Giải phương trình :
\(\dfrac{x-70}{130}\)+ \(\dfrac{x-25}{175}\)+\(\dfrac{x-50}{150}\)+\(\dfrac{x-275}{25}\)=0
giải phương trình sau
\(\dfrac{x-50}{50}+\dfrac{x-51}{49}+\dfrac{x-52}{48}+\dfrac{x-53}{47}+\dfrac{x-200}{25}=0\)
Ta có : \(\dfrac{x-50}{50}+\dfrac{x-51}{49}+\dfrac{x-52}{49}+\dfrac{x-53}{47}+\dfrac{x-200}{25}=0\)
\(\Leftrightarrow\dfrac{x-50}{50}-1+\dfrac{x-51}{49}-1+\dfrac{x-52}{49}-1+\dfrac{x-53}{47}-1+\dfrac{x-200}{25}+4=0\)
\(\Leftrightarrow\dfrac{x-100}{50}+\dfrac{x-100}{49}+\dfrac{x-100}{49}+\dfrac{x-100}{47}+\dfrac{x-100}{25}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}\right)=0\)
<=> x - 100 = 0
<=> x = 100
Vậy ..
Ta có: \(\dfrac{x-50}{50}+\dfrac{x-51}{49}+\dfrac{x-52}{48}+\dfrac{x-53}{47}+\dfrac{x-200}{25}=0\)
\(\Leftrightarrow\dfrac{x-50}{50}-1+\dfrac{x-51}{49}-1+\dfrac{x-52}{48}-1+\dfrac{x-53}{47}-1+\dfrac{x-200}{25}+4=0\)
\(\Leftrightarrow\dfrac{x-100}{50}+\dfrac{x-100}{49}+\dfrac{x-100}{48}+\dfrac{x-100}{47}+\dfrac{x-100}{25}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}\right)=0\)
mà \(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}>0\)
nên x-100=0
hay x=100
Vậy: S={100}
giải phương trình ; \(\dfrac{x-130}{20}\)+\(\dfrac{x-100}{25}\)+\(\dfrac{x-60}{30}\)+\(\dfrac{x-10}{35}\)=10
\(\dfrac{x-130}{20}\)+\(\dfrac{x-100}{25}\)+\(\dfrac{x-60}{30}\)+\(\dfrac{x-10}{35}\)=10
⇔\(\dfrac{2625\left(x-130\right)}{52500}\)+\(\dfrac{2100\left(x-100\right)}{52500}\)+\(\dfrac{1750\left(x-60\right)}{52500}\)+\(\dfrac{1500\left(x-10\right)}{52500}\)=\(\dfrac{525000}{52500}\)
⇔2625\(x\)-341250+2100\(x\)-210000+1750\(x\)-105000+1500\(x\)-15000=525000
⇔ 7975\(x\) = 1196250
⇔ \(x\) = \(\dfrac{1196250}{7975}\)
⇔\(x \) = 150
giải phương trình sau:
\(\dfrac{x+25}{75}+\dfrac{x+30}{70}=\dfrac{x+35}{65}+\dfrac{x+40}{60}\)
Lời giải:
PT $\Leftrightarrow \frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+35}{65}+1+\frac{x+40}{60}+1$
$\Leftrightarrow \frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}$
$\Leftrightarrow (x+100)(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60})=0$
Dễ thấy $\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}<0$
$\Rightarrow x+100=0$
$\Leftrightarrow x=-100$ (tm)
mọi người ơi giúp mình câu giải phương trình này với:
\(\dfrac{x-5}{1990}+\dfrac{x-15}{1980}+\dfrac{x-25}{1970}+\dfrac{x-1990}{5}+\dfrac{x-1980}{15}+\dfrac{x-1970}{25}=0\)
\(\Leftrightarrow\left(\dfrac{x-5}{1990}-1\right)+\left(\dfrac{x-15}{1980}-1\right)+\left(\dfrac{x-25}{1970}-1\right)\\ +\left(\dfrac{x-1990}{5}-1\right)+\left(\dfrac{x-1980}{15}-1\right)+\left(\dfrac{x-1970}{25}-1\right)=0\\ \Leftrightarrow\dfrac{x-1995}{1990}+\dfrac{x-1995}{1980}+\dfrac{x-1995}{1970}+\dfrac{x-1995}{5}\\ +\dfrac{n-1995}{15}+\dfrac{n-1995}{25}=0\\ \Rightarrow\left(x-1995\right)\left(\dfrac{1}{1990}+\dfrac{1}{1980}+\dfrac{1}{1970}+\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{25}\right)=0\)
\(\Rightarrow x-1995=0\\ \Rightarrow x=1995\)
giải phương trình sau
\(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=0\)
Sửa đề: \(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=-5\)Ta có: \(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=-5\)
\(\Leftrightarrow\dfrac{74-x}{26}+1+\dfrac{75-x}{25}+1+\dfrac{76-x}{24}+1+\dfrac{77-x}{23}+1+\dfrac{78-x}{22}+1=0\)
\(\Leftrightarrow\dfrac{100-x}{26}+\dfrac{100-x}{25}+\dfrac{100-x}{24}+\dfrac{100-x}{23}+\dfrac{100-x}{22}=0\)
\(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}\right)=0\)
mà \(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}>0\)
nên 100-x=0
hay x=100
Vậy: S={100}
Ta có : \(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=-5\)
\(\Leftrightarrow\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}+5=0\)
\(\Leftrightarrow\dfrac{74-x}{26}+1+\dfrac{75-x}{25}+1+\dfrac{76-x}{24}+1+\dfrac{77-x}{23}+1+\dfrac{78-x}{22}+1=0\)
\(\Leftrightarrow\dfrac{100-x}{26}+\dfrac{100-x}{25}+\dfrac{100-x}{24}+\dfrac{100-x}{23}+\dfrac{100-x}{22}=0\)
\(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}\right)=0\)
Thấy : \(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}\ne0\)
\(\Rightarrow100-x=0\)
\(\Leftrightarrow x=100\)
Vậy ...
Giải phương trình sau:
\(\dfrac{x}{50}\)+\(\dfrac{x-1}{49}\)+\(\dfrac{x-2}{48}\)+\(\dfrac{x-3}{47}\)+\(\dfrac{x-150}{25}\)= 0
Giải phương trình sau:
\(\dfrac{x}{50}\) +\(\dfrac{x_{ }-1}{49}\)+\(\dfrac{x-2}{48}\)+\(\dfrac{x-3}{47}\)+\(\dfrac{x-150}{25}\)= 0
⇔ \(\dfrac{\left(x-50\right)+50}{50}\)+\(\dfrac{\left(x-50\right)+49}{49}\)+\(\dfrac{\left(x-50\right)+48}{48}\)+\(\dfrac{\left(x-50\right)-100}{25}\)= 0
⇔\(\dfrac{x-50}{50}\)+ 1 + \(\dfrac{x-50}{49}\)+1+\(\dfrac{x-50}{48}\)+1+\(\dfrac{x-50}{47}\)+1+\(\dfrac{x-50}{25}\)-4 = 0
⇔\(\dfrac{x-50}{50}\)+\(\dfrac{x-50}{49}\)+\(\dfrac{x-50}{48}\)+\(\dfrac{x-50}{47}\)+\(\dfrac{x-50}{25}\)= 0
⇔ (x - 50 ) ( \(\dfrac{1}{50}\)+ \(\dfrac{1}{49}\)+\(\dfrac{1}{48}\)+\(\dfrac{1}{47}\)+\(\dfrac{1}{25}\)) = 0
⇔ x-50 =\(\dfrac{0}{\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}}\)
⇔ x- 50 = 0
⇔ x = 50
vậy S = \(\left\{50\right\}\)
Giải các phương trình:
\(\dfrac{x+24}{1996}+\dfrac{x+25}{1995}+\dfrac{x+26}{1994}+\dfrac{x+27}{1993}+\dfrac{x+2036}{4}=0\)
Mỗi số hạng của vế trái cộng thêm 1, vế phải = 5. Mỗi số hạng vế trái có mẫu số giống nhau, bạn đặt x+ 2020 làm nhân tử chung, phần còn lại tự làm nhé.
mấy bài còn lại bạn đăng cx làm tương tự
\(\frac{x+24}{1996}+\frac{x+25}{1995}+\frac{x+26}{1994}+\frac{x+27}{1993}+\frac{x+2036}{4}=0\)
\(\Leftrightarrow\left(\frac{x+24}{1996}+1\right)+\left(\frac{x+25}{1995}+1\right)+\left(\frac{x+26}{1994}+1\right)+\left(\frac{x+27}{1993}+1\right)+\left(\frac{x+2036}{4}-4\right)=0\)
\(\Leftrightarrow\frac{x+2020}{1996}+\frac{x+2020}{1995}+\frac{x+2020}{1994}+\frac{x+2020}{1993}+\frac{x+2020}{4}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{1996}+\frac{1}{1995}+\frac{1}{1994}+\frac{1}{1993}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\)
Vậy ....
Giải các phương trình sau: (TM ĐK)
1) \(\dfrac{11}{x}=\dfrac{9}{x+1}+\dfrac{2}{x-4}\)
2) \(\dfrac{14}{3x-12}-\dfrac{2+x}{x-4}=\dfrac{3}{8-2x}-\dfrac{5}{6}\)
3) \(\dfrac{x+5}{x^2-5x}-\dfrac{x+25}{2x^2-50}=\dfrac{x-5}{2x^2+10}\)
4) \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{16}{x^2-1}\)
5) \(\left(1-\dfrac{x-1}{x+1}\right)\left(x+2\right)=\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}\)
mng giúp mk bài này nha. Cảm ơn bạn nhiều
\(1,\left(dk:x\ne0,-1,4\right)\)
\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)
\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)
\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)
\(\Leftrightarrow-x=-44\)
\(\Leftrightarrow x=44\left(tm\right)\)
\(2,\left(đk:x\ne4\right)\)
\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)
\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)
\(\Leftrightarrow28-12-6x-9+5x-20=0\)
\(\Leftrightarrow-x=13\)
\(\Leftrightarrow x=-13\left(tm\right)\)
Giải phương trình:
\(\dfrac{158-x}{31}+\dfrac{185-x}{29}+\dfrac{208-x}{27}+\dfrac{227-x}{25}=10\)
`[158-x]/31+[185-x]/29+[208-x]/27+[227-x]/25=10`
`<=>[158-x]/31-1+[185-x]/29-2+[208-x]/27-3+[227-x]/25-4=0`
`<=>[127-x]/21+[127-x]/29+[127-x]/27+[127-x]/25=0`
`<=>(127-x)(1/21+1/29+1/27+1/25)=0`
`<=>127-x=0`
`<=>x=127`