Giải phương trình sau
(x- 49)/50 +(x- 50)/49=49/(x- 50)+50/(x-49)
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KN
10 tháng 3 2020
\(ĐKXĐ:x\ne49;x\ne50\)
Đặt \(x-49=u;x-50=v\)
Phương trình trở thành \(\frac{50}{u}+\frac{49}{v}=\frac{u}{50}+\frac{v}{49}\)
\(\Rightarrow\frac{50v+49u}{uv}=\frac{49u+50v}{2450}\)
\(\Rightarrow\orbr{\begin{cases}50v+49u=0\\uv=2450\end{cases}}\)
+) \(50v+49u=0\)
\(\Rightarrow50v=-49u\)
\(\Rightarrow\frac{v}{-49}=\frac{u}{50}=\frac{\left(x-50\right)-\left(x-49\right)}{-49-50}\)
\(=\frac{-1}{-99}=\frac{1}{99}\)
\(\Rightarrow\hept{\begin{cases}v=\frac{-49}{99}\\u=\frac{50}{99}\end{cases}}\Rightarrow x=\frac{4901}{99}\)(tm)
+) \(uv=2450\)
hay \(\left(x-49\right)\left(x-50\right)=2450\)
\(\Leftrightarrow x^2-99x+2450=2450\)
\(\Leftrightarrow x^2-99x=0\Leftrightarrow x\left(x-99\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=99\end{cases}}\left(tm\right)\)
Vậy phương trình có 3 nghiệm \(S=\left\{0;\frac{4901}{99};99\right\}\)
TN
0
KT
10 tháng 2 2018
10) \(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+116}{4}=0\)
\(\Leftrightarrow\)\(\frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+17}{83}+1+\frac{x+116}{4}-4=0\)
\(\Leftrightarrow\)\(\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\)\(x+100=0\) (vì 1/86 + 1/85 + 1/84 + 1/83 + 1/4 \(\ne\)0)
\(\Leftrightarrow\)\(x=-100\)
Vậy....
NP
22 tháng 2 2019
\(\dfrac{x-49}{50}+\dfrac{x-50}{49}=\dfrac{49}{x-50}+\dfrac{50}{x-49}\)
\(\Rightarrow\dfrac{x-49}{50}+\dfrac{x-50}{49}-\dfrac{49}{x-50}-\dfrac{50}{x-49}=0\)
\(\Leftrightarrow\left(\dfrac{x-49}{50}-\dfrac{50}{x-49}\right)+\left(\dfrac{x-50}{49}-\dfrac{49}{x-50}\right)=0\)
\(\Leftrightarrow\left(x-49\right)-50+\left(x-50\right)-49=0\)
\(\Leftrightarrow2x-198=0\)
\(\Leftrightarrow x=99\)
NK
0
KT
5 tháng 2 2018
7) \(\frac{x+25}{75}+\frac{x+30}{70}=\frac{x+35}{65}+\frac{x+40}{60}\)
\(\Leftrightarrow\)\(\frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+36}{65}+1+\frac{x+40}{60}+1\)
\(\Leftrightarrow\)\(\frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}\right)=0\)
\(\Leftrightarrow\)\(x+100=0\) (vì 1/75 + 1/70 - 1/65 - 1/60 \(\ne\)0)
\(\Leftrightarrow\)\(x=-100\)
Vậy.....
KT
7 tháng 2 2018
7) \(\frac{x+25}{75}+\frac{x+30}{70}=\frac{x+35}{65}+\frac{x+40}{60}\)
\(\Leftrightarrow\)\(\frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+35}{65}+1+\frac{x+40}{60}+1\)
\(\Leftrightarrow\)\(\frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}\right)=0\)
\(\Leftrightarrow\)\(x+100=0\) (1/75 + 1/70 - 1/65 - 1/60 \(\ne\)0)
\(\Leftrightarrow\)\(x=-100\)
Vậy...
18 tháng 6 2022
9: \(\dfrac{x-49}{50}+\dfrac{x-50}{49}=\dfrac{49}{x-50}+\dfrac{50}{x-49}\)
=>x-99=0
hay x=99
7: \(\Leftrightarrow\left(\dfrac{x+25}{75}+1\right)+\left(\dfrac{x+30}{70}+1\right)=\left(\dfrac{x+35}{65}+1\right)+\left(\dfrac{x+40}{60}+1\right)\)
=>x+100=0
hay x=-100
8:
Sửa đề: \(\dfrac{99-x}{101}+\dfrac{97-x}{103}+\dfrac{95-x}{105}+\dfrac{93-x}{107}=-4\)
\(\Leftrightarrow\left(\dfrac{99-x}{101}+1\right)+\left(\dfrac{97-x}{103}+1\right)+\left(\dfrac{95-x}{105}+1\right)+\left(\dfrac{93-x}{107}+1\right)=0\)
=>200-x=0
hay x=200
HT
3
17 tháng 2 2016
chuyen ve trai sang phai ta co x-49/50+x-50/49-50/x-49-49/x-50
=(x-49/50-1)+(x-50/49-1)+(-50/x-49+1)+(-49/x-50+1)
=(x-99)(1/50+1/49+1/x-49+1/x-50)
lý luận thi x=90 k nha
10 tháng 3 2016
nhưng mà bạn ơi! cái cụm bên phải(1/50+1/49+1/x-49+1/x-50) nó còn có chứa ẩn x mà bạn :( làm sao đẻ cm nó luôn lớn hơn ) đây????
ĐKXĐ : x khác 49 , x khác 50
Ta có :
\(\frac{x-49}{50}+\frac{x-50}{49}=\frac{49}{x-50}+\frac{50}{x-49}\)
\(\Leftrightarrow\frac{x-49}{50}-1+\frac{x-50}{49}-1=\frac{49}{x-50}-1+\frac{50}{x-49}-1\)
\(\Leftrightarrow\frac{x-99}{50}+\frac{x-99}{49}=\frac{99-x}{x-50}+\frac{99-x}{x-49}\)
\(\Leftrightarrow\left(x-99\right)\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{x-50}+\frac{1}{x-49}\right)=0\)
Mà 1/50 + 1/49 + 1/x-50 + 1/x-49 khác 0
\(\Leftrightarrow x-99=0\)
\(\Leftrightarrow x=99\)
@Kyo-kun
\(\Leftrightarrow\left(x-99\right)\left(\frac{1}{50}+\frac{1}{x-50}+\frac{1}{49}+\frac{1}{x-49}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-99=0\\\frac{x}{50\left(x-50\right)}+\frac{x}{49\left(x-49\right)}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=99\\x\left(\frac{1}{50\left(x-50\right)}+\frac{1}{49\left(x-49\right)}\right)=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=99\\x=0\end{cases}\left(t.m\right)}}\)
Vậy x = 99 hoặc x = 0