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Câu 1:
a. 3/4+1/8=6/8+1/8=7/8
x-5/7=3/5*1/2
x-5/7=3/10
x=3/10+5/7
x=61/70
b.4/5+3/10=8/10+3/10=11/10
x*2/3=8/8+3/4
x*2/3=7/4
x=7/4:2/3
x=21/8
Thế nha, bài này dài đó!!!! :))))
1) 1/3 x 1/2 x 3/7 = 3/42 = 1/14
2) 5/4 x 1/3 +1/7 = 5/12 + 1/7 = 35/84 + 12/84 = 47/84
3) 8 x ( 8/9 - 2/3 ) = 8 x 2/9 = 16/9
4) 5/6 x 48/20 x 1/2 = 240/240 = 1
5) ( 2/5 + 3/4 ) + 8 = 23/20 + 8 = 23//20 + 160/20 = 183/20
6) 10 x ( 1/2 - 1/5 ) = 10 x 3/10 = 10/1 x 3/10 = 30/10 = 3
a,C1
(5/6 + 5/8) x 2/3
= 35/24 x 2/3
=35/36
C2
(5/6 + 5/8) x 2/3
= 5/6x2/3 + 5/8x2/3
= 10/18 + 10/24
= 35/36
b,C1
7/5 x 3/4 - 1/2 x 3/4
= 21/20 - 3/8
= 27/40
C2
7/5 x 3/4 - 1/2 x 3/4
=3/4 x (7/5-1/2)
=3/4x9/10
=27/40
a, 5/6 x (1/8x2/3)
=5/6 x 1/12
=5/72
b,(7/5 - 1/2) x 3/4
=0,9 x 3/4
=3,06
c,(5/6+5/8) x 2/3
=70/48 x 2/3
=35/36
a, 5/6 x (1/8x2/3)
=5/6 x 1/12
=5/72
b,(7/5 - 1/2) x 3/4
=0,9 x 3/4
=3,06
c,(5/6+5/8) x 2/3
=70/48 x 2/3
=35/36
Bài 2:
a: 2/6x5/3=10/18=5/9
b: 11/9x5/10=55/90=11/18
c: 3/9x6/8=1/3x3/4=1/4
d: 4/9x12/16=48/144=1/3
e: 25/15x6/7=5/3x6/7=30/21=10/7
f: 6/10x15/20=90/200=9/20
Bài 1
4/5 x 6/7= 24/35
2/9 x 1/2= 2/18= 1/9
1/2 x 8/3= 8/6= 4/3
7/9 x 6/5= 42/45= 14/15
8/7 x 5/9= 40/63
10/11 x 22/15= 220/165= 4/3
Bài 2
2/6 x 5/3= 1/3 x 5/3=5/9
11/9 x 5/10= 11/9 x 1/2= 11/18
3/9 x 6/8= 1/3 x 3/4 =3/12= 1/4
4/9 x 12/16= 4/9 x 3/4= 12/36= 1/3
25/15 x 6/7= 5/3 x 6/7= 30/21= 10/7
6/10 x 15/20= 3/5 x 3/4= 9/20
1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)
\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)
=0
2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)
\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)
\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)
3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)
\(=\dfrac{52546}{15}\)
4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)
\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)
\(=\dfrac{4}{7}\)
\(C,\dfrac{1}{2}.\dfrac{3}{4}+\dfrac{1}{2}.\dfrac{1}{4}=\dfrac{1}{2}.\left(\dfrac{3}{4}+\dfrac{1}{4}\right)\)
\(=\dfrac{1}{2}.\dfrac{4}{4}=\dfrac{1}{2}.1=\dfrac{1}{2}\)
\(D,\dfrac{11}{3}.\dfrac{26}{7}-\dfrac{26}{7}.\dfrac{8}{3}=\dfrac{26}{7}.\left(\dfrac{11}{3}-\dfrac{8}{3}\right)\)
\(=\dfrac{26}{7}.\dfrac{3}{3}=\dfrac{26}{7}.1=\dfrac{26}{7}\)
a) \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+...-\dfrac{1}{128}+\dfrac{1}{128}-\dfrac{1}{256}\)
\(=1-\dfrac{1}{256}\)
\(=\dfrac{255}{256}\)
b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{13.14}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}\)
\(=1-\dfrac{1}{14}\)
\(=\dfrac{13}{14}\)
c) \(\dfrac{3}{15.18}+\dfrac{3}{18.21}+\dfrac{3}{21.24}+...+\dfrac{3}{87.90}\)
\(=3.\left(\dfrac{1}{15.18}+\dfrac{1}{18.21}+\dfrac{1}{21.24}+...+\dfrac{1}{87.90}\right)\)
\(=3.\left[\dfrac{1}{3}.\left(\dfrac{1}{15}-\dfrac{1}{18}\right)+\dfrac{1}{3}.\left(\dfrac{1}{18}-\dfrac{1}{21}\right)+\dfrac{1}{3}.\left(\dfrac{1}{21}-\dfrac{1}{24}\right)+...+\dfrac{1}{3}.\left(\dfrac{1}{87}-\dfrac{1}{90}\right)\right]\)
\(=3.\dfrac{1}{3}.\left(\dfrac{1}{15}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{24}+...+\dfrac{1}{87}-\dfrac{1}{90}\right)\)
\(=\dfrac{1}{15}-\dfrac{1}{90}\)
\(=\dfrac{6}{90}-\dfrac{1}{90}\)
\(=\dfrac{5}{90}=\dfrac{1}{18}\)
a: =5/48+10/18=95/144
b: =3/4(7/5-1/2)=3/4x9/10=27/40
c: =2/3x35/24=70/72=35/36
1 2 x 8 3 = 1 x 8 2 x 3 = 4 3