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8 tháng 3 2023
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: \(=\dfrac{-3}{31}-\dfrac{28}{31}-\dfrac{6}{17}-\dfrac{11}{17}+\dfrac{1}{25}-\dfrac{5}{25}=\dfrac{-4}{25}-2=-\dfrac{54}{25}\)
23 tháng 2 2023
\(1.678\simeq1.68;1.734\simeq1.73\)
\(2.014\simeq2.01;2.536\simeq2.54\)
\(2.834\simeq2.83;2.927\simeq2.93\)
18 tháng 3 2023
2:
a: AC=BC=6/2=3cm
b: Vì BD=BC
nên B là trung điểm của DC
c: AD=6+3=9cm
OP
28 tháng 1 2022
bài 3.
1.\(\dfrac{-51}{136}=\dfrac{-3}{8};\dfrac{-60}{108}=\dfrac{-5}{9};\dfrac{26}{-156}=\dfrac{-1}{6}\)
\(\dfrac{-3}{8}=\dfrac{-24}{72};\dfrac{-5}{9}=\dfrac{-40}{72};\dfrac{-1}{6}=\dfrac{-12}{72}\)
NN
3
H9
HT.Phong (9A5)
CTVHS
17 tháng 8 2023
a) \(\dfrac{x}{12}=\dfrac{6}{9}\)
\(\Rightarrow x=\dfrac{12\cdot6}{9}=8\)
b) \(\dfrac{9}{15}=x+\dfrac{11}{35}\)
\(\Rightarrow x=\dfrac{9}{15}-\dfrac{11}{35}\)
\(\Rightarrow x=\dfrac{2}{7}\)
c) \(x+\dfrac{6}{5}=\dfrac{x}{2}\)
\(\Rightarrow2\left(x+\dfrac{6}{5}\right)=x\)
\(\Rightarrow2x+\dfrac{12}{5}=x\)
\(\Rightarrow2x-x=-\dfrac{12}{5}\)
\(\Rightarrow x=-\dfrac{12}{5}\)
HQ
Hà Quang Minh
Giáo viên
17 tháng 8 2023
\(a,\dfrac{x}{12}=\dfrac{6}{9}\\ \Leftrightarrow x=12\cdot\dfrac{6}{9}\\ \Leftrightarrow x=8\\ b,\dfrac{9}{15}=x+\dfrac{11}{35}\\ \Leftrightarrow x=\dfrac{9}{15}-\dfrac{11}{35}\\ \Leftrightarrow x=\dfrac{2}{7}\\ c,x+\dfrac{6}{5}=\dfrac{x}{2}\\ \Leftrightarrow\dfrac{x}{2}=-\dfrac{6}{5}\\ \Leftrightarrow x=-\dfrac{12}{5}\)
\(a,\dfrac{-4}{5}+\dfrac{2}{7}-\dfrac{1}{5}-\dfrac{9}{7}=\left(\dfrac{-4}{5}-\dfrac{1}{5}\right)+\left(\dfrac{2}{7}-\dfrac{9}{7}\right)=-1+\left(-1\right)=-2\)
\(b,\dfrac{1}{12}-\left(\dfrac{1}{2}+\dfrac{7}{12}\right)=\dfrac{1}{12}-\left(\dfrac{6}{12}+\dfrac{7}{12}\right)=\dfrac{1}{12}-\dfrac{13}{12}=\dfrac{-12}{12}=-1\)
\(c,\dfrac{-12}{5}-\left(\dfrac{3}{5}-\dfrac{1}{8}\right)=\dfrac{-12}{5}-\left(\dfrac{24}{40}-\dfrac{5}{40}\right)=\dfrac{-12}{5}-\dfrac{19}{40}=\dfrac{-96}{40}-\dfrac{16}{40}=\dfrac{-100}{40}=\dfrac{-5}{2}\)
\(d,\dfrac{-5}{7}-\left(\dfrac{-7}{6}+\dfrac{2}{7}\right)+\dfrac{7}{-6}=\dfrac{-5}{7}+\dfrac{7}{6}-\dfrac{2}{7}+\dfrac{-7}{6}=\left(\dfrac{-5}{7}-\dfrac{2}{7}\right)+\left(\dfrac{7}{6}+\dfrac{-7}{6}\right)=-1\)
\(e,\dfrac{-6}{5}-\dfrac{2}{7}+\dfrac{1}{5}-\dfrac{5}{7}=\left(\dfrac{-6}{5}+\dfrac{1}{5}\right)+\left(\dfrac{-2}{7}-\dfrac{5}{7}\right)=-1+\left(-1\right)=-2\)