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1/2= 2/4 2/3=4/6 3/4=9/12 -4/9=-12/9 3/-2=6/-4 -4/5=20/-25 -7/13=-91/143
1. \(\frac{9}{30}=\frac{3}{10};\frac{98}{80}=\frac{49}{40};\frac{15}{1000}=\frac{3}{200}\)
Vì \(200⋮10;200⋮40\)
=> BCNN(10; 40; 200) = 200
200 : 10 = 20
200 : 40 = 5
=> \(\frac{3}{10}=\frac{3\cdot20}{10\cdot20}=\frac{60}{200}\), \(\frac{49}{40}=\frac{49\cdot5}{40\cdot5}=\frac{245}{200}\)
A= \(11\frac{3}{13}-\left(2\frac{4}{7}+5\frac{3}{13}\right)\)
= \(11\frac{3}{13}-2\frac{4}{7}-5\frac{3}{13}\)
= \(\left(11\frac{3}{13}-5\frac{3}{13}\right)-2\frac{4}{7}\)
= \(6-2\frac{4}{7}\)
= \(3\frac{3}{7}\)
B= \(\left(96\frac{4}{9}+3\frac{7}{11}\right)-4\frac{4}{9}\)
= \(96\frac{4}{9}+3\frac{7}{11}-4\frac{4}{9}\)
= \(\left(96\frac{4}{9}-4\frac{4}{9}\right)+3\frac{7}{11}\)
= \(92+3\frac{7}{11}\)
= \(95\frac{7}{11}\)
C= \(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+1\frac{5}{7}\)
= \(\frac{-5}{7}.\left(\frac{2}{11}+\frac{9}{11}\right)+1\frac{5}{7}\)
= \(\frac{-5}{7}.1+1\frac{5}{7}\)
= \(\frac{-5}{7}+1\frac{5}{7}\)
= 1
(ý D bn tự làm ha)
mình vieetts ra thì dài lám
nếu bạn đã chác chăn về kiến thức thì hãy phân tích rồi tìm cách lam
còn ngược lại ban hãy tính máy tính rồi đoán cách làm cũng được
Trả lời:
1/2 = 5/10 = 25/50 = 50/100
2/3 = 6/9 = 18/27 = 54/81
3/4 = 9/12 = 27/36 = 81/108
4/9 = 12/27 = 36/81 = 108/243
Có 2 câu 3/4 nên tớ chỉ trả lời 1 câu thôi
Chuc hok tốt nhé:)
Câu 6: Khôg có cau nào đúng
Câu 7: C
Câu 8: B
Câu 9: B
Câu 10: D
Bài 1:
a) \(\dfrac{-5}{6}\ne\dfrac{10}{-14}\left(\dfrac{10}{-14}=-\dfrac{5}{7}\right).\)
b) \(\dfrac{-15}{-60}\ne\dfrac{-3}{12}\left(\dfrac{-15}{-60}=\dfrac{1}{4}\right).\)
Bài 2:
a) \(\dfrac{20}{-140}=-\dfrac{1}{7}.\)
b) \(\dfrac{4.18}{9.12}=\dfrac{72}{108}=\dfrac{2}{3}.\)
c) \(\dfrac{17.25-17.3}{2.\left(-15\right)}=\dfrac{17.\left(25-3\right)}{-30}=-\dfrac{17.22}{30}=\dfrac{374}{30}=\dfrac{187}{15}.\)
Bài 3:
a) \(\dfrac{-3}{5}< \dfrac{4}{-7}.\)
b) \(\dfrac{-4}{21}>\dfrac{-7}{35}.\)
c) \(\dfrac{-7}{24}>\dfrac{-2}{3}.\)
d) \(\dfrac{-52}{167}< \dfrac{-3}{-4}.\)
Bài 7:
7.1: I là trung điểm của AB
=>\(AB=2\cdot IA=4\left(cm\right)\)
7.2:
C nằm giữa A và B
=>AC+CB=AB
=>CB=10-8=2(cm)
C là trung điểm của NB
=>NC=CB=2cm
C là trung điểm của NB
=>\(NB=2\cdot NC=2\cdot2=4\left(cm\right)\)
Bài 6:
a: \(\dfrac{4}{5}=\dfrac{4\cdot6}{5\cdot6}=\dfrac{24}{30}\)
\(\dfrac{8}{15}=\dfrac{8\cdot2}{15\cdot2}=\dfrac{16}{30}\)
\(-\dfrac{3}{2}=\dfrac{-3\cdot15}{2\cdot15}=-\dfrac{45}{30}\)
b: \(2=\dfrac{2\cdot45}{45}=\dfrac{90}{45}\)
\(\dfrac{-10}{5}=\dfrac{-10\cdot9}{5\cdot9}=\dfrac{-90}{45}\)
\(\dfrac{7}{-9}=\dfrac{-7}{9}=\dfrac{-7\cdot5}{9\cdot5}=\dfrac{-35}{45}\)
c: \(\dfrac{3}{-2}=\dfrac{-3}{2}=\dfrac{-3\cdot6}{2\cdot6}=\dfrac{-18}{12}\)
\(\dfrac{5}{-6}=\dfrac{-5}{6}=\dfrac{-5\cdot2}{6\cdot2}=\dfrac{-10}{12}\)
\(\dfrac{-6}{4}=\dfrac{-6\cdot3}{4\cdot3}=\dfrac{-18}{12}\)
d: \(-\dfrac{1}{2}=\dfrac{-1\cdot15}{2\cdot15}=\dfrac{-15}{30}\)
\(\dfrac{4}{3}=\dfrac{4\cdot10}{3\cdot10}=\dfrac{40}{30}\)
\(\dfrac{6}{-5}=\dfrac{-6}{5}=\dfrac{-6\cdot6}{5\cdot6}=\dfrac{-36}{30}\)
bài 5:
a: \(\dfrac{3}{4}=\dfrac{9}{12};\dfrac{-3}{12}=\dfrac{-3}{12};\dfrac{-2}{3}=-\dfrac{8}{12};\dfrac{-1}{-6}=\dfrac{1}{6}=\dfrac{2}{12}\)
mà -8<-3<2<9
nên \(-\dfrac{8}{12}< -\dfrac{3}{12}< \dfrac{2}{12}< \dfrac{9}{12}\)
=>\(\dfrac{-2}{3}< \dfrac{-3}{12}< \dfrac{-1}{-6}< \dfrac{3}{4}\)
b: Ta có: \(\dfrac{-7}{9}=\dfrac{-28}{36};\dfrac{-1}{3}=\dfrac{-12}{36};-1=-\dfrac{36}{36}\)
mà -36<-28<-12
nên \(-1< -\dfrac{28}{36}< -\dfrac{12}{36}\)
=>\(-1< \dfrac{-7}{9}< -\dfrac{1}{3}< 0\)
\(\dfrac{5}{12}=\dfrac{15}{36};\dfrac{-1}{-4}=\dfrac{1}{4}=\dfrac{9}{36}\)
mà 9<15
nên \(0< \dfrac{1}{4}< \dfrac{5}{12}\)
=>\(-1< -\dfrac{7}{9}< -\dfrac{1}{3}< 0< \dfrac{1}{4}< \dfrac{5}{12}\)
c: \(\dfrac{-1}{-2};0;\dfrac{3}{10};1;\dfrac{-2}{-5};\dfrac{3}{-4}\)
\(-\dfrac{3}{4}< 0\)
\(\dfrac{-1}{-2}=\dfrac{1}{2}=\dfrac{5}{10};\dfrac{3}{10}=\dfrac{3}{10};1=\dfrac{10}{10};\dfrac{-2}{-5}=\dfrac{4}{10}\)
mà 3<4<5<10
nên \(\dfrac{3}{10}< \dfrac{4}{10}< \dfrac{5}{10}< \dfrac{10}{10}\)
=>\(0< \dfrac{3}{10}< \dfrac{-2}{-5}< \dfrac{-1}{-2}< 1\)
=>\(-\dfrac{3}{4}< 0< \dfrac{3}{10}< \dfrac{-2}{-5}< \dfrac{-1}{-2}< 1\)
d: \(-\dfrac{37}{150}=\dfrac{-37}{150};\dfrac{17}{-50}=\dfrac{-17}{50}=\dfrac{-51}{150}\)
\(\dfrac{23}{-25}=\dfrac{-23}{25}=\dfrac{-138}{150};\dfrac{-7}{10}=\dfrac{-105}{150};\dfrac{-2}{5}=-\dfrac{60}{150}\)
mà -138<-105<-60<-51<-37
nên \(-\dfrac{138}{150}< -\dfrac{105}{150}< -\dfrac{60}{150}< -\dfrac{51}{150}< -\dfrac{37}{150}\)
=>\(\dfrac{23}{-25}< \dfrac{-7}{10}< \dfrac{-2}{5}< \dfrac{-17}{50}< \dfrac{37}{-150}\)
Tham khảo:
Như vậy,\(\dfrac{2}{5} = \dfrac{4}{{10}}\)
\(\dfrac{1}{3} = \dfrac{3}{9}\)
\(\frac{1}{2}=\frac{2}{4}=\frac{3}{6}\)
\(\frac{2}{3}=\frac{4}{6}=\frac{6}{9}\)
\(\frac{3}{4}=\frac{-6}{-8}=\frac{9}{12}\)
\(-\frac{4}{9}=-\frac{2}{3}=\frac{2}{-3}\)
\(-\frac{3}{4}=-\frac{6}{8}=\frac{3}{-4}\)
\(\frac{3}{-2}=\frac{6}{-4}=-\frac{3}{2}\)
\(-\frac{4}{5}=-\frac{8}{10}=\frac{4}{-5}\)
\(\frac{-7}{13}=-\frac{14}{26}=\frac{14}{-26}\)