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ính giá trị của các biểu thức sau:
A=827−(349+427)A=827−(349+427)
B=(1029+235)−629B=(1029+235)−629
Giải:
A=827−(349+427)A=827−(349+427)
=587−(319+307)=58−307−319=4−319=587−(319+307)=58−307−319=4−319
= 36−319=5936−319=59
B=(1029+235)−629B=(1029+235)−629
=1029−629+235=4+235=635
ính giá trị của các biểu thức sau:
A
=
8
2
7
−
(
3
4
9
+
4
2
7
)
A=827−(349+427)
B
=
(
10
2
9
+
2
3
5
)
−
6
2
9
B=(1029+235)−629
Giải:
A
=
8
2
7
−
(
3
4
9
+
4
2
7
)
A=827−(349+427)
=
58
7
−
(
31
9
+
30
7
)
=
58
−
30
7
−
31
9
=
4
−
31
9
=587−(319+307)=58−307−319=4−319
=
36
−
31
9
=
5
9
36−319=59
B
=
(
10
2
9
+
2
3
5
)
−
6
2
9
B=(1029+235)−629
=
10
2
9
−
6
2
9
+
2
3
5
=
4
+
2
3
5
=
6
3
5
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\(A=11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
\(A=11\dfrac{3}{13}-5\dfrac{3}{13}-2\dfrac{4}{7}\)
\(A=6-2\dfrac{4}{7}\)
\(A=5\dfrac{7}{7}-2\dfrac{4}{7}\)
\(A=3\dfrac{3}{7}\)
\(B=\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)
\(B=\left(6\dfrac{4}{9}-4\dfrac{4}{9}\right)+3\dfrac{7}{11}\)
\(B=2+3\dfrac{7}{11}\)
\(B=5\dfrac{7}{11}\)
\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+1\right)-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\dfrac{13}{11}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-65}{77}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{4}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{160}{11}\)
\(D=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(D=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{375}{1000}.\dfrac{5}{28}\)
\(D=\dfrac{7}{28}=\dfrac{5}{2}\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{12}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right).0\)
\(\Rightarrow E=0\)
\(A=\dfrac{6}{7}+\dfrac{1}{7}.\dfrac{2}{7}+\dfrac{1}{7}.\dfrac{5}{7}.\)
\(A=\dfrac{6}{7}+\dfrac{1}{7}\left(\dfrac{2}{7}+\dfrac{5}{7}\right).\)
\(A=\dfrac{6}{7}+\dfrac{1}{7}.1.\)
\(A=\dfrac{6}{7}+\dfrac{1}{7}=1.\)
Vậy \(A=1.\)
\(B=\dfrac{40}{9}.\dfrac{13}{3}-\dfrac{4}{3}.\dfrac{40}{9}.\)
\(B=\dfrac{4}{9}.\dfrac{13}{3}-\dfrac{4}{9}.\dfrac{40}{3}.\)
\(B=\dfrac{4}{9}\left(\dfrac{13}{3}-\dfrac{40}{3}\right).\)
\(B=\dfrac{4}{9}.\left(-9\right).\)
\(B=-4.\)
Vậy \(B=-4.\)
Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm thừa số chung ra ngoài.
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
3) \(\dfrac{5}{3}.\dfrac{11}{2}+\dfrac{1}{4}\)
\(=\dfrac{5.11}{3.2}+\dfrac{1}{4}\)\
\(=\dfrac{55}{6}+\dfrac{1}{4}\)
\(=\dfrac{220}{24}+\dfrac{6}{24}\)
\(=\dfrac{226}{24}\)
\(=\dfrac{113}{12}\)
4) \(\left(\dfrac{5}{2}-\dfrac{1}{3}\right).\dfrac{9}{2}-\dfrac{6}{7}\)
\(=\left(\dfrac{15}{6}-\dfrac{2}{6}\right).\dfrac{9}{2}-\dfrac{6}{7}\)
\(=\dfrac{13}{6}.\dfrac{9}{2}-\dfrac{6}{7}\)
\(=\dfrac{13.9}{6.2}-\dfrac{6}{7}\)
\(=\dfrac{117}{12}-\dfrac{6}{7}\)
\(=\dfrac{39}{4}-\dfrac{6}{7}\)
\(=\dfrac{273}{28}-\dfrac{24}{28}\)
\(=\dfrac{249}{28}\)
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)
\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)
A = \(\dfrac{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{9}}{\dfrac{4}{3}+\dfrac{4}{5}-\dfrac{4}{9}}\)
A = \(\dfrac{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{9}}{2.\left(\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{9}\right)}=\dfrac{1}{2}\)