Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,\dfrac{-4}{5}+\dfrac{2}{7}-\dfrac{1}{5}-\dfrac{9}{7}=\left(\dfrac{-4}{5}-\dfrac{1}{5}\right)+\left(\dfrac{2}{7}-\dfrac{9}{7}\right)=-1+\left(-1\right)=-2\)
\(b,\dfrac{1}{12}-\left(\dfrac{1}{2}+\dfrac{7}{12}\right)=\dfrac{1}{12}-\left(\dfrac{6}{12}+\dfrac{7}{12}\right)=\dfrac{1}{12}-\dfrac{13}{12}=\dfrac{-12}{12}=-1\)
\(c,\dfrac{-12}{5}-\left(\dfrac{3}{5}-\dfrac{1}{8}\right)=\dfrac{-12}{5}-\left(\dfrac{24}{40}-\dfrac{5}{40}\right)=\dfrac{-12}{5}-\dfrac{19}{40}=\dfrac{-96}{40}-\dfrac{16}{40}=\dfrac{-100}{40}=\dfrac{-5}{2}\)
\(d,\dfrac{-5}{7}-\left(\dfrac{-7}{6}+\dfrac{2}{7}\right)+\dfrac{7}{-6}=\dfrac{-5}{7}+\dfrac{7}{6}-\dfrac{2}{7}+\dfrac{-7}{6}=\left(\dfrac{-5}{7}-\dfrac{2}{7}\right)+\left(\dfrac{7}{6}+\dfrac{-7}{6}\right)=-1\)
\(e,\dfrac{-6}{5}-\dfrac{2}{7}+\dfrac{1}{5}-\dfrac{5}{7}=\left(\dfrac{-6}{5}+\dfrac{1}{5}\right)+\left(\dfrac{-2}{7}-\dfrac{5}{7}\right)=-1+\left(-1\right)=-2\)
a: \(=\dfrac{3}{13}\left(\dfrac{6}{11}+\dfrac{9}{11}-\dfrac{4}{11}\right)+1+\dfrac{10}{13}=\dfrac{3}{13}+\dfrac{10}{13}+1=2\)
b: \(=4\left(2.86+3.14\right)-25\cdot6.01+9\cdot0.75\)
=4*6-25*6,01+9*0,75
=-119,5
c: \(=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}+\dfrac{15}{4}=\dfrac{7}{8}+\dfrac{15}{4}=\dfrac{7+30}{8}=\dfrac{37}{8}\)
d: \(=\dfrac{-3}{31}-\dfrac{28}{31}-\dfrac{6}{17}-\dfrac{11}{17}+\dfrac{1}{25}-\dfrac{5}{25}=\dfrac{-4}{25}-2=-\dfrac{54}{25}\)
\(1.678\simeq1.68;1.734\simeq1.73\)
\(2.014\simeq2.01;2.536\simeq2.54\)
\(2.834\simeq2.83;2.927\simeq2.93\)
2:
a: AC=BC=6/2=3cm
b: Vì BD=BC
nên B là trung điểm của DC
c: AD=6+3=9cm
bài 3.
1.\(\dfrac{-51}{136}=\dfrac{-3}{8};\dfrac{-60}{108}=\dfrac{-5}{9};\dfrac{26}{-156}=\dfrac{-1}{6}\)
\(\dfrac{-3}{8}=\dfrac{-24}{72};\dfrac{-5}{9}=\dfrac{-40}{72};\dfrac{-1}{6}=\dfrac{-12}{72}\)
