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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 \(-\dfrac{138}{150}< -\dfrac{105}{150}< -\dfrac{60}{150}< -\dfrac{51}{150}< -\dfrac{37}{150}\)
=>\(\dfrac{23}{-25}< \dfrac{-7}{10}< \dfrac{-2}{5}< \dfrac{-17}{50}< \dfrac{37}{-150}\)
a: 3/5>-19/5
b: 8/7<8/3
c: 3/4=15/20
2/5=8/20
mà 15>8
nên 3/4>2/5
d: -3/5=-18/30
-4/6=-20/30
mà -18>-20
nên -3/5>-4/6
a: 3/5>-19/5
b: 8/7<8/3
c: 3/4=15/20
2/5=8/20
mà 15>8
nên 3/4>2/5
d: -3/5=-18/30
-4/6=-20/30
mà -18>-20
nên -3/5>-4/6
a) Ta có:
\(A=-3\cdot7\cdot\left(-2\right)\cdot\left(-13\right)\)
\(A=-21\cdot26\)
\(A=-546\)
\(B=-1\cdot\left(-2\right)\cdot\left(-3\right)\cdot\left(-4\right)\cdot5\)
\(B=2\cdot12\cdot5\)
\(B=2\cdot60\)
\(B=120\)
Mà: \(120>-546\)
\(\Rightarrow B>A\)
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 làm:
Ta có: \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}\)
\(A=\left(1+\frac{1}{3}+...+\frac{1}{9}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
\(A=\left[\left(1+\frac{1}{3}+...+\frac{1}{9}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\right]-\left[\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\right]\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)=B\)
Vậy A = B
a) 3 2 + − 4 3 = 1 6
1 10 + − 4 5 = − 7 10
Mà 1 6 > − 7 10 nên 3 2 + − 4 3 > 1 10 + − 4 5
b) 1 2 + 1 3 + 1 4 + 1 5 + 1 6 = 29 20 < 2
Nên 1 2 + 1 3 + 1 4 + 1 5 + 1 6 < 2