I= \(10101.\left(\dfrac{5}{111111} \dfrac{5}{222222}-\dfrac{4}{3.7.11.13.37}\right)\) B=\(71\dfrac{38}{45}-\left(43\dfrac{8}{45}-1\dfrac{17}{57}\right)\) D= \(\left(19\dfrac{5}{8}:\dfrac{7}{12}-13\d...

L

lqhiuu

31 tháng 7 2017

I= \(10101.\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{3.7.11.13.37}\right)\) B=\(71\dfrac{38}{45}-\left(43\dfrac{8}{45}-1\dfrac{17}{57}\right)\) D= \(\left(19\dfrac{5}{8}:\dfrac{7}{12}-13\dfrac{1}{4}:\dfrac{7}{12}\right).\dfrac{4}{5}\) F= \(\left(9.75.21\dfrac{3}{7}+\dfrac{39}{4}.18\dfrac{4}{7}\right).\dfrac{15}{78}\) Giups mk...

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#Toán lớp 6

7

NH

Nguyễn Hồng Khánh Như

31 tháng 7 2017

Trời ơi cái đề bài !!!

Thoy thì làm từng câu vậy

a) \(I=10101.\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{111111}\right)\)

\(I=10101.\left(\dfrac{10}{222222}+\dfrac{5}{222222}-\dfrac{8}{222222}\right)\)

\(I=10101.\left(\dfrac{15}{222222}-\dfrac{8}{222222}\right)\)

\(I=10101.\dfrac{7}{222222}\)

\(I=\dfrac{7}{22}\)

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N

Nguyên

31 tháng 7 2017

1.

a) \(A=10101\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{3.7.11.13.37}\right)\)\(A=\dfrac{10101.5}{10101.11}+\dfrac{10101.5}{10101.22}-\dfrac{10101.4}{10101.11}\)

\(A=\dfrac{5}{11}+\dfrac{5}{22}-\dfrac{4}{11}=\dfrac{7}{22}\)

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SN

Sir Nghi

12 tháng 7 2023

1.

b) \(B=8\dfrac{1}{5}.\left(11\dfrac{94}{1591}-6\dfrac{38}{1517}\right):8\dfrac{11}{43}\)

c) \(C=10101.\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{3.7.11.13.37}\right)\)

#Toán lớp 7

1

NL

Nguyễn Lê Phước Thịnh

12 tháng 7 2023

b: \(=8.2\left(11+\dfrac{94}{1591}-6-\dfrac{38}{1517}\right):\left(8+\dfrac{11}{43}\right)\)
\(=\dfrac{41}{5}\cdot\left(5+\dfrac{60}{1763}\right):\dfrac{355}{43}\)

\(=\dfrac{41}{5}\cdot\dfrac{8875}{1763}\cdot\dfrac{43}{355}\)

\(=5\)

c: \(=10101\cdot\left(\dfrac{10+5}{222222}-\dfrac{4}{111111}\right)\)

\(=10101\cdot\dfrac{7}{222222}=\dfrac{7}{22}\)

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TK

Trần Kim Cường

7 tháng 3 2022

Bài 3: Tính hợp lí các giá trị biểu thức sau

A= \(49\dfrac{8}{23}=-\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)

B=\(71\dfrac{38}{45}-\left(43\dfrac{8}{45}-1\dfrac{17}{57}\right)\)

#Toán lớp 6

2

KA

KIỀU ANH

7 tháng 3 2022

biểu thức đâu

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TK

Trần Kim Cường

7 tháng 3 2022

biểu thức đó

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TK

Trần Kim Cường

7 tháng 3 2022

Bài 3: Tính hợp lí các giá trị biểu thức sau:

A=\(49\dfrac{8}{23}-\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)

B=\(71\dfrac{38}{45}-\left(43\dfrac{8}{45}-1\dfrac{17}{57}\right)\)

#Toán lớp 6

2

HV

HACKER VN2009

7 tháng 3 2022

khó quá

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NL

Nguyễn Lê Phước Thịnh

7 tháng 3 2022

a: \(A=49+\dfrac{8}{23}-14-\dfrac{8}{23}-5-\dfrac{7}{32}=30-\dfrac{7}{32}=\dfrac{953}{32}\)

b:

Sửa đề: \(B=71\dfrac{38}{45}-\left(43\dfrac{8}{45}-1\dfrac{17}{51}\right)\)

\(B=71+\dfrac{38}{45}-43-\dfrac{8}{45}+1+\dfrac{17}{51}\)

\(=71-43+1+1\)

=28+2=30

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SN

Sir Nghi

12 tháng 7 2023

...

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#Toán lớp 7

1

NL

Nguyễn Lê Phước Thịnh

12 tháng 7 2023

loading...

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NN

no no

29 tháng 3 2017

Bài 1: Thực hiện phép tính a) \(\left(\dfrac{3}{8}+\dfrac{-3}{4}+\dfrac{7}{12}\right):\dfrac{5}{6}+\dfrac{1}{2}\) b) \(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\) c)...

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#Toán lớp 6

2

PN

Phạm Ngân Hà

28 tháng 4 2017

Bài 1:

a) \(\left(\dfrac{3}{8}+\dfrac{-3}{4}+\dfrac{7}{12}\right):\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\left(\dfrac{9}{24}+\dfrac{-18}{24}+\dfrac{14}{24}\right):\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\dfrac{5}{24}:\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\dfrac{5}{24}.\dfrac{6}{5}+\dfrac{1}{2}\)

\(=\dfrac{1}{4}+\dfrac{1}{2}\)

\(=\dfrac{1}{4}+\dfrac{2}{4}\)

\(=\dfrac{3}{4}\)

b) \(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\)

\(=\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\)

\(=\left(\dfrac{1}{2}+\dfrac{4}{5}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)\)

\(=\dfrac{1}{2}+\dfrac{4}{5}\)

\(=\dfrac{5}{10}+\dfrac{8}{10}\)

\(=\dfrac{9}{5}\)

c) \(6\dfrac{5}{12}:2\dfrac{3}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{77}{12}:\dfrac{11}{4}+\dfrac{42}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{77}{12}.\dfrac{4}{11}+\dfrac{42}{4}.\left(\dfrac{5}{15}+\dfrac{3}{15}\right)\)

\(=\dfrac{7}{3}+\dfrac{42}{4}.\dfrac{8}{15}\)

\(=\dfrac{7}{3}+\dfrac{14.2}{1.3}\)

\(=\dfrac{7}{3}+\dfrac{28}{3}\)

\(=\dfrac{35}{3}\)

d) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{7}.\left(3,5\right)^2\)

\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{7}.12\dfrac{1}{4}\)

\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{7}.\dfrac{49}{4}\)

\(=\dfrac{1}{6}-\dfrac{7}{2}\)

\(=\dfrac{1}{6}-\dfrac{21}{6}\)

\(=\dfrac{-10}{3}\)

e) \(\left(\dfrac{3}{5}+0,415-\dfrac{3}{200}\right).2\dfrac{2}{3}.0,25\)

\(=\left(\dfrac{3}{5}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)

\(=\left(\dfrac{120}{200}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)

\(=1.\dfrac{8}{3}.\dfrac{1}{4}\)

\(=\dfrac{2}{3}\)

f) \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)

\(=\dfrac{5}{16}:\dfrac{1}{8}-\left(\dfrac{9}{4}-\dfrac{3}{5}\right).\dfrac{10}{11}\)

\(=\dfrac{5}{16}.\dfrac{8}{1}-\left(\dfrac{45}{20}-\dfrac{12}{20}\right).\dfrac{10}{11}\)

\(=\dfrac{5}{2}-\dfrac{33}{20}.\dfrac{10}{11}\)

\(=\dfrac{5}{2}-\dfrac{3}{2}\)

\(=\dfrac{2}{2}=1\)

g) \(0,25:\left(10,3-9,8\right)-\dfrac{3}{4}\)

\(=\dfrac{1}{4}:\dfrac{1}{2}-\dfrac{3}{4}\)

\(=\dfrac{1}{4}.\dfrac{2}{1}-\dfrac{3}{4}\)

\(=\dfrac{1}{2}-\dfrac{3}{4}\)

\(=\dfrac{2}{4}-\dfrac{3}{4}\)

\(=\dfrac{-1}{4}\)

h) \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+20\%\right):\dfrac{7}{3}\)

\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{5}\right):\dfrac{7}{3}\)

\(=\dfrac{7}{5}-\left(\dfrac{11}{20}+\dfrac{4}{20}\right):\dfrac{7}{3}\)

\(=\dfrac{7}{5}-\dfrac{3}{4}:\dfrac{7}{3}\)

\(=\dfrac{7}{5}-\dfrac{9}{28}\)

\(=\dfrac{196}{140}-\dfrac{45}{140}\)

\(=\dfrac{151}{140}\)

i) \(\dfrac{\left(\dfrac{1}{2-0,75}\right).\left(0,2-\dfrac{2}{5}\right)}{\dfrac{5}{9}-1\dfrac{1}{12}}\)

\(=\dfrac{\left(\dfrac{1}{1,25}\right).\left(\dfrac{1}{5}-\dfrac{2}{5}\right)}{\dfrac{5}{9}-\dfrac{13}{12}}\)

\(=\dfrac{\dfrac{1}{1,25}.\dfrac{-1}{5}}{\dfrac{20}{36}-\dfrac{39}{36}}\)

\(=\dfrac{\dfrac{-1}{6,25}}{\dfrac{-19}{36}}\)

k) \(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)

\(=\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{28}}{-\dfrac{3}{3}-\dfrac{3}{7}+\dfrac{3}{28}}\)

\(=\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}{\left(-3\right)\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}\)

\(=-\dfrac{2}{3}\)

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PN

Phạm Ngân Hà

29 tháng 4 2017

\(A=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)

\(A=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)

\(A=\left(\dfrac{7}{10}.\dfrac{5}{28}\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).20\)

\(A=\dfrac{1}{8}.1.20\)

\(A=\dfrac{20}{8}=\dfrac{5}{2}\)

\(B=\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)

\(B=\left(9\dfrac{3}{8}+7\dfrac{5}{8}\right)+4,03\)

\(B=\left[\left(9+7\right)+\left(\dfrac{3}{8}+\dfrac{5}{8}\right)\right]+4,03\)

\(B=\left(16+1\right)+4,03\)

\(B=17+4,03\)

\(B=21,03\)

\(C=\left(9,75.21\dfrac{3}{7}+\dfrac{39}{4}.18\dfrac{4}{7}\right).\dfrac{15}{78}\)

\(C=\left(\dfrac{39}{4}.\dfrac{150}{7}+\dfrac{39}{4}.\dfrac{130}{7}\right).\dfrac{15}{78}\)

\(C=\dfrac{39}{4}.\left(\dfrac{150}{7}+\dfrac{130}{7}\right).\dfrac{15}{78}\)

\(C=\dfrac{39}{4}.40.\dfrac{15}{78}\)

\(C=390.\dfrac{15}{78}\)

\(C=75\)

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HL

hunh lê

18 tháng 5 2017

Tính nhanh:

\(M=\dfrac{120-\dfrac{1}{2}.40.5.\dfrac{1}{5}.20.\dfrac{1}{4}-20}{1+5+9+...+33+37+41}\)

\(N=10101.\left(\dfrac{6}{111111}+\dfrac{6}{222222}-\dfrac{7}{3.7.11.13.37}\right)\)

#Toán lớp 6

1

NT

Ngô Tấn Đạt

18 tháng 5 2017

\(M=\dfrac{120-\dfrac{1}{2}.40.5.\dfrac{1}{5}.20.\dfrac{1}{4}-20}{1+5+....+41}\\ =\dfrac{120-20.5-20}{1+5+...+41}\\ =\dfrac{0}{1+5+...+41}\\ =0\)

\(N=10101\left(\dfrac{6}{3.7.11.13.37}+\dfrac{6}{2.3.7.11.13.37}-\dfrac{7}{3.7.11.13.37}\right)\\ =10101\left(-\dfrac{2}{2.3.7.11.13.37}+\dfrac{6}{2.3.7.11.13.37}\right)\\ =3.7.13.37\left(\dfrac{4}{2.3.7.11.13.37}\right)=\dfrac{4}{2.11}=\dfrac{2}{11}\)

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BT

Bùi Tấn Sỹ

21 tháng 5 2017

giỏi z đạt - thánh toán học

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NT

Nguyễn Thị Diệu Ly

17 tháng 4 2021

bài...

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#Toán lớp 6

2

YC

Ye Chi-Lien

17 tháng 4 2021

ngu

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NT

Nguyễn Thị Diệu Ly

17 tháng 4 2021

ko giải đc thì đi ra chỗ khác

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PT

pham thi ngoc

16 tháng 9 2017

B= 0,25 +3/5 - (1/8-2/5+\(1\dfrac{1}{4}\) )

C=\(\dfrac{2}{3}-\left(\dfrac{-1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(\dfrac{-5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{35}\)

D=\(\left(5-\dfrac{3}{4}+\dfrac{1}{5}\right)-\left(6+\dfrac{7}{4}-\dfrac{8}{5}\right)-\left(2-\dfrac{5}{7}+\dfrac{16}{5}\right)\)

#Toán lớp 7

1

TT

thám tử

16 tháng 9 2017

\(B=0,25+3,5-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)\)

\(=\dfrac{17}{20}-\left(\dfrac{39}{40}\right)\)

\(=\dfrac{-1}{8}\)

\(C=\dfrac{2}{3}-\left(\dfrac{-1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(\dfrac{-5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{35}\)

\(=\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{12}+\dfrac{1}{35}\)

\(=\dfrac{71}{35}\)

\(D=\left(5-\dfrac{3}{4}+\dfrac{1}{5}\right)-\left(6+\dfrac{7}{4}-\dfrac{8}{5}\right)-\left(2-\dfrac{5}{7}+\dfrac{16}{5}\right)\)

\(=5-\dfrac{3}{4}+\dfrac{1}{5}-6-\dfrac{7}{4}+\dfrac{8}{5}-2+\dfrac{5}{7}-\dfrac{16}{5}\)

\(=\left(5-6-2\right)+\left(\dfrac{-3}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{5}+\dfrac{8}{5}-\dfrac{16}{5}\right)+\dfrac{5}{7}\)

\(=\left(-3\right)+\left(\dfrac{-5}{2}\right)+\left(\dfrac{-7}{5}\right)+\dfrac{5}{7}\)

\(=\dfrac{-433}{70}\)

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PT

pham thi ngoc

17 tháng 9 2017

bạn ơi mk thấy đây đâu có j là hợp lí đâu

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TT

Trần Thiên Ngọc Tú

24 tháng 6 2017

Tính hợp lí giá trị các biểu thức sau

I = \(10101.\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{3.7.11.13.37}\right)\)

giúp miình ới

#Toán lớp 6

3

Q

qwerty

24 tháng 6 2017

\(I=10101\cdot\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{3\cdot7\cdot11\cdot13\cdot37}\right)\)

\(=10101\cdot\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{111111}\right)\)

\(=10101\cdot\dfrac{1}{31746}\)

\(=7\cdot\dfrac{1}{22}\)

\(=\dfrac{7}{22}\)

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DH

Đức Hiếu

24 tháng 6 2017

\(I=10101.\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{3.7.11.13.37}\right)\)

\(I=10101.\left(\dfrac{5}{111111}+\dfrac{1}{2}.\dfrac{5}{111111}-\dfrac{5}{111111}.\dfrac{4}{5}\right)\)

\(I=10101.\dfrac{5}{111111}.\left(1+\dfrac{1}{2}-\dfrac{4}{5}\right)\)

\(I=\dfrac{5}{11}.\dfrac{7}{10}=\dfrac{7}{22}\)

Chúc bạn học tốt!!!

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