Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
\(\frac{x+32}{11}+\frac{x+33}{12}=\frac{x+34}{13}+\frac{x+35}{14}\)
\(\Leftrightarrow\left(\frac{x+32}{11}-1\right)+\left(\frac{x+33}{12}-1\right)=\left(\frac{x+34}{13}-1\right)+\left(\frac{x+35}{14}-1\right)\)
\(\Leftrightarrow\frac{x-21}{11}+\frac{x-21}{12}=\frac{x-21}{13}+\frac{x-21}{14}\)
\(\Leftrightarrow\left(x-21\right)\left(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Vì \(\left(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)
\(\Rightarrow x-21=0\Rightarrow x=21\)
\(\frac{x+32}{11}+\frac{x+33}{12}=\frac{x+34}{13}+\frac{x+35}{14}\)
\(\Leftrightarrow\left(\frac{x+32}{11}-1\right)+\left(\frac{x+33}{12}-1\right)=\left(\frac{x+34}{13}-1\right)+\left(\frac{x+35}{14}-1\right)\)( 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 \(x=21\)
\(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)+\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
\(\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
\(\frac{1}{x+3}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
\(\frac{\left(x+34\right)-\left(x+3\right)}{\left(x+3\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
\(\frac{31}{\left(x+3\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
\(\Rightarrow x=31\)
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
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/
\(\frac{x}{3}=\frac{y}{4}=\frac{3y}{12}\Rightarrow\frac{x}{9}=\frac{y}{12}\)
\(\frac{z}{4}=\frac{y}{3}=\frac{4y}{12}\Rightarrow\frac{z}{16}=\frac{y}{12}\)
\(\Rightarrow\frac{x}{9}=\frac{y}{12}=\frac{z}{16}=\frac{x+y+z}{9+12+16}=\frac{37}{37}=1\)
\(\Rightarrow x=9;y=12;z=16\)