Tính: a.25/16-7/20 b.25/4-3 c.16/24-1/6 d.2+1/4 g.7-5/9
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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
13 tháng 10 2024
a; \(\dfrac{9}{27}\) + \(\dfrac{7}{-49}\)
= \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\)
= \(\dfrac{7}{21}\) - \(\dfrac{3}{21}\)
= \(\dfrac{4}{21}\)
b; - \(\dfrac{12}{10}\) + \(\dfrac{-25}{30}\)
= - \(\dfrac{6}{5}\) - \(\dfrac{5}{6}\)
= -\(\dfrac{36}{30}\) - \(\dfrac{25}{30}\)
= \(\dfrac{-61}{30}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
13 tháng 10 2024
c; \(\dfrac{-20}{35}\) + \(\dfrac{-16}{-24}\)
= - \(\dfrac{4}{7}\) + \(\dfrac{2}{3}\)
= - \(\dfrac{12}{21}\) + \(\dfrac{14}{21}\)
= \(\dfrac{2}{21}\)
d; - \(\dfrac{21}{77}\) + \(\dfrac{10}{-35}\)
= - \(\dfrac{3}{11}\) - \(\dfrac{2}{7}\)
= - \(\dfrac{21}{77}\) - \(\dfrac{22}{77}\)
= - \(\dfrac{43}{77}\)
18 tháng 6 2023
a: =-1+17/20=-3/20
b: =(28/60-33/60)*(-25/3)
=(-1/12)*(-25/3)=1/12*25/3=25/36
c: \(=\dfrac{1}{3}\cdot\dfrac{1}{3}=\dfrac{1}{9}\)
BT
10 tháng 9 2017
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!
10 tháng 9 2017
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}
25 tháng 2 2019
\(a)9+\frac{24}{25}=\frac{9\cdot25+24}{25}=\frac{249}{25}\)
\(b)\frac{1}{20}+\frac{1}{4}+\frac{2}{5}=\frac{1}{20}+\frac{5}{20}+\frac{8}{20}=\frac{1+5+8}{20}=\frac{14}{20}=\frac{7}{10}\)
\(c)\frac{7}{12}+4=\frac{7}{12}+\frac{4}{1}=\frac{7+48}{12}=\frac{55}{12}\)
\(d)10+\frac{9}{8}=\frac{10}{1}+\frac{9}{8}=\frac{80+9}{8}=\frac{89}{8}\)
\(e)\frac{1}{4}+\frac{2}{25}+\frac{9}{100}=\frac{25}{100}+\frac{8}{100}+\frac{9}{100}=\frac{25+8+9}{100}=\frac{42}{100}=\frac{21}{50}\)
Còn 3 câu làm tương tự
25 tháng 2 2019
Ba câu còn lại mình làm luôn :D
\(f)\frac{28}{90}+6=\frac{14}{45}+6=\frac{14}{45}+\frac{6}{1}=\frac{14+270}{45}=\frac{284}{45}\)
\(g)\frac{3}{7}+\frac{4}{9}+\frac{4}{7}+\frac{7}{11}=(\frac{3}{7}+\frac{4}{7})+\frac{4}{9}+\frac{4}{11}=1+\frac{4}{9}+\frac{4}{11}\)
\(=\frac{7}{7}+\frac{4}{9}+\frac{4}{11}=\frac{693+308+252}{693}=\frac{1253}{693}=\frac{179}{99}\)
\(h)\frac{1}{8}+\frac{1}{12}+\frac{3}{8}+\frac{5}{12}=(\frac{1}{8}+\frac{3}{8})+(\frac{1}{12}+\frac{5}{12})=\frac{4}{8}+\frac{6}{12}=\frac{1}{2}+\frac{1}{2}=1\)
S
6 tháng 3 2023
câu a :
\(\dfrac{-8}{24}+\dfrac{-4}{12}=\dfrac{-1}{3}+\dfrac{-1}{3}=\dfrac{-2}{3}\)
câu b :
\(\dfrac{-20}{35}+\dfrac{16}{24}=\dfrac{-4}{7}+\dfrac{2}{3}=\dfrac{2}{21}\)
câu c :
\(\dfrac{-3}{9}+\dfrac{-6}{15}=\dfrac{-1}{3}+\dfrac{-2}{5}=\dfrac{-11}{15}\)
câu d :
\(\dfrac{3}{13}-\dfrac{4}{10}=\dfrac{3}{13}-\dfrac{2}{5}=\dfrac{-11}{65}\)
câu e :
\(\dfrac{5}{17}-\dfrac{9}{15}=\dfrac{5}{17}-\dfrac{3}{5}=\dfrac{-26}{85}\)
câu g :
\(\dfrac{9}{18}-\dfrac{6}{15}+\dfrac{3}{-9}=\dfrac{9}{18}-\dfrac{6}{15}+\dfrac{-3}{9}\\ =\dfrac{1}{2}-\dfrac{2}{5}+\dfrac{-1}{3}=\dfrac{-7}{30}\)
câu h :
\(\dfrac{5}{4}-\dfrac{1}{2}+\dfrac{-7}{8}=\dfrac{10}{8}-\dfrac{4}{8}+\dfrac{-7}{8}=\dfrac{-1}{8}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 7 2024
a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8
= 1 + 1 + 8
= 2 + 8
= 10
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 7 2024
b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{10}{20}\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (\(\dfrac{2}{2}\) + \(\dfrac{3}{3}\) + \(\dfrac{4}{4}\) + \(\dfrac{5}{5}\)+ \(\dfrac{6}{6}+\dfrac{7}{7}+\dfrac{8}{8}\) + \(\dfrac{10}{10}\))
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (1 + 1 +1 + 1+ 1+ 1+ 1 +1)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x 1 x 8
= \(\dfrac{1}{2}\) + \(\)\(\dfrac{1}{2}\) x 8
= \(\dfrac{1}{2}\) + 4
= \(\dfrac{9}{2}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 9 2024
a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + 8
= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8
= 1 + 1 + 8
= 2 + 8
= 10
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 9 2024
b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{9}{18}\) + \(\dfrac{10}{20}\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)
= \(\dfrac{1}{2}\) x 10
= 5
H9
HT.Phong (9A5)
CTVHS
10 tháng 10 2023
a) \(4^8\cdot4^4=\left(2^2\right)^8\cdot\left(2^2\right)^4=2^{16}\cdot2^8=2^{16+8}=2^{24}\)
b) \(5^{12}\cdot7-5^{11}\cdot10\)
\(=5^{11}\cdot\left(5\cdot7-10\right)=5^{11}\cdot\left(35-10\right)=5^{11}\cdot25\)
\(=5^{11}\cdot5^2=5^{11+2}=5^{13}\)
d) \(27^{16}:9^{10}\)
\(=\left(3^3\right)^{16}:\left(3^2\right)^{10}=3^{48}:3^{20}=3^{48-20}=3^{28}\)
e) \(125^3:25^4=\left(5^3\right)^3:\left(5^2\right)^4=5^9:5^8=5^{9-8}=5\)
f) \(24^4:3^4-32^{12}:16^{12}\)
\(=\left(24:4\right)^4-\left(32:16\right)^{12}\)
\(=6^4-2^{12}\)
\(=2^4\cdot\left(3^4-2^8\right)=2^4\cdot-175=-2800\)
2 tháng 7 2018
b, 3/5 + 4/7 + 2/8 + 10/25 + 9/21 + 28/16
= 3/5 + 4/7 + 2/8 + 2/5 + 3/7 + 14/8
= (3/5 + 2/5) + ( 4/7 + 3/7) + ( 2/8 + 14/8)
= 1 + 1 + 7/4
= 2 + 7/4 = 15/4
c , 8/7 + 7/6 + 5/8 + 10/12 + 24/28 + 6/16
= c , 8/7 + 7/6 + 5/8 + 5/6 + 6/7 + 1/2
= (8/7 + 6/7) + (7/6 + 5/6) + 5/8 + 1/2
= 14/7 + 12/6 + 5/8 + 1/2
= 2 + 2 + 5/8 + 1/2
= 4 + 9/8 = 41/8
a: \(\dfrac{25}{16}-\dfrac{7}{20}=\dfrac{25\cdot5-7\cdot4}{80}=\dfrac{125-28}{80}=\dfrac{97}{80}\)
b: \(\dfrac{25}{4}-3=\dfrac{25-12}{4}=\dfrac{13}{4}\)
c: \(\dfrac{16}{24}-\dfrac{1}{6}=\dfrac{2}{3}-\dfrac{1}{6}=\dfrac{4}{6}-\dfrac{1}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)
d: \(2+\dfrac{1}{4}=\dfrac{2\cdot4+1}{4}=\dfrac{9}{4}\)
g: \(7-\dfrac{5}{9}=\dfrac{7\cdot9-5}{9}=\dfrac{63-5}{9}=\dfrac{58}{9}\)