\(\frac{x-11}{37}+\frac{x+11}{39}=\frac{x+34}{41}\)

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9 tháng 8 2016

Cộng 3 vào cả 2 vế và chuyển vế xong đặt nhân tử x+50 ra ngoài ta được :

(x+50)(1/39+1/37+1/35-1/33-1/31-1/29)=0

vì (1/39+1/37+1/35-1/33-1/31-1/29) khác 0

=> x+50=0

=> x=-50

Nhớ k mình nhé

Chúc bạn học  tốt

2 tháng 5 2017

x = -50 nha

12 tháng 1 2020

a,=1

b,=7+3,5=10,5

c,(11/37+26/37)-(5/41+36/41)-0,5

=1-1-0,5

=-0,5

29 tháng 6 2018

\(\frac{x+32}{11}+\frac{x+33}{12}=\frac{x+34}{13}+\frac{x+35}{14}\)

\(\Leftrightarrow\left(\frac{x+32}{11}-1\right)+\left(\frac{x+33}{12}-1\right)=\left(\frac{x+34}{13}-1\right)+\left(\frac{x+35}{14}-1\right)\)

\(\Leftrightarrow\frac{x-21}{11}+\frac{x-21}{12}=\frac{x-21}{13}+\frac{x-21}{14}\)

\(\Leftrightarrow\left(x-21\right)\left(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

\(\left(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)

\(\Rightarrow x-21=0\Rightarrow x=21\)

29 tháng 6 2018

\(\frac{x+32}{11}+\frac{x+33}{12}=\frac{x+34}{13}+\frac{x+35}{14}\)

\(\Leftrightarrow\left(\frac{x+32}{11}-1\right)+\left(\frac{x+33}{12}-1\right)=\left(\frac{x+34}{13}-1\right)+\left(\frac{x+35}{14}-1\right)\)( trừ cả hai vế cho 2 )

\(\Leftrightarrow\frac{x-21}{11}+\frac{x-21}{12}=\frac{x-21}{13}+\frac{x-21}{14}\)

\(\Leftrightarrow\frac{x-21}{11}+\frac{x-21}{12}-\frac{x-21}{13}-\frac{x-21}{14}=0\)

\(\Leftrightarrow\left(x-21\right)\left(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Mà  \(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)

\(\Rightarrow x-21=0\)

\(\Leftrightarrow x=21\)

Vậy  \(x=21\)

30 tháng 6 2017

\(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)+\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\frac{1}{x+3}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\frac{\left(x+34\right)-\left(x+3\right)}{\left(x+3\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\frac{31}{\left(x+3\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Rightarrow x=31\)

3 tháng 7 2017

a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\) 

\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)

\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)

\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)

Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)

Nên x + 1 = 0

=> x = -1

3 tháng 7 2017

còn b vs c thì sao ạ

16 tháng 11 2021

a/

 \(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\)

\(3y=z\Rightarrow y=\frac{z}{3}\Rightarrow\frac{y}{2}=\frac{z}{6}\)

\(\Rightarrow\frac{x}{3}=\frac{y}{2}=\frac{z}{6}=\frac{x+y+z}{3+2+6}=\frac{11}{11}=1\)

\(\frac{x}{3}=1\Rightarrow x=3;\frac{y}{2}=1\Rightarrow y=2;\frac{z}{6}=1\Rightarrow z=6\)

b/

\(\frac{x}{3}=\frac{y}{4}=\frac{3y}{12}\Rightarrow\frac{x}{9}=\frac{y}{12}\)

\(\frac{z}{4}=\frac{y}{3}=\frac{4y}{12}\Rightarrow\frac{z}{16}=\frac{y}{12}\)

\(\Rightarrow\frac{x}{9}=\frac{y}{12}=\frac{z}{16}=\frac{x+y+z}{9+12+16}=\frac{37}{37}=1\)

\(\Rightarrow x=9;y=12;z=16\)