+\frac{4}{27}}\)
\(B=\frac{\frac{8}{9}.\left(1-\frac{1}{3}-\frac{1}{9}+\frac{1}{27}\right)}{4.\left(1-\frac{1}{3}-\frac{1}{9}+\frac{1}{27}\right)}\)
\(B=\frac{8}{9}:4=\frac{2}{9}\)
c)\(C=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}-\frac{3}{293}}\)\(C=\frac{2.\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{239}\right)}{3.\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{239}\right)}\)
C=\(\frac{2}{3}\)
Bài 1: Tính (hợp lý nếu có thể)
a) \(\frac{8}{40}+\frac{-4}{20}-\frac{3}{5}\)
b) \(\frac{-7}{12}+\frac{-2}{12}-\frac{-3}{36}\)
c) \((\frac{1}{6}+\frac{-4}{13})-(\frac{-17}{6}-\frac{30}{13})\)
d) \(-\frac{-5}{4}+\frac{7}{4}-\frac{-11}{7}+\frac{2}{7}\)
e) \(-\frac{1}{8}+\frac{-7}{9}+\frac{-7}{8}+\frac{6}{7}+\frac{2}{14}\)
f) \(\frac{-2}{9}-\frac{11}{-9}+\frac{5}{7}-\frac{-6}{-7}\)
Bài 1:
a) Ta có: \(\frac{8}{40}+\frac{-4}{20}-\frac{3}{5}\)
\(=\frac{1}{5}+\frac{-1}{5}-\frac{3}{5}\)
\(=\frac{-3}{5}\)
b) Ta có: \(\frac{-7}{12}+\frac{-2}{12}-\frac{-3}{36}\)
\(=\frac{-7}{12}+\frac{-2}{12}-\frac{-1}{12}\)
\(=\frac{-9+1}{12}=\frac{-8}{12}=\frac{-2}{3}\)
c) Ta có: \(\left(\frac{1}{6}+\frac{-4}{13}\right)-\left(-\frac{17}{6}-\frac{30}{13}\right)\)
\(=\frac{1}{6}+\frac{-4}{13}+\frac{17}{6}+\frac{30}{13}\)
\(=3+2=5\)
d) Ta có: \(-\frac{-5}{4}+\frac{7}{4}-\frac{-11}{7}+\frac{2}{7}\)
\(=\frac{5}{4}+\frac{7}{4}+\frac{11}{7}+\frac{2}{7}\)
\(=3+\frac{13}{7}=\frac{21}{7}+\frac{13}{7}=\frac{34}{7}\)
e) Ta có: \(-\frac{1}{8}+\frac{-7}{9}+\frac{-7}{8}+\frac{6}{7}+\frac{2}{14}\)
\(=-1+1+\frac{-7}{9}\)
\(=-\frac{7}{9}\)
f) Ta có: \(\frac{-2}{9}-\frac{11}{-9}+\frac{5}{7}-\frac{-6}{-7}\)
\(=\frac{-2-\left(-11\right)}{9}+\frac{5-6}{7}\)
\(=1+\frac{-1}{7}=\frac{7}{7}+\frac{-1}{7}=\frac{6}{7}\)
Tính:
a) \(A=\frac{\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{7}}{\frac{8}{2}+\frac{8}{3}+\frac{8}{7}}}{\frac{\frac{1}{7}+\frac{1}{8}+\frac{1}{92}}{\frac{4}{7}+\frac{4}{8}+\frac{4}{92}}}\) ;b) \(B=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+.........+\frac{1}{120}\)
Giúp mình vs mình tick cho.
Cho:
\(A=4+\frac{1}{7^6}+\frac{3}{7}+\frac{4}{7^2}+\frac{-441}{7^6}+\frac{27}{7^5}\)
\(B=\frac{147}{7}+4+\frac{35}{7^7}+\frac{4}{7^2}+\frac{27}{7^5}+\frac{-9}{7^9}\)
Hãy so sánh A với B
1. tính nhanh các tổng sau
a) A=\(\frac{20}{39}+\frac{22}{37}+\frac{18}{43}vàB=\frac{14}{39}+\frac{22}{39}+\frac{18}{41^{ }}\)
b) A=\(\frac{3}{8^3}+\frac{7}{8^4}\)và B=\(\frac{7}{8^3}+\frac{3}{8^4}\)
so sánh A và B :
a) A = \(\frac{20}{39}+\frac{22}{27}+\frac{18}{43}\) ; B = \(\frac{14}{39}+\frac{22}{29}+\frac{18}{41}\)
b) A = \(\frac{3}{8^3}+\frac{7}{8^4}\) , B= \(\frac{7}{8^3}+\frac{3}{8^4}\)
c) A = \(\frac{10^7+5}{10^7-8}\) , B = \(\frac{10^8+6}{10^8-7}\)
d) A = \(\frac{10^{1992}+1}{10^{1991}+1}\), B = \(\frac{10^{1933}+1}{10^{1992}+1}\)
b/ Ta có
\(A-B=\frac{3}{8^3}+\frac{7}{8^4}-\frac{7}{8^3}-\frac{3}{8^4}\)
\(=\frac{4}{8^4}-\frac{4}{8^3}< 0\)
Vậy A < B
c/ Đặt \(10^7=a\)thì ta có
\(A=\frac{a+5}{a-8};B=\frac{10a+6}{10a-7}\)
Giả sử A>B thì ta có
\(\frac{a+5}{a-8}>\frac{10a+6}{10a-7}\)
\(\Leftrightarrow10a^2+43a-35>10a^2-574a-348\)
\(\Leftrightarrow617a+313>0\)(đúng)
Vậy A>B
c/ Đặt \(10^{1991}=a\)thì ta có
\(A=\frac{10a+1}{a+1};B=\frac{100a+1}{10a+1}\)
Giả sử A>B thì ta có
\(\frac{10a+1}{a+1}>\frac{100a+1}{10a+1}\)
\(\Leftrightarrow\left(10a+1\right)^2>\left(100a+1\right)\left(a+1\right)\)
\(\Leftrightarrow-81a>0\)(sai)
Vậy A < B
a/ Thì quy đồng là ra nhé
a,b,c,d giống nhau cùng nhân A và B với 1 số nào đấy tách ra r` so sạmh
mọi người giúp tớ nhanh nhanh với nhé, 1 h tớ phải nộp rồi
Bài 1: tính nhanh
a)\(6\frac{4}{5}-\left(1\frac{2}{3}+3\frac{4}{5}\right)\)
b)\(\left(\frac{-4}{5}+\frac{4}{3}\right)+\left(\frac{-5}{4}+\frac{14}{5}\right)-\frac{7}{3}\)
c)\(\frac{8}{3}.\frac{2}{5}.\frac{3}{8}.10\frac{19}{92}\)
d)\(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{14}+1\frac{5}{7}\)
e)\(\frac{12}{19}.\frac{7}{15}.\frac{-13}{17}.\frac{19}{12}.\frac{17}{13}\)
\(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(B=6\frac{3}{8}+5\frac{1}{2}\)
\(C=\frac{-3}{7}.\frac{5}{9}+\frac{4}{9}.\frac{-3}{7}+2\frac{3}{7}\)
\(D=8\frac{2}{7}-\left(3\frac{4}{9}+4\frac{2}{7}\right)\)