Bài 1 : TÍNH
a) \(\frac{1}{2} + \frac{3}{8} : 3 - \frac{3}{16} . 2^3 \) b) \((33\frac{1}{17}+67\frac{2}{193}-100\frac{7}{929}):(\frac{1}{17}+\frac{2}{193}-\frac{7}{929} \)
Bài 2 : TÌM x:
a) \(\frac{9}{4}.|x|-4=\frac{1}{4}.|x|\) b) \(3x-1,5=5-0,5\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right):\left(\left(\frac{7}{1931}+\frac{11}{3862}\right)\cdot\frac{1931}{25}+\frac{9}{2}\right)\)
= \(\left(\left(\frac{4}{386}-\frac{3}{386}\right)\cdot\frac{193}{17}+\frac{33}{34}\right):\left(\left(\frac{14}{3862}+\frac{11}{3862}\right)\cdot\frac{1931}{25}+\frac{9}{2}\right)\)
= \(\left(\frac{1}{186}\cdot\frac{193}{17}+\frac{33}{34}\right):\left(\frac{25}{3862}\cdot\frac{1931}{25}+\frac{9}{2}\right)\)
= \(\left(\frac{1}{34}+\frac{33}{34}\right):\left(\frac{1}{2}+\frac{9}{2}\right)\)
= \(1:5\)
= \(\frac{1}{5}=0,2\)
\(M=\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{2001}+\frac{11}{4002}\right).\frac{2001}{25}+\frac{9}{2}\right] \)
\(=\left(\frac{2}{17}-\frac{3}{34}+\frac{33}{34}\right):\left(\frac{7}{25}+\frac{11}{50}+\frac{9}{2}\right)\)
\(=\frac{4-3+33}{34}:\frac{14+11+225}{50}=1:5=0.2\)
\(=\left(\frac{1}{386}-\frac{193}{17}+\frac{33}{34}\right):\left(\frac{25}{3862}\cdot\frac{1931}{25}+\frac{9}{2}\right)\)
\(=\left[\frac{1}{386}-\left(\frac{193}{17}-\frac{33}{34}\right)\right]:\left(\frac{1}{2}+\frac{9}{2}\right)\)
\(=\left(\frac{1}{386}-\frac{386}{34}\right)\div5\)
\(=\frac{1}{386}\cdot\frac{1}{5}-\frac{386}{34}\cdot\frac{1}{5}=\frac{1}{1930}-\frac{386}{170}=\)là 1 phân số âm.
bài 1:
\(\frac{7}{4}\left(\frac{33}{42}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
\(=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(=\frac{7}{4}.33\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=\frac{231}{4}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{231}{4}\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(=\frac{231}{4}\cdot\frac{4}{21}=11\)
\(\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{1931}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)
= \(\left[\frac{193}{17}.\frac{2}{193}-\frac{193}{17}.\frac{3}{386}+\frac{33}{34}\right]:\left[\frac{1931}{25}.\frac{7}{1931}+\frac{1931}{25}.\frac{11}{3862}+\frac{9}{2}\right]\)
= \(\left[\frac{2}{17}-\frac{3}{17}+\frac{33}{34}\right]:\left[\frac{7}{25}+\frac{11}{50}+\frac{9}{2}\right]\)
= \(\left[\frac{4}{34}-\frac{6}{34}+\frac{33}{34}\right]:\left[\frac{14}{50}+\frac{11}{50}+\frac{225}{50}\right]\)
= \(\frac{31}{34}:2\)
= \(\frac{31}{68}\)
\(\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{17}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{17}}.\frac{\frac{3}{4}-\frac{3}{16}+\frac{3}{256}-\frac{3}{4}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}-\frac{-5}{8}\)
= \(\frac{1.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{17}\right)}{2.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{17}\right)}.\frac{3.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{256}+\frac{1}{4}\right)}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
= \(\frac{1}{2}.\left(\frac{3.\left(\frac{3}{4}+\frac{63}{256}\right)}{\frac{3}{4}+\frac{3}{64}}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\left(\frac{3.\left(\frac{3}{4}+\frac{63}{256}\right)}{\frac{3}{4}+\frac{12}{256}}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\left(\frac{3.3.\left(\frac{1}{4}+\frac{21}{256}\right)}{3.\left(\frac{1}{4}+\frac{1}{64}\right)}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\left(\frac{3.\left(\frac{1}{4}+\frac{1}{64}+\frac{17}{256}\right)}{\frac{1}{4}+\frac{1}{64}}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\left(\frac{3.\left(\frac{1}{4}+\frac{1}{64}\right)+3.\frac{17}{256}:\left(\frac{1}{4}+\frac{1}{64}\right)}{1.\left(\frac{1}{4}+\frac{1}{64}\right)}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\left(3+\frac{51}{256}:\frac{17}{64}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\left(3+\frac{3}{4}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\frac{15}{4}+\frac{5}{8}\)
= \(\frac{15}{8}+\frac{5}{8}\)
= \(\frac{5}{2}\)
\(\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{17}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{17}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{4}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}-\frac{-5}{8}\)
\(=\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{17}}{2.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{17}\right)}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{4}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{111}{68}+\frac{5}{8}\)
\(=\frac{49}{34}\)
A = \(\frac{1}{2}-\frac{3}{4}+\frac{5}{6}-\frac{7}{12}\)
A = \(\left(-\frac{1}{4}\right)+\frac{5}{6}-\frac{7}{12}\)
A = \(\frac{7}{12}-\frac{7}{12}\)
A = \(0\).
Mình làm câu A thôi nhé.
Chúc bạn học tốt!
\(\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{1931}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)
\(=\left[\left(\frac{4}{386}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{14}{3862}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)
\(=\left(\frac{1}{386}.\frac{193}{17}+\frac{33}{34}\right):\left(\frac{25}{3862}.\frac{1931}{25}+\frac{9}{2}\right)\)
\(=\left(\frac{1}{34}+\frac{33}{34}\right):\left(\frac{1}{2}+\frac{9}{2}\right)\)
\(=1:5\)
\(=\frac{1}{5}\)
\(A=\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]\div\left[\left(\frac{7}{1931}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)
\(=\left[\left(\frac{4}{386}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]\div\left[\left(\frac{14}{3862}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)
\(=\left[\frac{1}{386}.\frac{193}{17}+\frac{33}{34}\right]\div\left[\frac{25}{3862}.\frac{1931}{25}+\frac{9}{2}\right]\)
\(=\left[\frac{1}{34}+\frac{33}{34}\right]\div\left[\frac{1}{2}+\frac{9}{2}\right]\)
\(=1\div5=0,2\)
Vậy A = 0,2
Bài 1
a) \(\frac{1}{2}+\frac{3}{8}:3-\frac{3}{16}.2^3 \)
\(=\frac{1}{2}+\frac{3}{8}.\frac{1}{3}-\frac{3}{16}.8 \)
\(=\frac{1}{2}+\frac{1}{8}-\frac{3}{2} \)
\(=\frac{4}{8}+\frac{1}{8}-\frac{12}{8} \)
\(=\frac{-7}{8}\)