Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a) \(0,5-\frac{5}{41}+\frac{1}{2}-\frac{36}{41}\)
\(=\frac{1}{2}-\frac{5}{41}+\frac{1}{2}-\frac{36}{41}\)
\(=\left(\frac{1}{2}+\frac{1}{2}\right)-\left(\frac{5}{41}+\frac{36}{41}\right)\)
\(=1-1\)
\(=0.\)
b) \(\left(-\frac{2}{3}+\frac{3}{7}\right):\frac{4}{5}+\left(-\frac{1}{3}+\frac{4}{7}\right):\frac{4}{5}\)
\(=-\frac{2}{3}+\frac{3}{7}:\frac{4}{5}-\frac{1}{3}+\frac{4}{7}:\frac{4}{5}\)
\(=\left[\left(-\frac{2}{3}\right)-\frac{1}{3}\right]+\left(\frac{3}{7}+\frac{4}{7}\right):\frac{4}{5}\)
\(=\left(-1\right)+1:\frac{4}{5}\)
\(=\left(-1\right)+\frac{5}{4}\)
\(=\frac{1}{4}.\)
c) \(\left(-\frac{3}{4}\right).\sqrt{\frac{16}{9}+3.\sqrt{49}}\)
\(=\left(-\frac{3}{4}\right).\sqrt{\frac{16}{9}+3.7}\)
\(=\left(-\frac{3}{4}\right).\sqrt{\frac{16}{9}+21}\)
\(=\left(-\frac{3}{4}\right).\sqrt{\frac{205}{9}}\)
\(=\left(-\frac{3}{4}\right).\frac{\sqrt{205}}{3}\)
\(=-\frac{\sqrt{205}}{4}.\)
d) \(\left(-\frac{1}{3}\right)^2.\frac{4}{11}+1\frac{5}{11}.\left(\frac{1}{3}\right)^2\)
\(=\frac{1}{9}.\frac{4}{11}+\frac{16}{11}.\frac{1}{9}\)
\(=\frac{1}{9}.\left(\frac{4}{11}+\frac{16}{11}\right)\)
\(=\frac{1}{9}.\frac{20}{11}\)
\(=\frac{20}{99}.\)
Chúc bạn học tốt!
\(A=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-1\frac{15}{17}+\frac{2}{3}=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-\frac{64}{34}+\frac{14}{21}=\left(\frac{15}{34}+\frac{9}{34}-\frac{64}{34}\right)+\left(\frac{7}{21}+\frac{14}{21}\right)=\frac{30}{34}+\frac{21}{21}=\frac{15}{17}+1=\frac{32}{17}\)
a) \(\frac{17}{9}-\frac{17}{9}:\left(\frac{7}{3}+\frac{1}{2}\right)\)
= \(\frac{17}{9}-\frac{17}{9}:\frac{17}{6}\)
= \(\frac{17}{9}-\frac{2}{3}\)
= \(\frac{11}{9}\)
b) \(\frac{4}{3}.\frac{2}{5}-\frac{3}{4}.\frac{2}{5}\)
= \(\frac{2}{5}.\left(\frac{4}{3}-\frac{3}{4}\right)\)
= \(\frac{2}{5}.\frac{7}{12}\)
= \(\frac{7}{30}\)
Mình lười làm quá, hay mình nói kết quả cho bn thôi nha
c) -6
d) 3
e) 3
g) 12
h) \(\frac{23}{18}\)
i) \(\frac{-69}{20}\)
k) \(\frac{-1}{2}\)
l) \(\frac{49}{5}\)
Ta có: A = \(\left|\frac{4}{9}-\left(\frac{\sqrt{2}}{2}\right)^2\right|+\left|0,\left(4\right)+\frac{\frac{1}{3}-\frac{2}{5}-\frac{3}{7}}{\frac{2}{3}-\frac{4}{5}-\frac{6}{7}}\right|\)
= \(\left|\frac{4}{7}-\frac{\sqrt{2}^2}{2^2}\right|+\left|0,\left(1\right).4+\frac{\frac{1}{3}-\frac{2}{5}-\frac{3}{7}}{2\left(\frac{1}{3}-\frac{2}{5}-\frac{3}{7}\right)}\right|\)
= \(\left|\frac{4}{7}-\frac{1}{2}\right|+\left|\frac{1}{9}.4+\frac{1}{2}\right|\)
= \(\left|\frac{8-7}{14}\right|+\left|\frac{8+9}{18}\right|\)
= \(\left|\frac{1}{14}\right|+\left|\frac{17}{18}\right|\)
= 1/14 + 17/18 = 64/63
A = \(\left|\frac{4}{9}-\left(\frac{\sqrt{2}}{2}\right)^2\right|+\left|0,\left(4\right)+\frac{\frac{1}{3}-\frac{2}{5}-\frac{3}{7}}{\frac{2}{3}-\frac{4}{5}-\frac{6}{7}}\right|\)
= \(\left|\frac{4}{9}-\left(\frac{\sqrt{2}^2}{2^2}\right)\right|+\left|0,\left(1\right).4+\frac{\frac{1}{3}-\frac{2}{5}-\frac{3}{7}}{2.\left(\frac{1}{3}-\frac{2}{5}-\frac{3}{7}\right)}\right|\)
= \(\left|\frac{4}{9}-\frac{1}{2}\right|+\left|\frac{1}{9}.4+\frac{1}{2}\right|\)
= \(\left|\frac{8-9}{18}\right|+\left|\frac{4}{9}+\frac{1}{2}\right|\)
= \(\left|-\frac{1}{18}\right|+\left|\frac{8+9}{18}\right|\)
= \(\frac{1}{18}+\frac{17}{18}=1\)
a) \(\frac{3}{5}+\frac{1}{10}-\frac{6}{5}\)
\(=\left(\frac{3}{5}-\frac{6}{5}\right)+\frac{1}{10}\)
\(=\left(-\frac{3}{5}\right)+\frac{1}{10}\)
\(=-\frac{1}{2}.\)
b) \(1\frac{3}{4}.\frac{2}{7}+1\frac{3}{4}.\frac{5}{7}\)
\(=1\frac{3}{4}.\left(\frac{2}{7}+\frac{5}{7}\right)\)
\(=1\frac{3}{4}.1\)
\(=\frac{7}{4}.1\)
\(=\frac{7}{4}.\)
c) Sao lại có dấu chấm phẩy thế kia?
Chúc bạn học tốt!
a) \(\frac{3}{5}+\frac{1}{10}-\frac{6}{5}=\frac{6+1-12}{10}=\frac{-5}{10}=\frac{-1}{2}\)
b) \(1\frac{3}{4}.\frac{2}{7}+1\frac{3}{4}.\frac{5}{7}=1\frac{3}{4}\left(\frac{2}{7}+\frac{5}{7}\right)=1\frac{3}{4}.1=1\frac{3}{4}=\frac{7}{4}\)
c)\(\left(\frac{3}{4}\right)^2.\sqrt{16}+\left[\left(-2\right)^3:\left(-8\right)-1^{2019}\right]=\frac{9}{16}.4+\left[\left(-8\right):\left(-8\right)-1\right]=\frac{9}{16}.4=\frac{9}{4}\)
a) \(\frac{1}{5}-\frac{1}{3}+\frac{2}{7}=\frac{21}{105}-\frac{35}{105}+\frac{30}{105}=\frac{16}{105}\)
b) \(\frac{1}{5}.\frac{3}{7}+\frac{4}{5}.\frac{3}{7}=\frac{3}{35}+\frac{12}{35}=\frac{15}{35}=\frac{3}{7}\)
c) \(\sqrt{16}:\sqrt{4}+\sqrt{9}=4:2+3=5\)
d)\(\left(\frac{-1}{4}\right)^4:\left(-\frac{1}{4}\right)=\left(\frac{-1}{4}\right)^3=\frac{-1}{64}\)