Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: -51/136=-3/8
-60/108=-5/9
26/-156=-26/156=-1/6
b: -165/270=-11/18
-91/-156=7/12
-210/1134=-5/27
f, \(\dfrac{165}{270}\) = \(\dfrac{165:15}{270:15}\) = \(\dfrac{11}{18}\) = \(\dfrac{11\times6}{18\times6}\) = \(\dfrac{66}{108}\)
\(\dfrac{91}{156}\) = \(\dfrac{91:13}{156:13}\) = \(\dfrac{7}{12}\) = \(\dfrac{7\times9}{12\times9}\) = \(\dfrac{63}{108}\)
\(\dfrac{210}{1134}\) = \(\dfrac{210:42}{1134:42}\) = \(\dfrac{5}{27}\) = \(\dfrac{5\times4}{27\times4}\) = \(\dfrac{20}{108}\)
g, \(\dfrac{21}{9}\) = \(\dfrac{21:3}{9:3}\) = \(\dfrac{7}{3}\) = \(\dfrac{7\times10}{3\times10}\) = \(\dfrac{70}{30}\)
\(\dfrac{120}{50}\) = \(\dfrac{120:10}{50:10}\) = \(\dfrac{12}{5}\) = \(\dfrac{12\times6}{5\times6}\) = \(\dfrac{72}{30}\)
\(\dfrac{63}{54}\) = \(\dfrac{63:9}{54:9}\) = \(\dfrac{7}{6}\) = \(\dfrac{7\times5}{6\times5}\) = \(\dfrac{35}{30}\)
\(\frac{-51}{136};\frac{-60}{108};\frac{26}{-156}\)
Rút gọn : \(\frac{-51}{136}=\frac{-51:17}{136:17}=\frac{-3}{8}\)
\(\frac{-60}{108}=\frac{-60:12}{108:12}=\frac{-5}{9}\)
\(\frac{26}{-156}=\frac{-26}{156}=\frac{-26:26}{156:26}=\frac{-1}{6}\)
Như vậy ta được ba phân số mới đó là : \(\frac{-3}{8};\frac{-5}{9};\frac{-1}{6}\)
Ta phải có : \(MSC=BCNN(8,9,6)=72\)
Quy đồng : \(\frac{-3}{8}=\frac{-3\cdot9}{8\cdot9}=\frac{-27}{72}\)
\(\frac{-5}{9}=\frac{-5\cdot8}{9\cdot8}=\frac{-40}{72}\)
\(\frac{-1}{6}=\frac{-1\cdot12}{6\cdot12}=\frac{-12}{72}\)
Vậy : ....
\(\frac{5.6+5.7}{5.8+20}=\frac{5.\left(6+7\right)}{5.8+5.4}=\frac{5.\left(6+7\right)}{5.\left(8+4\right)}=\frac{13}{12}\)
\(\frac{8.9.4-15}{12.7-180}=\frac{273}{-96}=\frac{-273}{96}\)
Quy đồng: MSC = 96
Ta có: \(\frac{13}{12}=\frac{13.8}{12.8}=\frac{104}{96}\)
lúc trước mik viết nhầm đề , bh mik sửa lại rồi dó , giúp mik vs nhé
a: -5/5=-1=-7/7
8/7=8/7
b: -3/15=-1/5=-6/30
5/6=25/30
c: -34/136=-1/4=-9/36
-12/108=-1/9=-4/36
26/-156=-1/6=-6/36
a, \(\frac{51}{60};\frac{52}{60};\frac{41}{60}\)
b, \(\frac{1}{3};\frac{1}{2};\frac{11}{12}=\frac{4}{12};\frac{6}{12};\frac{11}{12}\)