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Bài 1:
\(\frac{3}{5}+\frac{4}{15}=\frac{9}{15}+\frac{4}{15}=\frac{13}{15}\)
\(\frac{5}{6}:\frac{-7}{12}=\frac{5}{6}.\frac{-12}{7}=\frac{-60}{42}=\frac{-10}{7}\)
\(\frac{-21}{24}:\frac{-14}{8}=\frac{-21}{24}.\frac{-8}{14}=\frac{168}{336}=\frac{1}{2}\)
\(\frac{4}{5}:\frac{-8}{15}=\frac{4}{5}.\frac{-15}{8}=\frac{-60}{40}=\frac{-3}{2}\)
\(\frac{5}{12}-\frac{-7}{6}=\frac{5}{12}+\frac{7}{6}=\frac{5}{12}+\frac{14}{12}=\frac{19}{12}\)
\(\frac{-15}{16}.\frac{8}{25}=\frac{-120}{400}=\frac{-3}{10}\)
Bài 2 :
\(6\frac{4}{5}-\left(1\frac{2}{3}+3\frac{4}{5}\right)\)
\(=\frac{34}{5}-\left(\frac{5}{3}+\frac{19}{5}\right)\)
\(=\frac{34}{5}-\frac{5}{3}-\frac{19}{5}\)
\(=\left(\frac{34}{5}-\frac{19}{5}\right)-\frac{5}{3}\)
\(=3-\frac{5}{3}\)
\(=\frac{4}{3}\)
\(6\frac{5}{7}-\left(1\frac{2}{3}+2\frac{5}{7}\right)\)
\(=\frac{47}{7}-\left(\frac{5}{3}+\frac{19}{7}\right)\)
\(=\frac{47}{7}-\frac{5}{3}-\frac{19}{7}\)
\(=\left(\frac{47}{7}-\frac{19}{7}\right)-\frac{5}{3}\)
\(=4-\frac{5}{3}\)
\(=\frac{7}{3}\)
\(\frac{4}{19}.\frac{-3}{7}+\frac{-3}{7}.\frac{15}{19}+\frac{5}{7}\)
\(=\left(\frac{4}{19}+\frac{15}{19}\right).\frac{-3}{7}+\frac{5}{7}\)
\(=1.\frac{-3}{7}+\frac{5}{7}\)
\(=\frac{-3}{7}+\frac{5}{7}\)
\(=\frac{2}{7}\)
\(\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}\)
\(=\frac{5}{9}.\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)\)
\(=\frac{5}{9}.1\)
\(=\frac{5}{9}\)
a: \(=\dfrac{-1}{4}+\dfrac{7}{33}-\dfrac{5}{3}-\dfrac{5}{4}-\dfrac{6}{11}+\dfrac{48}{49}\)
\(=\dfrac{-3}{2}+\dfrac{7}{33}-\dfrac{18}{33}-\dfrac{5}{3}+\dfrac{48}{49}\)
\(=\dfrac{-9-10}{6}+\dfrac{-11}{33}+\dfrac{48}{49}\)
\(=\dfrac{-19}{6}+\dfrac{-1}{3}+\dfrac{48}{49}=-\dfrac{247}{98}\)
b: \(=\dfrac{11}{125}-\dfrac{17}{18}+\dfrac{4}{9}-\dfrac{5}{7}+\dfrac{17}{14}\)
\(=\dfrac{11}{125}+\dfrac{-17+8}{9}+\dfrac{-10+17}{14}\)
\(=\dfrac{11}{125}-1+\dfrac{7}{14}=\dfrac{11}{125}+\dfrac{1}{2}=\dfrac{22+125}{250}=\dfrac{147}{250}\)
A= 13;21;34
B= 37;70;135
C= 64;128;256
D= 22;29;37
E= 53;68;75
F= 127;255;511
G= 49;64;81
H= 324;841;2209
I= chịu
k cho mk nha!
a, A={x thuộc các số nguyên tố |2<hoặc bằng x<hoặc bằng 7}
oặc A={x thuộc R |(x^2-5*x+6)*(x^2-12*x+35)=0}
b,B={x thuộc Z | -3<hoặc bằng x<hoặc bằng 3}
c,C={5*x thuộc Z |-1<hoặc bằng x<hoặc bằng 3}
dễ mk bn cho mình hỏi nhé câu 4 là \(\frac{1}{2\cdot3}\)hay là\(\frac{1}{2}\cdot3\)
Tính giá trị biểu thứa
A=(6÷3/5-1 1/6×6/7)÷(4 1/5×10/11+5 2/11)
B=5 9/10÷3/2-(2 1/3×4 1/2-2×2 1/3)÷7/4