\(\dfrac{48}{25}\)x\(\dfrac{27}{55}\) 2\(\dfrac{4}{9}\)x\(\dfrac{14}{33}\) 1\(\dfrac{19}{22}\)x(\(\dfrac{47}{77}\)-\(\dfrac{16}{55}\)) (3\(\dfrac{10}{99}\) 4\(\dfrac{11}{99}\)-5\(\dfrac{8}{299}\))-...

NT

nguyen thi quynh

11 tháng 4 2017

\(\dfrac{48}{25}\)x\(\dfrac{27}{55}\) +2\(\dfrac{4}{9}\)x\(\dfrac{14}{33}\) 1\(\dfrac{19}{22}\)x(\(\dfrac{47}{77}\)-\(\dfrac{16}{55}\)) (3\(\dfrac{10}{99}\)+4\(\dfrac{11}{99}\)-5\(\dfrac{8}{299}\))-(\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)-\(\dfrac{1}{6}\)) đề bài là thực hiện phép...

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#Toán lớp 6

1

ND

Nguyễn Đắc Định

11 tháng 4 2017

\(\dfrac{48}{25}\cdot\dfrac{27}{55}+2\dfrac{4}{9}\cdot\dfrac{14}{33}\)

\(=\dfrac{1296}{1375}+\dfrac{22}{9}\cdot\dfrac{14}{33}\\ =\dfrac{1296}{1375}+\dfrac{28}{27}\\ =\dfrac{34992}{37125}+\dfrac{38500}{37125}\\ =\dfrac{73492}{37125}\)

\(1\dfrac{19}{22}\cdot\left(\dfrac{47}{77}-\dfrac{16}{15}\right)\\ =\dfrac{41}{22}\cdot\dfrac{-527}{1155}\\ =\dfrac{-21607}{25410}\)

\(\left(3\dfrac{10}{99}+4\dfrac{11}{99}-5\dfrac{8}{299}\right)-\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\\ =\left(\dfrac{307}{99}+\dfrac{37}{99}-\dfrac{1503}{299}\right)-0\\ =\dfrac{344}{99}-\dfrac{1053}{299}\\ =-\dfrac{107}{2277}\)

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TN

Trần Nguyễn Lam My

18 tháng 3 2023

\(A=\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{65}+...+\dfrac{1}{99}=\dfrac{16}{x}\)
\(B=\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+\dfrac{29}{30}+\dfrac{41}{42}+\dfrac{55}{56}+\dfrac{x}{16}\)\(C=\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{9}\right)\times\left(1-\dfrac{1}{16}\right)\times...\times\left(1-\dfrac{1}{100}\right)=\dfrac{22}{x}\)
m.ng giúp em với

#Toán lớp 5

1

NL

Nguyễn Lê Phước Thịnh

18 tháng 3 2023

a: \(\Leftrightarrow\dfrac{32}{x}=\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{99}\)

=>32/x=1/3-1/5+1/5-1/7+...+1/9-1/11

=>32/x=1/3-1/11=8/33

=>x=32:8/33=132

b: \(\Leftrightarrow1-\dfrac{1}{6}+1-\dfrac{1}{12}+...+1-\dfrac{1}{56}=\dfrac{x}{16}\)
\(\Leftrightarrow6-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\right)=\dfrac{x}{16}\)

=>x/16=6-1/2+1/8=11/2+1/8=45/8=90/16

=>x=90

c: \(\Leftrightarrow\dfrac{22}{x}=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{10}\right)\left(1+\dfrac{1}{10}\right)\)

=>22/x=1/2*2/3*...*9/10*3/2*4/3*...*11/10

=>22/x=1/10*11/2=11/20=22/40

=>x=40

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VM

Vương Minh Phong

25 tháng 3 2022

\(\dfrac{1}{5}+\dfrac{2}{11} \dfrac{x}{55} \dfrac{2}{5}+\dfrac{1}{5}\) \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6} x\le\dfrac{13}{4}+\dfrac{14}{8}\) \(\dfrac{1}{4}+\dfrac{5}{12}+\dfrac{-1}{13} x \dfrac{7}{5}+\dfrac{2}{10}+\dfrac{1}{2}\) \(\dfrac{79}{15}+\dfrac{7}{5}+\dfrac{-8}{3}\le...

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#Toán lớp 6

3

HG

Hương Giang Vũ

25 tháng 3 2022

\(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)

\(\dfrac{11+10}{55}< \dfrac{x}{55}< \dfrac{3}{5}\)

\(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{33}{55}\)

Vậy \(x\in\left\{22;23;24;...\right\}\)

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VM

Vương Minh Phong

25 tháng 3 2022

\(\dfrac{????????}{????????????}\)

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JM

Ju Moon Adn

8 tháng 2 2018

Baif 1 : giải phương trình a. \(\dfrac{x-23}{24}+\dfrac{x-23}{25}=\dfrac{x-23}{26}+\dfrac{x-23}{27}\) b. \(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)\left(\dfrac{x+5}{95}+1\right)\) c. \(\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}\) d. \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\) e.\(\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53}{47}\) f....

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#Toán lớp 8

4

ND

Nhã Doanh

8 tháng 2 2018

h.

\(\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)

\(\Leftrightarrow\dfrac{2-x}{2002}+1-2=\dfrac{1-x}{2003}+1+1-\dfrac{x}{2004}-2\)

\(\Leftrightarrow\dfrac{2004-x}{2002}=\dfrac{2004-x}{2003}+\dfrac{2004-x}{2004}\)

\(\Leftrightarrow\dfrac{2004-x}{2002}-\dfrac{2004-x}{2003}-\dfrac{2004-x}{2004}=0\)

\(\Leftrightarrow\left(2004-x\right)\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\right)=0\)

Vì: \(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\ne0\)

Suy ra: 2004 - x = 0

Vậy x = 2004

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LT

lê thị hương giang

8 tháng 2 2018

\(a,\dfrac{x-23}{24}+\dfrac{x-23}{25}=\dfrac{x-23}{26}+\dfrac{x-23}{27}\)

\(\Leftrightarrow\dfrac{x-23}{24}+\dfrac{x-23}{25}-\dfrac{x-23}{26}-\dfrac{x-23}{27}=0\)

\(\Leftrightarrow\left(x-23\right)\left(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\right)=0\)

\(\Leftrightarrow x-23=0\) ( vì \(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\ne0\) )

\(\Leftrightarrow x=23\)

Vậy pt có tập nghiệm S = { 23 }

\(b,\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)

\(\Leftrightarrow\dfrac{x+2+98}{98}+\dfrac{x+3+97}{97}-\dfrac{x+4+96}{96}-\dfrac{x+5+95}{95}=0\)

\(\Leftrightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}-\dfrac{x+100}{96}-\dfrac{x+100}{95}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)

\(\Leftrightarrow x+100=0\)

\(\Leftrightarrow x=-100\)

Vậy pt có tập nghiệm S = { 100 }

\(c,\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}\)

\(\Leftrightarrow\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1=\dfrac{x+3}{2002}+1+\dfrac{x+4}{2001}+1\)

\(\Leftrightarrow\dfrac{x+1+2004}{2004}+\dfrac{x+2+2003}{2003}-\dfrac{x+3+2002}{2002}-\dfrac{x+4+2001}{2001}=0\)

\(\Leftrightarrow\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}-\dfrac{x+2005}{2002}-\dfrac{x+2005}{2001}=0\)

\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\right)=0\)

\(\Leftrightarrow x+2005=0\)

\(\Leftrightarrow x=-2005\)

Vậy pt có tập nghiệm S = { 2005 }

\(d,\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)

\(\Leftrightarrow\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)

\(\Leftrightarrow\dfrac{201-x+99}{99}+\dfrac{203-x+97}{97}+\dfrac{205-x+95}{95}=0\)

\(\Leftrightarrow\dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\)

\(\Leftrightarrow\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)

\(\Leftrightarrow300-x=0\)

\(\Leftrightarrow x=300\)

Vậy pt có tập nghiệm S = { 300 }

\(e,\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53}{47}\)

\(\Leftrightarrow\dfrac{x-45}{55}-1+\dfrac{x-47}{53}-1=\dfrac{x-55}{45}-1+\dfrac{x-53}{47}-1\)

\(\Leftrightarrow\dfrac{x-45-55}{55}+\dfrac{x-47-53}{53}-\dfrac{x-55-45}{45}-\dfrac{x-53-47}{47}=0\)

\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}-\dfrac{x-100}{45}-\dfrac{x-100}{47}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\right)=0\)

\(\Leftrightarrow x-100=0\)

\(\Leftrightarrow x=100\)

Vậy pt có tập nghiệm S = { 100 }

\(f,\dfrac{x+1}{9}+\dfrac{x+2}{8}=\dfrac{x+3}{7}+\dfrac{x+4}{6}\)

\(\Leftrightarrow\dfrac{x+1}{9}+1+\dfrac{x+2}{8}+1=\dfrac{x+3}{7}+1+\dfrac{x+4}{6}+1\)

\(\Leftrightarrow\dfrac{x+10}{9}+\dfrac{x+10}{8}-\dfrac{x+10}{7}-\dfrac{x+10}{6}=0\)

\(\Leftrightarrow\left(x+10\right)\left(\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{7}-\dfrac{1}{6}\right)=0\)

\(\Leftrightarrow x+10=0\)

\(\Leftrightarrow x=-10\)

Vậy pt có tập nghiệm S = { 10 }

\(h,\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)

\(\Leftrightarrow\dfrac{2-x}{2002}=\dfrac{1-x}{2003}+\dfrac{-x}{2004}+1\)

\(\Leftrightarrow\dfrac{2-x}{2002}+1=\dfrac{1-x}{2003}+1+\dfrac{-x}{2004}+1\)

\(\Leftrightarrow\dfrac{2-x+2002}{2002}-\dfrac{1-x+2003}{2003}-\dfrac{2004-x}{2004}=0\)

\(\Leftrightarrow\dfrac{2004-x}{2002}-\dfrac{2004-x}{2003}-\dfrac{2004-x}{2004}=0\)

\(\Leftrightarrow\left(2004-x\right)\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\right)=0\)

\(\Leftrightarrow2004-x=0\)

\(\Leftrightarrow x=2004\)

Vậy pt có tập nghiệm S = { 2004 }

\(g,\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+\dfrac{x+8}{92}\)

\(\Leftrightarrow\dfrac{x+2}{98}+1+\dfrac{x+4}{96}+1=\dfrac{x+6}{94}+1+\dfrac{x+8}{92}+1\)

\(\Leftrightarrow\dfrac{x+100}{98}+\dfrac{x+100}{96}-\dfrac{x+100}{94}-\dfrac{x+100}{92}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\right)=0\)

\(\Leftrightarrow x+100=0\)

\(\Leftrightarrow x=-100\)

Vậy pt có tập nghiệm S = { -100 }

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DP

Đoàn Phương Linh

13 tháng 2 2018

Tính một cách hợp lý: a\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)...\left(\dfrac{1}{100}-1\right)\)) \(x:\dfrac{99}{100}:\dfrac{98}{99}:...:\dfrac{2}{3}:\dfrac{1}{2}\) b) \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}:\dfrac{15-\dfrac{15}{11}+\dfrac{15}{121}}{16-\dfrac{16}{11}+\dfrac{16}{121}}\) c) \(\dfrac{\dfrac{1}{9}-\dfrac{5}{6}-4}{\dfrac{7}{12}-\dfrac{1}{36}-10}\) d)...

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#Toán lớp 6

1

R

Ruby

30 tháng 3 2018

b) \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}=\dfrac{5}{8}\)

Vì không có thời gian nên mình chỉ làm câu khó nhất thôi, tick mình nhé

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DP

Đoàn Phương Linh

30 tháng 3 2018

cảm ơn bạn

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PT

Phạm Tuấn Long

3 tháng 2 2019

I : Giải các phương trình sau : a) \(\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}\) b) \(\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53}{47}\) c) \(\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+\dfrac{x+8}{92}\) d) \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\) e) \(\dfrac{x+1}{9}+\dfrac{x+2}{8}=\dfrac{x+3}{7}+\dfrac{x+4}{6}\) f) \(\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\) g)...

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#Toán lớp 8

11

NT

Nguyễn Thành Trương

3 tháng 2 2019

Câu a)

Giáº£i phÆ°Æ¡ng trÃ¬nh,(x + 1)/2004 + (x + 2)/2003 = (x + 3)/2002 + (x + 4)/2001,ToÃ¡n há»c Lá»p 8,bÃ i táºp ToÃ¡n há»c Lá»p 8,giáº£i bÃ i táºp ToÃ¡n há»c Lá»p 8,ToÃ¡n há»c,Lá»p 8

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NT

Nguyễn Thành Trương

3 tháng 2 2019

b) x-45/55 + x-47/53 = x-55/45 + x-53/47
<=>x-45/55 -1 + x-47/53 -1= x-55/45 -1 + x-53/47 - 1
<=>x-100/55 + x-100/53 = x-100/45 + x-100/47
<=>(x-100)(1/55+1/53-1/45-1/47)=0
<=>x-100=0
<=>x=100

Vậy x = 100

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FS

FAN ST - Hiha

9 tháng 5 2022

98775 - 32 x 85 / 67500 - 24 x 236 / 568 + 101598 : 287 / 6875 + 980 -180 \(\dfrac{2}{5}+\dfrac{3}{10}-\dfrac{1}{2}\) / \(\dfrac{8}{11}+\dfrac{8}{33}\) x \(\dfrac{3}{4}\) / 7/9 x 3/14 :5/8 / \(\dfrac{5}{12}-\dfrac{7}{32}:\dfrac{21}{16}\) ÉT O...

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#Toán lớp 4

4

VQ

Vũ Quang Huy

9 tháng 5 2022

98775 - 32 x 85

=98775 -2720

=96055

67500 - 24 x 236

= 67500 -5664

=61836

568 + 101598 : 287

= 568 +354

=922

6875 + 980 -180

=7855 -180

=7675

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VQ

Vũ Quang Huy

9 tháng 5 2022

\(\dfrac{2}{5}+\dfrac{3}{10}-\dfrac{1}{2}\)

\(=\dfrac{7}{10}-\dfrac{1}{2}\)

= \(\dfrac{1}{5}\)

\(\dfrac{8}{11}+\dfrac{8}{33}x\dfrac{3}{4}\)

\(=\dfrac{8}{11}+\dfrac{2}{11}\)

\(=\dfrac{10}{11}\)

\(\dfrac{7}{9}x\dfrac{3}{14}:\dfrac{5}{8}\)

\(=\dfrac{1}{6}:\dfrac{5}{8}\)

\(=\dfrac{1}{6}x\dfrac{8}{5}\)

\(=\dfrac{8}{30}\)

\(=\dfrac{4}{15}\)

\(\dfrac{5}{12}-\dfrac{7}{32}:\dfrac{21}{16}\)

\(=\dfrac{5}{12}-\dfrac{7}{32}x\dfrac{16}{21}\)

\(=\dfrac{5}{12}-\dfrac{1}{6}\)

\(=\dfrac{5}{12}-\dfrac{2}{12}\)

\(=\dfrac{3}{12}=\dfrac{1}{4}\)

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NQ

Nguyễn Quốc Toàn

25 tháng 3 2018

\(A.\dfrac{-15}{28}x\dfrac{7}{25}\\ B.\dfrac{-5}{14}x\dfrac{7}{-3}\\ C.\dfrac{-1}{5}-\dfrac{7}{15}x\dfrac{9}{35}\\ D.\dfrac{-3}{4}-(\dfrac{-1}{2})^2\\ E.\dfrac{-4}{5}-\dfrac{-4}{5}x\dfrac{15}{16}\\F.(\dfrac{3}{4}+\dfrac{-7}{2})x(\dfrac{2}{11}+\dfrac{12}{22})\)

#Toán lớp 6

1

NL

Nguyễn Lê Phước Thịnh

26 tháng 6 2022

a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)

b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)

c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)

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

e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)

f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)

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C

Cihce

22 tháng 10 2021

Bài 1 : Thực hiện phép tính a) \(\dfrac{19}{12}\) + | \(\dfrac{-5}{2}\) | + ( \(\dfrac{3}{2}\) )2b) \(\dfrac{2}{11}\) . \(\dfrac{16}{9}\) - \(\dfrac{2}{11}\) . \(\dfrac{7}{9}\)Bài 2 : Tìm x , biết \(\dfrac{a}{8}\) = \(\dfrac{b}{3}\) và a - b =...

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#Toán lớp 7

1

LL

Lấp La Lấp Lánh

22 tháng 10 2021

Bài 1:

a) \(\dfrac{19}{12}+\left|\dfrac{-5}{2}\right|+\left(\dfrac{3}{2}\right)^2=\dfrac{19}{12}+\dfrac{5}{2}+\dfrac{9}{4}\)

\(=\dfrac{19+5.6+9.3}{12}=\dfrac{76}{12}=\dfrac{19}{3}\)

b) \(\dfrac{2}{11}.\dfrac{16}{9}-\dfrac{2}{11}.\dfrac{7}{9}=\dfrac{2}{11}\left(\dfrac{16}{9}-\dfrac{7}{9}\right)=\dfrac{2}{11}.1=\dfrac{2}{11}\)

Bài 2:

Áp dụng t/c dtsbn:

\(\dfrac{a}{8}=\dfrac{b}{3}=\dfrac{a-b}{8-3}=\dfrac{55}{5}=11\)

\(\Rightarrow\left\{{}\begin{matrix}x=11.8=88\\b=11.3=33\end{matrix}\right.\)

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R

Rip_mochi

2 tháng 5 2023

(x+\(\dfrac{2}{3}\))+\(\dfrac{1}{11}\)=(\(\dfrac{10}{-33}\).\(\dfrac{11}{8}\)).\(\dfrac{24}{-55}\)

#Toán lớp 6

2

H9

HT.Phong (9A5)

CTVHS

2 tháng 5 2023

\(\left(x+\dfrac{2}{3}\right)+\dfrac{1}{11}=\left(\dfrac{10}{-33}.\dfrac{11}{8}\right).\dfrac{24}{-55}\)

\(\Rightarrow\left(x+\dfrac{2}{3}\right)+\dfrac{1}{11}=\dfrac{-5}{12}.\dfrac{24}{-55}\)

\(\Rightarrow\left(x+\dfrac{2}{3}\right)+\dfrac{1}{11}=\dfrac{2}{11}\)

\(\Rightarrow x+\dfrac{2}{3}=\dfrac{2}{11}-\dfrac{1}{11}\)

\(\Rightarrow x+\dfrac{2}{3}=\dfrac{1}{11}\)

\(\Rightarrow x=\dfrac{1}{11}-\dfrac{2}{3}\)

\(\Rightarrow x=-\dfrac{19}{33}\)

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Y

YangSu

2 tháng 5 2023

mắc cái j dậy sớm z

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