\(A=\frac{3}{2}-(-3)-(2,5-3)-[1,5+(-2,5)]\)
\(B=18-(-13)-16+(-27)+17-19\)
\(C=[3-(a+b)-(a-b+3)]-[(a-5)-(a+5)]\)
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1.
\(a,\dfrac{3}{4}+\dfrac{5}{2}=\dfrac{3}{4}+\dfrac{10}{4}=\dfrac{13}{4}\)
\(b,3+\dfrac{6}{11}+\dfrac{33}{11}+\dfrac{6}{11}=\dfrac{39}{11}\)
\(c,2-\dfrac{5}{7}=\dfrac{14}{7}-\dfrac{5}{7}=\dfrac{9}{7}\)
\(\dfrac{11}{3}-2=\dfrac{11}{3}-\dfrac{6}{3}=\dfrac{5}{3}\)
2.
\(a,\dfrac{13}{19}+\dfrac{18}{19}+\dfrac{17}{19}=\dfrac{13}{19}+\dfrac{17}{19}+\dfrac{18}{19}=\dfrac{30}{19}+\dfrac{18}{19}=\dfrac{48}{19}\)
\(b,\dfrac{3}{5}+\dfrac{3}{16}+13+16=\dfrac{63}{80}+29=\dfrac{2383}{80}\)
3.
Nửa chu vi :
\(4:2=2\left(m\right)\)
Chiều rộng :
\(2-\dfrac{5}{6}=\dfrac{7}{6}\left(m\right)\)
A = - 1 + 3 - 5 + 7 - 9 + 11 - 13 + 15 - 17
A = (-1 + 11) + (3 - 13) + (7- 17) + (-5 + 15) - 9
= - 10 + 10 - 10 + 10 - 9
= (-10 + 10) + (-10 + 10) - 9
= 0 + 0 - 9
= - 9
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 x 5
B = -20
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 2:
a, \(\dfrac{5}{23}\) \(\times\) \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) \(\times\) \(\dfrac{9}{26}\)
= \(\dfrac{5}{23}\) \(\times\) ( \(\dfrac{17}{26}\) + \(\dfrac{9}{26}\))
= \(\dfrac{5}{23}\) \(\times\) \(\dfrac{26}{26}\)
= \(\dfrac{5}{23}\)
b, \(\dfrac{3}{4}\) \(\times\) \(\dfrac{7}{9}\) + \(\dfrac{7}{4}\) \(\times\) \(\dfrac{3}{9}\)
= \(\dfrac{7}{12}\) + \(\dfrac{7}{12}\)
= \(\dfrac{14}{12}\)
= \(\dfrac{7}{6}\)
\(A=\frac{3}{2}+3-2,5+3+1\)
\(A=6\)
\(B=18-\left(-13\right)-16+\left(-27\right)+17-19\)
\(B=18+13-16-27+17-19\)
\(B=-14\)
\(C=[3-\left(A+B\right)-\left(A-B+3\right)]-[\left(A-5\right)-\left(A+5\right)]\)'
\(C=\left(3-A-B-A+B-3\right)-\left(A+5-A-5\right)\)
\(C=-2A-A-5+A+5\)
\(C=-2A\)
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