Tính giá trị của biểu thức:
a, 6 : 5/2 - 3/10
b, 4/6 : 4/3 + 5 : 4/3
K
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Những câu hỏi liên quan
NK
5
SC
5
NK
2 tháng 4 2023
\(a,6:\dfrac{5}{2}-\dfrac{3}{10}=6\times\dfrac{2}{5}-\dfrac{3}{10}=\dfrac{12}{5}-\dfrac{3}{10}=\dfrac{120}{50}-\dfrac{15}{50}=\dfrac{105}{50}=\dfrac{21}{10}\)
\(b,\dfrac{4}{6}:\dfrac{4}{3}+5:\dfrac{4}{3}=\left(\dfrac{4}{6}+5\right):\dfrac{4}{3}=\left(\dfrac{4}{6}+\dfrac{30}{6}\right):\dfrac{4}{3}=\dfrac{34}{6}\times\dfrac{3}{4}=\dfrac{102}{24}=\dfrac{17}{4}\)
TZ
1
5 tháng 11 2023
a: \(2\dfrac{3}{5}+1\dfrac{2}{5}\cdot\dfrac{31}{2}\)
\(=\dfrac{13}{5}+\dfrac{7}{5}\cdot\dfrac{31}{2}\)
\(=\dfrac{26}{10}+\dfrac{217}{10}=\dfrac{243}{10}\)
b: \(4\dfrac{3}{4}-3\dfrac{2}{3}:1\dfrac{1}{6}\)
\(=\dfrac{19}{4}-\dfrac{11}{3}:\dfrac{7}{6}\)
\(=\dfrac{19}{4}-\dfrac{11}{3}\cdot\dfrac{6}{7}\)
\(=\dfrac{19}{4}-\dfrac{22}{7}\)
\(=\dfrac{19\cdot7-22\cdot4}{28}=\dfrac{45}{28}\)
MN
26 tháng 3 2021
a) 32 . 53 + 92 = 9 . 125 + 81
= 1 125 + 81 = 1 206
b) 83 : 42 - 52 = 512 : 16 - 25 = 32 - 25 = 7
c) 33 . 92 - 52.9 + 18 : 6 = 27 . 81 - 25 . 9 + 3
= 2 187 - 225 + 3 = 1 962 + 3 = 1 965
VN
2
17 tháng 3 2023
=(1-2-3+4)+(5-6-7+8)+...+(2017-2018-2019+2020)+2021-2022-2023
=0+0+...+0-1-2023
=-2024
29 tháng 6 2021
\(a,A=2\sqrt{2}-9\sqrt{2}+16\sqrt{2}-5\sqrt{2}\)
\(=4\sqrt{2}\)
\(b,B=\left|1-\sqrt{5}\right|+\sqrt{5+2\sqrt{5}+1}\)
\(=\left|1-\sqrt{5}\right|+\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\left|1-\sqrt{5}\right|+\left|\sqrt{5}+1\right|=\sqrt{5}-1+\sqrt{5}+1=2\sqrt{5}\)
\(c,C=\dfrac{2+\sqrt{6}+2-\sqrt{6}}{\left(2+\sqrt{6}\right)\left(2-\sqrt{6}\right)}=\dfrac{4}{4-6}=-2\)
AH
Akai Haruma
Giáo viên
29 tháng 6 2021
Lời giải:
a.
\(A=2\sqrt{2}-3\sqrt{18}+4\sqrt{32}-\sqrt{50}=2\sqrt{2}-9\sqrt{2}+16\sqrt{2}-5\sqrt{2}\)
\(=(2-9+16-5)\sqrt{2}=4\sqrt{2}\)
b.
\(B=\sqrt{(1-\sqrt{5})^2}+\sqrt{(\sqrt{5}+1)^2}=|1-\sqrt{5}|+|\sqrt{5}+1|=\sqrt{5}-1+\sqrt{5}+1=2\sqrt{5}\)
c.
\(C=\frac{2+\sqrt{6}+2-\sqrt{6}}{(2-\sqrt{6})(2+\sqrt{6})}=\frac{4}{2^2-6}=-2\)
8 tháng 10 2023
2:
a: \(=\dfrac{1}{3}\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)=-\dfrac{1}{3}\cdot2=-\dfrac{2}{3}\)
1:
\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)
\(=-4-\dfrac{1}{4}=-\dfrac{17}{4}\)
H9
HT.Phong (9A5)
CTVHS
8 tháng 10 2023
Bài 1:
\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)
\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)
\(A=\left(7-6-5\right)-\left(\dfrac{3}{4}+\dfrac{5}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\right)\)
\(A=-4-\dfrac{3+5-7}{4}+\dfrac{1+4-5}{3}\)
\(A=-4-\dfrac{1}{4}+\dfrac{0}{3}\)
\(A=-\dfrac{16}{4}-\dfrac{1}{4}+0\)
\(A=\dfrac{-16-1}{4}\)
\(A=-\dfrac{17}{4}\)
Bài 2:
\(\dfrac{1}{3}\cdot-\dfrac{4}{5}+\dfrac{1}{3}\cdot-\dfrac{6}{5}\)
\(=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{-4-6}{5}\)
\(=\dfrac{1}{3}\cdot\dfrac{-10}{5}\)
\(=\dfrac{1}{3}\cdot-2\)
\(=-\dfrac{2}{3}\)
O
7 tháng 5 2022
a) \(\dfrac{5}{3}+\dfrac{4}{9}:\dfrac{1}{2}=\dfrac{5}{3}+\dfrac{4}{9}\times2=\dfrac{5}{3}+\dfrac{8}{9}=\dfrac{23}{9}\)
b) \(\dfrac{11}{10}-\dfrac{2}{5}:\dfrac{2}{3}=\dfrac{11}{10}-\dfrac{2}{5}\times\dfrac{3}{2}=\dfrac{11}{10}-\dfrac{3}{5}=\dfrac{11}{10}-\dfrac{6}{10}=\dfrac{5}{10}=\dfrac{1}{2}\)
Câu 4
\(\dfrac{12\times15\times20}{10\times16\times25}=\dfrac{3\times4\times3\times5\times4\times5}{5\times2\times4\times4\times5\times5}=\dfrac{3\times3}{5\times2}=\dfrac{9}{10}\)
7 tháng 5 2022
Câu 3:
\(a.\dfrac{5}{3}+\dfrac{4}{9}:\dfrac{1}{2}=\dfrac{5}{3}+\dfrac{8}{9}=\dfrac{15}{9}+\dfrac{8}{9}=\dfrac{23}{9}\)
\(b.\dfrac{11}{10}-\dfrac{2}{5}:\dfrac{2}{3}=\dfrac{11}{10}-\dfrac{3}{5}=\dfrac{11}{10}-\dfrac{6}{10}=\dfrac{5}{10}=\dfrac{1}{2}\)
Câu 4:
\(\dfrac{12\times15\times20}{10\times16\times25}=\dfrac{3\times3\times1}{2\times1\times5}=\dfrac{9}{10}\)
a)
`6:5/2-3/10`
`=6xx2/5-3/10`
`=12/5-3/10`
`=24/10-3/10`
`=21/10`
b)
`4/6:4/3+5:4/3`
`=2/3xx3/4+5xx3/4`
`=3/4xx(2/3+5)`
`=3/4xx(2/3+15/3)`
`=3/4xx17/3`
`=17/4`
\(a,6:\dfrac{5}{2}-\dfrac{3}{10}\\ =\dfrac{6}{1}:\dfrac{5}{2}-\dfrac{3}{10}\\ =\dfrac{6}{1}\times\dfrac{5}{2}-\dfrac{3}{10}\\ =\dfrac{12}{5}-\dfrac{3}{10}\\ =\dfrac{12\times2}{5\times2}-\dfrac{3}{10}\\ =\dfrac{24}{10}-\dfrac{3}{10}\\ =\dfrac{21}{10}\)
\(b,\dfrac{4}{6}:\dfrac{4}{3}+5:\dfrac{4}{3}\\ =\dfrac{4}{6}:\dfrac{4}{3}+\dfrac{5}{1}:\dfrac{4}{3}\\ =\dfrac{4}{3}:\left(\dfrac{4}{6}+\dfrac{5}{1}\right)\\ =\dfrac{4}{3}:\left(\dfrac{4}{6}+\dfrac{5\times6}{1\times6}\right)\\ =\dfrac{4}{3}:\left(\dfrac{4}{6}+\dfrac{30}{6}\right)\\ =\dfrac{4}{3}:\dfrac{34}{6}\\ =\dfrac{4}{3}:\dfrac{17}{3}\\ =\dfrac{4}{3}\times\dfrac{3}{17}\\=\dfrac{12}{51} \\ =\dfrac{4}{17}.\)