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2 tháng 2 2017

( 1 + 1 +1 ) ! = 6

2 + 2+ 2 = 6

3.3-3 = 6

5+(5:5) = 6

6+6-6 = 6

7- ( 7 : 7) = 6

2 tháng 2 2017

2 + 2 + 2 = 6

3 x 3 - 3 = 6

5 : 5 + 5 = 6

6 x 6 : 6 = 6

7 - 7 : 7 = 6

12 tháng 3 2020

793476480

14 tháng 3 2020

2.9241805e+26

hihi

30 tháng 9 2015

Gọi tổng là X, ta có :

X = 1+2+3+4+5+6+7+8+9+10+11+12+1+1345+6+544+435+22+00+34+567+13+7+6+9+8+0+6+5+4+3+4+5+6+66+6+2+3+8888+666+1000+87+98 !

Đây là bài toán hay đấy !

a: 14/5-7/5=7/5

b: 7/8-1/3+5/4

=21/24-8/24+30/24

=43/24

c; =7/6+5/6+2/15+13/15

=2+1

=3

d: =4*5/3*11=20/33

e: =2/9*1/6*1/4=2/9*1/24=1/108

2:
a: \(=\dfrac{3}{9}\cdot\dfrac{4}{4}\cdot\dfrac{5}{5}\cdot\dfrac{6}{6}\cdot\dfrac{7}{7}=\dfrac{1}{3}\)

b: \(=\dfrac{1}{6}\left(\dfrac{22}{3}-\dfrac{2}{3}\right)=\dfrac{10}{3}\cdot\dfrac{1}{6}=\dfrac{10}{18}=\dfrac{5}{9}\)

c; \(=\dfrac{1}{3}\left(9-\dfrac{2}{5}-\dfrac{3}{5}\right)=\dfrac{8}{3}\)

29 tháng 6 2023

14/5-7/5=7/5 

7/8-1/3+5/4=13/24+5/4=43/24 

7/6+2/15+5/6+13/15=13/10+5/6+13/15=32/15+13/15=3

4/3*5/11=20/35

2/9:6*1/4=1/27*1/4=1/108

7 tháng 2 2022

bạn viết cái này nó dễ hơn đó undefined

a: \(\dfrac{6}{7}:\left(\dfrac{2}{5}\cdot\dfrac{6}{7}\right)\)

\(=\dfrac{6}{7}:\dfrac{12}{35}\)

\(=\dfrac{6}{7}\cdot\dfrac{35}{12}=\dfrac{6}{12}\cdot\dfrac{35}{7}=\dfrac{5}{2}\)

b: \(\dfrac{6}{7}+\dfrac{5}{7}:5-\dfrac{8}{9}\)

\(=\dfrac{6}{7}+\dfrac{1}{7}-\dfrac{8}{9}\)

\(=1-\dfrac{8}{9}=\dfrac{1}{9}\)

c: \(\dfrac{6}{7}+\dfrac{5}{8}\cdot\dfrac{1}{5}-\dfrac{3}{16}\cdot4\)

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

\(=\dfrac{48+7-42}{56}=\dfrac{13}{56}\)

d: \(\dfrac{-1}{6}+\dfrac{2}{3}\cdot\dfrac{-3}{4}+\dfrac{4}{5}\)

\(=-\dfrac{1}{6}-\dfrac{1}{2}+\dfrac{4}{5}\)

\(=\dfrac{-5-15+24}{30}=\dfrac{4}{30}=\dfrac{2}{15}\)

4: \(=\dfrac{2+3}{7}+\dfrac{1+6}{9}-\dfrac{5}{6}=\dfrac{5}{7}+\dfrac{7}{9}-\dfrac{5}{6}=\dfrac{83}{126}\)

5: \(=\dfrac{-5-2}{7}+\dfrac{3+1}{4}-\dfrac{1}{5}=-\dfrac{1}{5}\)

6: \(=\dfrac{-3-28}{31}+\dfrac{-6-1}{17}+\dfrac{1-5}{25}=-1-\dfrac{7}{17}-\dfrac{4}{25}=-\dfrac{668}{425}\)

19 tháng 1 2022

4. \(\dfrac{2}{7}+\dfrac{1}{9}+\dfrac{3}{7}+\dfrac{5}{9}+\dfrac{-5}{6}\)

\(=\left(\dfrac{2}{7}+\dfrac{3}{7}\right)+\left(\dfrac{1}{9}+\dfrac{5}{9}\right)+\dfrac{-5}{6}\)

\(=\dfrac{5}{7}+\dfrac{2}{3}+\dfrac{-5}{6}\)

\(=\dfrac{30+28+\left(-35\right)}{42}=\dfrac{23}{42}\)
 

5. \(\dfrac{-5}{7}+\dfrac{3}{4}+\dfrac{-1}{5}+\dfrac{-2}{7}+\dfrac{1}{4}\)

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

\(=\dfrac{-7}{7}+\dfrac{4}{4}+\dfrac{-1}{5}\)

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

\(=-\dfrac{1}{5}\)
 

6. \(\dfrac{-3}{31}+\dfrac{-6}{17}+\dfrac{1}{25}+\dfrac{-28}{31}+\dfrac{-1}{17}+\dfrac{-1}{5}\)

\(=\left(\dfrac{-3}{31}+\dfrac{-28}{31}\right)+\left(\dfrac{-6}{17}+\dfrac{-1}{17}\right)+\dfrac{1}{25}+\dfrac{-1}{5}\)

\(=\dfrac{-31}{31}+\dfrac{-7}{17}+\dfrac{1}{25}+\dfrac{-1}{5}\)

\(=-1+\dfrac{-7}{17}+\dfrac{1}{25}+\dfrac{-1}{5}\)

\(=\dfrac{-425+\left(-175\right)+17+\left(-85\right)}{425}=\dfrac{-668}{425}\)

19 tháng 1 2022

chx làm :)))

19 tháng 1 2022

à nhìn nhầm

b: \(x+\dfrac{5}{6}=\dfrac{3}{8}\)

=>\(x=\dfrac{3}{8}-\dfrac{5}{6}\)

=>\(x=\dfrac{9}{24}-\dfrac{20}{24}=-\dfrac{11}{24}\)

c: \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{5}{6}\)

=>\(\dfrac{2}{3}x=\dfrac{5}{6}+\dfrac{1}{2}=\dfrac{8}{6}=\dfrac{4}{3}\)

=>2x=4

=>x=4/2=2

d: \(\dfrac{x}{7}=\dfrac{6}{-21}\)

=>\(\dfrac{x}{7}=\dfrac{-2}{7}\)

=>x=-2

e: \(\left(\dfrac{7}{3}x-0,6\right):\dfrac{3^2}{5}=1\)

=>\(\dfrac{7}{3}x-0,6=\dfrac{3^2}{5}=1,8\)

=>\(\dfrac{7}{3}x=2,4\)

=>\(x=2,4:\dfrac{7}{3}=2.4\cdot\dfrac{3}{7}=\dfrac{7.2}{7}=\dfrac{36}{35}\)

f: \(\dfrac{x}{45}=\dfrac{5}{6}+\dfrac{-29}{30}\)

=>\(\dfrac{x}{45}=\dfrac{25}{30}-\dfrac{29}{30}=-\dfrac{4}{30}=-\dfrac{2}{15}\)

=>\(x=-\dfrac{2}{15}\cdot45=-6\)

g: \(\left(4,5-2x\right)\cdot\left(-\dfrac{1^4}{7}\right)=\dfrac{11}{14}\)

=>\(4,5-2x=\dfrac{11}{14}:\dfrac{-1}{7}=\dfrac{-11}{2}\)

=>\(2x=4,5+\dfrac{11}{2}=\dfrac{20}{2}=10\)

=>x=10/2=5

h: \(-\dfrac{2}{7}+\dfrac{4}{7}x=\dfrac{5}{7}\)

=>\(\dfrac{4}{7}x=\dfrac{5}{7}+\dfrac{2}{7}=\dfrac{7}{7}\)

=>4x=7

=>\(x=\dfrac{7}{4}\)