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9 tháng 3 2022

93/46

9 tháng 3 2022

 94/93-93/92+1/8556

=1/8556+1/8556

=1/4278

9 tháng 2 2021

Ta có : \(\dfrac{2x+5}{95}+\dfrac{2x+6}{94}+\dfrac{2x+7}{93}=\dfrac{2x+93}{7}+\dfrac{2x+94}{6}+\dfrac{2x+95}{5}\)

\(\Leftrightarrow\dfrac{2x+5}{95}+\dfrac{2x+6}{94}+\dfrac{2x+7}{93}-\dfrac{2x+93}{7}-\dfrac{2x+94}{6}-\dfrac{2x+95}{5}=0\)

\(\Leftrightarrow\dfrac{2x+5}{95}+1+\dfrac{2x+6}{94}+1+\dfrac{2x+7}{93}+1-\dfrac{2x+93}{7}-1-\dfrac{2x+94}{6}-1-\dfrac{2x+95}{5}-1=0\)

\(\Leftrightarrow\dfrac{2x+100}{95}+\dfrac{2x+6}{94}+\dfrac{2x+7}{93}-\dfrac{2x+100}{7}-\dfrac{2x+100}{6}-\dfrac{2x+100}{5}=0\)

\(\Leftrightarrow\left(2x+100\right)\left(\dfrac{1}{95}+\dfrac{1}{94}+\dfrac{1}{93}-\dfrac{1}{7}-\dfrac{1}{6}-\dfrac{1}{5}\right)=0\)

Thấy : \(\dfrac{1}{95}+\dfrac{1}{94}+\dfrac{1}{93}-\dfrac{1}{7}-\dfrac{1}{6}-\dfrac{1}{5}\ne0\)

\(\Rightarrow2x+100=0\)

\(\Leftrightarrow x=-50\)

Vậy ...

 

 

 

 

 

 

9 tháng 2 2021

 -1 ở đâu vậy bạn,giải thích hộ mik đc ko

 

26 tháng 1 2018
\(\dfrac{x+1}{94}+\dfrac{x+2}{93}+\dfrac{x+3}{92}=\dfrac{x+4}{91}+\dfrac{x+5}{90}+\dfrac{x+6}{89}\) <=>\(\dfrac{x+1}{94}+1+\dfrac{x+2}{93}+1+\dfrac{x+3}{92}+1=\dfrac{x+4}{91}+1+\dfrac{x+5}{90}+1+\dfrac{x+6}{89}+1\) <=>\(\dfrac{x+95}{94}+\dfrac{x+95}{93}+\dfrac{x+95}{92}=\dfrac{x+95}{91}+\dfrac{x+95}{90}+\dfrac{x+95}{89}\) <=>\(\dfrac{x+95}{94}+\dfrac{x+95}{93}+\dfrac{x+95}{92}-\dfrac{x+95}{91}-\dfrac{x+95}{90}-\dfrac{x+95}{89}=0\) <=> (x+95)(\(\dfrac{1}{94}+\dfrac{1}{93}+\dfrac{1}{92}-\dfrac{1}{91}-\dfrac{1}{90}-\dfrac{1}{89}\)) vì \(\dfrac{1}{94}+\dfrac{1}{93}+\dfrac{1}{92}-\dfrac{1}{91}-\dfrac{1}{90}-\dfrac{1}{89}\) khác 0 =>x+95=0=>x=-95 chức bạn học tốt ^ ^
6 tháng 1 2018

\(\dfrac{x+1}{94}+\dfrac{x+2}{93}+\dfrac{x+3}{92}=\dfrac{x+4}{91}+\dfrac{x+5}{90}+\dfrac{x+6}{89}\)

\(\Rightarrow\dfrac{x+1}{94}+1+\dfrac{x+2}{93}+1+\dfrac{x+3}{92}+1=\dfrac{x+4}{91}+1+\dfrac{x+5}{90}+1+\dfrac{x+6}{89}+1\)

\(\Rightarrow\dfrac{x+95}{94}+\dfrac{x+95}{93}+\dfrac{x+95}{92}=\dfrac{x+95}{91}+\dfrac{x+95}{90}+\dfrac{x+95}{89}\)

\(\Rightarrow\dfrac{x+95}{94}+\dfrac{x+95}{93}+\dfrac{x+95}{92}-\dfrac{x+95}{91}-\dfrac{x+95}{90}-\dfrac{x+95}{89}=0\)

\(\Rightarrow\left(x+95\right)\left(\dfrac{1}{94}+\dfrac{1}{93}+\dfrac{1}{92}-\dfrac{1}{91}-\dfrac{1}{90}-\dfrac{1}{89}\right)=0\)

\(\dfrac{1}{94}+\dfrac{1}{93}+\dfrac{1}{92}-\dfrac{1}{91}-\dfrac{1}{90}-\dfrac{1}{89}\ne0\) nên \(x+95=0\Leftrightarrow x=-95\)

6 tháng 1 2018

Mk làm luôn nhé , không chép lại đề đâu !!! Ahihi

\(\dfrac{x+1}{94}+1+\dfrac{x+2}{93}+1+\dfrac{x+3}{92}+1=\dfrac{x+4}{91}+1+\dfrac{x+5}{90}+1+\dfrac{x+6}{89}+1\)\(\dfrac{x+95}{94}+\dfrac{x+95}{93}+\dfrac{x+95}{92}-\dfrac{x+95}{91}-\dfrac{x+95}{90}-\dfrac{x+95}{89}=0\)

\(\left(x+95\right)\)\(\left(\dfrac{1}{94}+\dfrac{1}{93}+\dfrac{1}{92}-\dfrac{1}{91}-\dfrac{1}{90}-\dfrac{1}{89}\right)\) = 0

\(x+95=0\)

\(x=-95\)

Vậy , ......

31 tháng 12 2022

a: \(\Leftrightarrow4x+4+9\left(2x+1\right)=2\left(5x+3\right)+12x+7\)

=>4x+4+18x+9=10x+6+12x+7

=>22x+13=22x+13(luôn đúng)

b: \(\Leftrightarrow\left(\dfrac{x+1}{94}+1\right)+\left(\dfrac{x+2}{93}+1\right)+\left(\dfrac{x+3}{92}+1\right)=\left(\dfrac{x+4}{91}+1\right)+\left(\dfrac{x+5}{90}+1\right)+\left(\dfrac{x+6}{89}+1\right)\)

=>x+95=0

=>x=-95

\(=\dfrac{9^3\left(2^3\cdot9+45\right)}{9^2\cdot9}=8\cdot9+45=117\)

1 tháng 7 2018

\(\dfrac{1909-x}{91}+\dfrac{1907-x}{93}+\dfrac{1905-x}{95}+\dfrac{1903}{97}+4=0\) ( Sửa đề)

\(\dfrac{1909-x}{91}+1+\dfrac{1907-x}{93}+1+\dfrac{1905-x}{95}+1+\dfrac{1903}{97}+1=0\)\(\dfrac{2000-x}{91}+\dfrac{2000-x}{93}+\dfrac{2000-x}{95}+\dfrac{2000-x}{97}=0\)

\(\left(2000-x\right)\left(\dfrac{1}{91}+\dfrac{1}{93}+\dfrac{1}{95}+\dfrac{1}{97}\right)=0\)

Do : \(\dfrac{1}{91}+\dfrac{1}{93}+\dfrac{1}{95}+\dfrac{1}{97}>0\)

\(\text{⇔}2000-x=0\)

\(\text{⇔}x=2000\)

Vậy ,....

29 tháng 12 2019

Where 4

17 tháng 4 2022

\(\dfrac{6}{15}+\dfrac{11}{15}=\dfrac{17}{15}\)

\(\dfrac{22}{77}-\dfrac{14}{77}=\dfrac{8}{77}\)

\(\dfrac{11}{13}\times\dfrac{26}{31}=11\times\dfrac{2}{31}=\dfrac{22}{31}\)

\(\dfrac{1}{2}\times3\times\dfrac{2}{5}=\dfrac{3}{5}\)

17 tháng 4 2022

mình cảm ơn

11 tháng 3 2021

1) PT \(\Leftrightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)

\(\Leftrightarrow\dfrac{x+36}{35}+\dfrac{x+36}{33}=\dfrac{x+36}{31}+\dfrac{x+36}{29}\)

\(\Leftrightarrow\left(x+36\right)\left(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}\right)=0\)

\(\Leftrightarrow x+36=0\) (Do \(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}>0\))

\(\Leftrightarrow x=-36\).

Vậy nghiệm của pt là x = -36.

17 tháng 7

2) x(x+1)(x+2)(x+3)= 24

⇔ x.(x+3)  .   (x+2).(x+1)  = 24

⇔(\(x^2\) + 3x) . (\(x^2\) + 3x + 2) = 24

Đặt \(x^2\)+ 3x = b

⇒ b . (b+2)= 24

Hay: \(b^2\) +2b = 24

\(b^2\) + 2b + 1 = 25

\(\left(b+1\right)^2\)= 25

+ Xét b+1 = 5 ⇒ b=4 ⇒  \(x^2\)+ 3x = 4 ⇒ \(x^2\)+4x-x-4=0 ⇒x(x+4)-(x+4)=0

⇒(x-1)(x+4)=0⇒x=1 và x=-4

+ Xét b+1 = -5 ⇒ b=-6 ⇒ \(x^2\)+3x=-6 ⇒\(x^2\) + 3x + 6=0

\(x^2\) + 2.x.\(\dfrac{3}{2}\) + (\(\dfrac{3}{2}\))2 = - \(\dfrac{15}{4}\)  Hay ( \(x^2\) +\(\dfrac{3}{2}\) )2= -\(\dfrac{15}{4}\) (vô lí)

⇒x= 1 và x= 4

\(\Leftrightarrow\left(\dfrac{x-1}{99}-1\right)+\left(\dfrac{x-99}{1}-1\right)+\left(\dfrac{x-3}{97}-1\right)+\left(\dfrac{x-7}{93}-1\right)+\left(\dfrac{x-5}{95}-1\right)+\left(\dfrac{x-95}{5}-1\right)=0\)=>x-100=0

hay x=100