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A = \(\frac{\frac{\frac{5+3.3-1.12}{12}}{3.6-5+2.2}+}{6}+\frac{16\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)}{17\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)}=\frac{\frac{\frac{5+9-12}{12}}{18-5+4}}{6}+\frac{16}{17}=\frac{2}{12}.\frac{6}{17}+\frac{16}{17}=\frac{1}{17}.\frac{6}{17}=1\)
A=\(\frac{\frac{5}{12}+\frac{9}{12}-\frac{12}{12}}{\frac{18}{6}-\frac{5}{6}+\frac{4}{6}}+\frac{16.\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)}{17.\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)}\)
=\(\frac{\frac{1}{6}}{\frac{17}{6}}+\frac{16}{17}\)
=\(\frac{1}{17}+\frac{16}{17}\)
=1
a) Ta có: \(\left|5\cdot0.6+\dfrac{2}{3}\right|-\dfrac{1}{3}\)
\(=\left|3+\dfrac{2}{3}\right|-\dfrac{1}{3}\)
\(=3+\dfrac{2}{3}-\dfrac{1}{3}\)
\(=3+\dfrac{1}{3}=\dfrac{10}{3}\)
b) Ta có: \(\left(0.25-1\dfrac{1}{4}\right):5-\dfrac{1}{5}\cdot\left(-3\right)^2\)
\(=\left(\dfrac{1}{4}-\dfrac{5}{4}\right)\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)
\(=\dfrac{-4}{4}\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)
\(=\dfrac{1}{5}\cdot\left(-1-9\right)\)
\(=-10\cdot\dfrac{1}{5}=-2\)
c) Ta có: \(\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}:\dfrac{5}{7}\)
\(=\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}\cdot\dfrac{7}{5}\)
\(=\dfrac{7}{5}\cdot\left(\dfrac{14}{17}+\dfrac{3}{17}\right)\)
\(=\dfrac{7}{5}\cdot1=\dfrac{7}{5}\)
d) Ta có: \(\dfrac{7}{16}+\dfrac{-9}{25}+\dfrac{9}{16}+\dfrac{-16}{25}\)
\(=\left(\dfrac{7}{16}+\dfrac{9}{16}\right)-\left(\dfrac{9}{25}+\dfrac{16}{25}\right)\)
\(=\dfrac{16}{16}-\dfrac{25}{25}\)
\(=1-1=0\)
e) Ta có: \(\dfrac{5}{6}+2\sqrt{\dfrac{4}{9}}\)
\(=\dfrac{5}{6}+2\cdot\dfrac{2}{3}\)
\(=\dfrac{5}{6}+\dfrac{4}{3}\)
\(=\dfrac{5}{6}+\dfrac{8}{6}=\dfrac{13}{6}\)
Ta có:\(\dfrac{\dfrac{5}{12}+\dfrac{3}{4}-1}{3-\dfrac{5}{6}+\dfrac{2}{3}}=\dfrac{\dfrac{5}{12}+\dfrac{9}{12}-\dfrac{12}{12}}{\dfrac{18}{6}-\dfrac{5}{6}+\dfrac{4}{6}}=\dfrac{1}{6}:\dfrac{17}{6}=\dfrac{1}{17}\)
Ta có\(\dfrac{\dfrac{16}{5}+\dfrac{16}{7}-\dfrac{16}{9}}{\dfrac{17}{5}+\dfrac{17}{7}-\dfrac{17}{9}}=\dfrac{16\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}\right)}{17\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}\right)}=\dfrac{16}{17}\)
Ta có:A=\(\dfrac{1}{17}+\dfrac{16}{17}=\dfrac{17}{17}=1\)
Vậy gt bt A=1
a; A = -1 + 3 - 5 + 7 - 9 + 11 - 13 + 15 - 17
A = (-1 + 11) + ( 3 - 13) + (-5 + 15) + (7 - 17) - 9
A = 10 - 10 + 10 - 10 - 9
A = (10 - 10) + (10 - 10) - 9
A = 0 + 0 - 9
A = -9
b; B = 1+2-3-4+5+6-7-8+9+10-11-12+13+14-15-16+17+18-19-20
B = (1+2-3-4) + (5+6-7-8)+(9+10-11-12)+(13+14-15-16)+(17+18-19-20)
B= -4+(-4)+(-4)+(-4)+(-4)
B= -4 . 5
B= -20
Bài 1:
A=\(\dfrac{\dfrac{5}{12}+\dfrac{3}{4}-1}{3-\dfrac{5}{6}+\dfrac{2}{3}}+\dfrac{\dfrac{16}{5}+\dfrac{16}{7}-\dfrac{16}{9}}{\dfrac{17}{5}+\dfrac{17}{7}-\dfrac{17}{9}}\)
A=\(\dfrac{\dfrac{1}{6}}{\dfrac{17}{6}}+\dfrac{16\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}\right)}{17\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}\right)}\)
A=\(\dfrac{1.\dfrac{1}{6}}{17.\dfrac{1}{6}}+\dfrac{16}{17}\)
A=\(\dfrac{1}{17}+\dfrac{16}{17}=\dfrac{17}{17}=1\)
Bài 2 mk chưa có câu trả lời, sorry nha!
bn ghép cac so lai voi nhau nhe