\(501\frac{1}{3}.10.\frac{7}{35}.0,75\%\)
câu này là câu lớp 6 nhưng mà hơi dễ đấy ^-^
Tính nhanh
\(501\frac{1}{3}.10.\left(\frac{7}{25}\right).0,75\%\)
cố gắng làm đc nha :)
\(=1670.\frac{7}{25}.\frac{3}{4}\)
\(=350,7\)
giup em lam bai voi
\(E=50\%.1\frac{1}{3}.10.\frac{7}{35}.0,75\)
\(\frac{1x2x3+3x6x9+5x10x15}{2x4x6+6x12x18+10x20x30}\)
\(\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{21}+\frac{1}{35}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{21}+\frac{2}{35}}\)
\(\frac{0,25+\frac{1}{9}-\frac{1}{11}}{0,75+\frac{1}{3}-\frac{3}{11}}+\frac{2}{3}\)
M=\(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}1\frac{5}{7}\)
N=\(\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}.\left(-2\right)^2\)
P=\(50\%.1\frac{1}{3}.10.\frac{7}{35}.0,75\)
thuc hiện phép tính
GIÚP MK VS =-=
a, \(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}.\frac{12}{7}\)
\(=\frac{-5}{7}.\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{12}{7}\)
\(=\frac{-5}{7}.1+\frac{12}{7}=\frac{-5}{7}+\frac{12}{7}=\frac{7}{7}=1\)
a) \(\frac{-1}{3}-\frac{8}{35}+\frac{-2}{9}-\frac{1}{135}+\frac{4}{5}+\frac{-4}{9}+\frac{3}{7}\)
b) \(\frac{0,75+0,6-\frac{3}{7}-\frac{3}{13}}{2,75+2,2-\frac{11}{7}-\frac{11}{13}}\)
\(a,15\frac{3}{13}-\left(3\frac{4}{7}+8\frac{3}{13}\right)\)
\(b,\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)
\(c,\frac{-7}{9}.\frac{4}{11}+\frac{-7}{9}.\frac{7}{11}+5\frac{7}{9}\)
\(d,50\%.1\frac{1}{3}.10.\frac{7}{35}.0,75\)
\(e,\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{40.43}\)
\(a,15\frac{3}{13}-\left(3\frac{4}{7}+8\frac{3}{13}\right)\\ =15\frac{3}{13}-3\frac{4}{7}-8\frac{3}{13}\\ =7\frac{3}{13}-3\frac{4}{7}\\ =\frac{94}{13}-\frac{25}{7}\\ =\frac{94\cdot7-25\cdot13}{13\cdot7}\\ =\frac{333}{91}\)
\(b,\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\\ =7\frac{4}{9}-3\frac{4}{9}+4\frac{7}{11}\\ =4\frac{4}{9}+4\frac{7}{11}\\ =\frac{40}{9}+\frac{51}{11}\\ =\frac{40\cdot11+51\cdot9}{11\cdot9}\\ =\frac{899}{99}\)
\(c,=-\frac{7}{9}\cdot\left(\frac{4}{11}+\frac{7}{11}\right)+\frac{52}{9}\\ =-\frac{7}{9}\cdot1+\frac{52}{9}\\ =\frac{-7+52}{9}\\ =\frac{45}{9}=5\)
\(d,=\frac{50}{100}\cdot\frac{4}{3}\cdot10\cdot\frac{7}{35}\cdot\frac{75}{100}\\ =\frac{50\cdot2\cdot2\cdot2\cdot5\cdot7\cdot25\cdot3}{50\cdot2\cdot3\cdot5\cdot7\cdot25\cdot2\cdot2}=1\)
\(e,=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\\ =1-\frac{1}{43}\\ =\frac{42}{43}\)
a) Ta có: \(15\frac{3}{13}-\left(3\frac{4}{7}+8\frac{3}{13}\right)\)
\(=15+\frac{3}{13}-3-\frac{4}{7}-8-\frac{3}{13}\)
\(=4-\frac{4}{7}=\frac{24}{7}\)
b) Ta có: \(\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)
\(=7+\frac{4}{9}+4+\frac{7}{11}-3-\frac{4}{9}\)
\(=8+\frac{7}{11}=\frac{95}{11}\)
c) Ta có: \(\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+5\frac{7}{9}\)
\(=\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+\frac{-7}{9}\cdot\frac{-52}{7}\)
\(=\frac{-7}{9}\cdot\left(\frac{4}{11}+\frac{7}{11}-\frac{52}{7}\right)\)
\(=\frac{-7}{9}\cdot\frac{45}{-7}=5\)
d) Ta có: \(50\%\cdot1\frac{1}{3}\cdot10\cdot\frac{7}{35}\cdot0.75\)
\(=\frac{1}{2}\cdot\frac{4}{3}\cdot10\cdot\frac{7}{35}\cdot\frac{3}{4}\)
\(=5\cdot\frac{7}{35}=1\)
e) Ta có: \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=1-\frac{1}{43}=\frac{43}{43}-\frac{1}{43}\)
\(=\frac{42}{43}\)
a,\(\frac{1}{2}-\left(\frac{-2}{5}\right)+\frac{1}{3}+\frac{5}{7}-\left(\frac{-1}{6}\right)+\frac{-4}{35}+\frac{1}{41}\)
b,\(0,75.\frac{3}{37}-\frac{3}{4}.\frac{29}{37}-\frac{15}{20}.\frac{11}{37}-0.25\)
Tìm x biết:
\(\frac{2,75-2,2+\frac{11}{7}+\frac{11}{3}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}\) -x-\(\frac{1}{9}=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)
Ai làm đc trước mik cho 2 tick nha
Sửa đề \(\frac{11}{13}\)chứ không phải \(\frac{11}{3}\)
\(\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}-x-\frac{1}{9}=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)
+) Đặt \(A=\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}\)
\(A=\frac{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}\)
\(A=\frac{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
\(A=\frac{11}{3}\)(1)
+) Đặt \(B=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)
\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}\)
\(B=\frac{2}{2}\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}\right)\)
\(B=\frac{2}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\right)\)
\(B=\frac{2}{2}\left(1-\frac{1}{9}\right)=1\cdot\frac{8}{9}=\frac{8}{9}\)(2)
Từ (1) và (2) => \(A-x-\frac{1}{9}=B\)
=> \(\frac{11}{3}-x-\frac{1}{9}=\frac{8}{9}\)
=> \(\frac{11}{3}-x=1\)
=> \(x=\frac{11}{3}-1=\frac{8}{3}\)
Vậy x = 8/3
12)\(0,2.\frac{15}{36}-\left(\frac{2}{5}+\frac{2}{3}\right):1\frac{1}{5}\)
13)\(1\frac{13}{15}.0,75-\left(\frac{8}{15}+0,25\right).\frac{24}{27}\)
16)\(\frac{1}{2}+\frac{3}{4}-\left(\frac{3}{4}-\frac{4}{5}\right).\left(-2,75\right)\)
18)\(\left(\frac{7}{8}-\frac{3}{4}\right).1\frac{1}{3}-\frac{2}{7}.\left(3.5\right)^2\)
22)\(1\frac{13}{15}.0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{3}\)
23)\(\frac{\left(\frac{2}{3}-0,75\right).\left(0,2-\frac{2}{5}\right)}{\frac{5}{9}-1\frac{1}{12}}\)
33)\(\left(25\%+\frac{1}{3}+0,75\right):\left(4\frac{3}{4}-3\frac{1}{2}\right)\)
35)\(\left(\frac{377}{-231}-\frac{123}{89}+\frac{34}{791}\right).\left(\frac{1}{6}-\frac{1}{8}-\frac{1}{24}\right)\)