Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\Rightarrow\dfrac{59-x}{41}+1+\dfrac{57-x}{43}+1+\dfrac{55-x}{45}+1+\dfrac{53-x}{47}+1+\dfrac{51-x}{49}+1=0\)\(\Rightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\)
\(\Rightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\)
\(\Rightarrow100-x=0\Rightarrow x=100\)
1. \(\Leftrightarrow\frac{59-x}{41}+1+\frac{57-x}{43}+1+\frac{55-x}{45}+1+\frac{51-x}{49}+1=-5+5\)
\(\Leftrightarrow\frac{100-x}{41}+\frac{100-x}{43}+\frac{100-x}{45}+\frac{100-x}{47}+\frac{100-x}{49}=0\)
\(\Leftrightarrow\left(100-x\right)\left(\frac{1}{41}+\frac{1}{43}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}\right)=0\)
\(\Leftrightarrow x-100=0\Leftrightarrow x=100\)
2. \(\Leftrightarrow\frac{x-5}{1990}+1+\frac{x-15}{1980}+1+\frac{x-25}{1970}=\frac{x-1990}{5}+1+\frac{x-1980}{15}+1+\frac{x-1970}{25}+1\)
\(\Leftrightarrow\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}=\frac{x-1995}{5}+\frac{x-1995}{15}+\frac{x-1995}{25}\)
\(\Leftrightarrow\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}-\frac{x-1995}{5}-\frac{x-1995}{15}-\frac{x-1995}{25}=0\)
\(\Leftrightarrow\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}-\frac{1}{5}-\frac{1}{15}-\frac{1}{25}\right)=0\)
\(\Leftrightarrow x-1995=0\Leftrightarrow x=1995\)
(Lần sau, "bé" nhớ sử dụng công cụ công thức trực quan nhé, dịch của "bé" mình mệt ghê :V)
a,
\(\frac{x+1}{55}+\frac{x+3}{53}+\frac{x+5}{51}=\frac{x+7}{49}+\frac{x+9}{47}+\frac{x+11}{45}\\ \Leftrightarrow\frac{x+1}{55}+1+\frac{x+3}{53}+1+\frac{x+5}{51}+1=\frac{x+7}{49}+1+\frac{x+9}{47}+1+\frac{x+11}{45}+1\\ \Leftrightarrow\frac{x+56}{55}+\frac{x+56}{53}+\frac{x+56}{51}=\frac{x+56}{49}+\frac{x+56}{47}+\frac{x+56}{45}\\ \Leftrightarrow\left(x+56\right)\left(\frac{1}{55}+\frac{1}{53}+\frac{1}{51}-\frac{1}{49}-\frac{1}{47}-\frac{1}{45}\right)=0\)
And... u know dat right? ( ͡~ ͜ʖ ͡°)
b,
\(\frac{x+29}{31}-\frac{x+27}{33}=\frac{x+17}{43}-\frac{x+15}{45}\\ \Leftrightarrow\frac{x+29}{31}+1-\frac{x+27}{33}+1=\frac{x+17}{43}+1-\frac{x+15}{45}+1\\ \Leftrightarrow\frac{x+60}{31}-\frac{x+60}{33}=\frac{x+60}{43}-\frac{x+60}{45}\\ \Leftrightarrow\left(x+60\right)\left(\frac{1}{31}-\frac{1}{33}-\frac{1}{43}+\frac{1}{45}\right)=0\)
Ok, phần còn lại là của bạn nhé. :)
Chúc bạn học tốt nha
.
\(\dfrac{x+35}{65}+\dfrac{x+39}{61}=\dfrac{x+43}{57}+\dfrac{x+47}{53}\)
\(\Leftrightarrow\dfrac{x+35}{65}+1+\dfrac{x+39}{61}+1=\dfrac{x+43}{57}+1+\dfrac{x+47}{53}+1\)
\(\Leftrightarrow\dfrac{x+100}{65}+\dfrac{x+100}{61}-\dfrac{x+100}{57}-\dfrac{x+100}{53}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{65}+\dfrac{1}{61}-\dfrac{1}{57}-\dfrac{1}{53}\ne0\right)=0\Leftrightarrow x=-100\)
Ta có:
\(\dfrac{x+35}{65}+\dfrac{x+39}{61}=\dfrac{x+43}{57}+\dfrac{x+47}{53}\\ \Rightarrow\left(\dfrac{x+35}{65}+1\right)+\left(\dfrac{x+39}{61}+1\right)=\left(\dfrac{x+43}{57}+1\right)+\left(\dfrac{x+47}{53}+1\right)\\ \Rightarrow\dfrac{x+100}{53}+\dfrac{x+100}{61}=\dfrac{x+100}{57}+\dfrac{x+100}{53}\\ \Rightarrow\left(x+100\right)\left(\dfrac{1}{65}+\dfrac{1}{61}-\dfrac{1}{57}-\dfrac{1}{53}\right)=0\)
Ta thấy:
\(\dfrac{1}{65}< \dfrac{1}{57}\\ \dfrac{1}{61}< \dfrac{1}{53}\\ \Rightarrow\left(\dfrac{1}{65}+\dfrac{1}{62}\right)-\left(\dfrac{1}{57}+\dfrac{1}{53}\right)< 0\)
Hay \(\dfrac{1}{65}+\dfrac{1}{62}-\dfrac{1}{57}-\dfrac{1}{53}\ne0\)
\(\Rightarrow x+100=0\\ \Rightarrow x=0-100\\ \Rightarrow x=-100\)
Vậy \(x=-100\)