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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) \(\dfrac{5}{6};\dfrac{2}{3};\dfrac{1}{2}\)
b) \(2\dfrac{7}{8};1\dfrac{4}{5};1\dfrac{3}{4}\)
c) \(\dfrac{11}{12};\dfrac{5}{6};\dfrac{3}{8}\)
Bài 1:
a: 3/4=54/72
-1/9=-8/72
-5/8=-45/72
b: -1/7=-8/56
-1/-8=1/8=7/56
3/4=42/56
Bài 1:
\(a.-5;-3;-2;0;1;2;4\)
\(b.-36;-8;-6;-5;-4;0;6;8;12;15\)
\(c.-129;-98;0;3;27;35\)
Bài 2:
\(a.15;14;9;0;-3;-7;-16\)
\(b.100;17;5;0;-1;-2;-3;-13;-99\)
a) \(4\dfrac{3}{8}+5\dfrac{2}{3}\)
\(=\dfrac{35}{8}+\dfrac{17}{3}\)
\(=\dfrac{105}{24}+\dfrac{136}{24}\)
\(=\dfrac{241}{24}\)
b) \(2\dfrac{3}{8}+1\dfrac{1}{4}+3\dfrac{6}{7}\)
\(=\dfrac{19}{8}+\dfrac{5}{4}+\dfrac{27}{7}\)
\(=\dfrac{29}{8}+\dfrac{27}{7}\)
\(=\dfrac{419}{56}\)
c) \(2\dfrac{3}{8}-1\dfrac{1}{4}+5\dfrac{1}{3}\)
\(=\dfrac{19}{8}-\dfrac{5}{4}+\dfrac{16}{3}\)
\(=\dfrac{9}{8}+\dfrac{16}{3}\)
\(=\dfrac{155}{24}\)
d) \(\left(\dfrac{5}{2}+\dfrac{1}{3}\right):\left(1-\dfrac{1}{2}\right)\)
\(=\dfrac{17}{6}:\dfrac{1}{2}\)
\(=\dfrac{17}{6}\cdot2\)
\(=\dfrac{17}{3}\)
e) \(\left(\dfrac{5}{2}-\dfrac{1}{3}\right)\cdot\dfrac{9}{2}-\dfrac{6}{7}\)
\(=\dfrac{13}{6}\cdot\dfrac{9}{2}-\dfrac{6}{7}\)
\(=\dfrac{39}{4}-\dfrac{6}{7}\)
\(=\dfrac{249}{28}\)
a: =4+3/8+5+2/3
=9+9/24+16/24
=9+25/24
=216/24+25/24=241/24
b: \(=\dfrac{19}{8}+\dfrac{5}{4}+\dfrac{27}{7}=\dfrac{19+10}{8}+\dfrac{27}{7}\)
=27/7+29/8
=419/56
c: =2+3/8-1-1/4+5+1/3
=6+3/8-1/4+1/3
=6+3/8+1/12
=144/24+9/24+2/24
=155/24
d: =(15/6+2/6):1/2
=17/6*2
=17/3
e: =(15/6-2/6)*9/2-6/7
=13/6*9/2-6/7
=117/12-6/7
=249/28
Mình làm mẫu 1 bài rùi bạn tự giải những bài còn lại nha
1, 7A = 7+7^2+7^3+....+7^2008
6A = 7A - A = (7+7^2+7^3+....+7^2008)-(1+7+7^2+....+7^2007) = 7^2008-1
=> A = (7^2008-1)/6
Tk mk nha
\(A=1+7+7^2+7^3+...+7^{2007}\)
\(\Rightarrow7A=7+7^2+7^3+7^4+...+7^{2008}\)
\(\Rightarrow7A-A=\left(7+7^2+7^3+...+7^{2008}\right)-\left(1+7+7^2+...+7^{2007}\right)\)
\(\Rightarrow6A=7^{2008}-1\)
\(\Rightarrow A=\frac{7^{2008}-1}{6}\)
Có: \(\dfrac{1}{2}=0,5;\dfrac{5}{6}=0,8\left(3\right);\dfrac{4}{5}=0,8\)
\(\dfrac{7}{8}=0,875;\dfrac{6}{7}=0,\left(857142\right)\)
Vì 0,5 < 0,8 < 0,8(3) < 0,(857142) < 0,875
nên \(\dfrac{1}{2}< \dfrac{4}{5}< \dfrac{5}{6}< \dfrac{6}{7}< \dfrac{7}{8}\)
=> Các phân số đó khi viết theo thứ tự từ bé đến lớn là: \(\dfrac{1}{2};\dfrac{4}{5};\dfrac{5}{6};\dfrac{6}{7};\dfrac{7}{8}\)
Đáp án B
Ta có: 1 3 = 6 18 ; 1 2 = 6 12 ; 3 8 = 6 16
Vì: 6 18 < 6 16 < 6 12 < 6 7 ⇒ 6 7 > 1 2 > 3 8 > 1 3 .
Vậy các phân số trên được sắp xếp theo thứ tự từ lớn đến bé là: 6 7 ; 1 2 ; 3 8 ; 1 3