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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}
Ta có : \(\dfrac{5x-150}{50}+\dfrac{5x-102}{49}+\dfrac{5x-56}{48}+\dfrac{5x-12}{47}+\dfrac{5x-660}{46}=0\)
\(\Leftrightarrow\dfrac{5x-150}{50}-1+\dfrac{5x-102}{49}-2+\dfrac{5x-56}{48}-3+\dfrac{5x-12}{47}-4+\dfrac{5x-660}{46}+10=0\)
\(\Leftrightarrow\dfrac{5x-200}{50}+\dfrac{5x-200}{49}+\dfrac{5x-200}{48}+\dfrac{5x-200}{47}+\dfrac{5x-200}{46}=0\)
\(\Leftrightarrow\left(5x-200\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\right)=0\)
\(\Leftrightarrow5x-200=0\)
\(\Leftrightarrow x=40\)
Vậy ...
Ta có: \(\dfrac{5x-150}{50}+\dfrac{5x-102}{49}+\dfrac{5x-56}{48}+\dfrac{5x-12}{47}+\dfrac{5x-660}{46}=0\)
\(\Leftrightarrow\dfrac{5x-150}{50}-1+\dfrac{5x-102}{49}-2+\dfrac{5x-56}{48}-3+\dfrac{5x-12}{47}-4+\dfrac{5x-660}{46}+10=0\)
\(\Leftrightarrow\dfrac{5x-200}{50}+\dfrac{5x-200}{49}+\dfrac{5x-200}{48}+\dfrac{5x-200}{47}+\dfrac{5x-200}{46}=0\)
\(\Leftrightarrow\left(5x-200\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\right)=0\)
mà \(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}>0\)
nên 5x-200=0
\(\Leftrightarrow5x=200\)
hay x=40
Vậy: S={40}
\(PT\Leftrightarrow\left(\dfrac{x-70}{130}-1\right)+\left(\dfrac{x-25}{175}-1\right)+\left(\dfrac{x-50}{150}-1\right)+\left(\dfrac{x-275}{25}+3\right)=0\)
\(\Leftrightarrow\left(x-200\right)\left(\dfrac{1}{130}+\dfrac{1}{175}+\dfrac{1}{150}+\dfrac{1}{25}\right)=0\Leftrightarrow x=200\).
Vậy...
ĐKXĐ: \(x\neq 0\).
Đặt \(\dfrac{x}{3}-\dfrac{4}{x}=t\).
PT đã cho tương đương:
\(3t^2+8-10t=0\)
\(\Leftrightarrow\left(t-2\right)\left(3t-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=\dfrac{4}{3}\end{matrix}\right.\).
Với t = 2 ta có \(\dfrac{x}{3}-\dfrac{4}{x}=2\Leftrightarrow\dfrac{x^2-12}{3x}=2\Leftrightarrow x^2-6x-12=0\Leftrightarrow x=\pm\sqrt{21}+3\).
Với t = \(\frac{4}{3}\) ta có \(\dfrac{x}{3}-\dfrac{4}{x}=\dfrac{4}{3}\Leftrightarrow\dfrac{x^2-12}{3x}=\dfrac{4}{3}\Leftrightarrow x^2-12=4x\Leftrightarrow x^2-4x-12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\).
Vậy...
b) \(\dfrac{5x-150}{50}+\dfrac{5x-102}{49}+\dfrac{5x-56}{48}+\dfrac{5x-12}{47}+\dfrac{5x-660}{46}=0\)
\(\Leftrightarrow\dfrac{5x-150}{50}-1+\dfrac{5x-102}{49}-2+\dfrac{5x-56}{48}-3+\dfrac{5x-12}{47}-4+\dfrac{5x-660}{46}+10=0\)
\(\Leftrightarrow\dfrac{5x-200}{50}+\dfrac{5x-200}{49}+\dfrac{5x-200}{48}+\dfrac{5x-200}{47}+\dfrac{5x-200}{46}=0\)
\(\Leftrightarrow\left(5x-200\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\right)=0\)
\(\Leftrightarrow5x-200=0\)
\(\Leftrightarrow x=40\)
b)
\(\dfrac{5x-150}{50}+\dfrac{5x-102}{49}+\dfrac{5x-56}{48}+\dfrac{5x-12}{47}+\dfrac{5x-660}{46}=0\)
\(\Rightarrow\left(\dfrac{5x-150}{50}-1\right)+\left(\dfrac{5x-102}{49}-2\right)+\left(\dfrac{5x-56}{48}-3\right)+\left(\dfrac{5x-12}{47}-4\right)\)
\(+\left(\dfrac{5x-660}{46}+10\right)=0\)
\(\Rightarrow\dfrac{5x-200}{50}+\dfrac{5x-200}{49}+\dfrac{5x-200}{48}+\dfrac{5x-200}{47}+\dfrac{5x-200}{46}=0\)
\(\Rightarrow\left(5x-200\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\right)=0\)
\(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\ne0\)
\(\Rightarrow5x-200=0\Rightarrow x=40\)
\(1,\left(dk:x\ne0,-1,4\right)\)
\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)
\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)
\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)
\(\Leftrightarrow-x=-44\)
\(\Leftrightarrow x=44\left(tm\right)\)
\(2,\left(đk:x\ne4\right)\)
\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)
\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)
\(\Leftrightarrow28-12-6x-9+5x-20=0\)
\(\Leftrightarrow-x=13\)
\(\Leftrightarrow x=-13\left(tm\right)\)
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
b)\(\dfrac{x+14}{86}+\dfrac{x+15}{85}+\dfrac{x+16}{84}+\dfrac{x+17}{83}+\dfrac{x+116}{4}=0\)
\(\Leftrightarrow\dfrac{x+14}{86}+1+\dfrac{x+15}{85}+1+\dfrac{x+16}{84}+1+\dfrac{x+17}{83}+1+\dfrac{x+116}{4}-4=0\)
\(\Leftrightarrow\dfrac{x+100}{86}+\dfrac{x+100}{85}+\dfrac{x+100}{84}+\dfrac{x+100}{83}+\dfrac{x+100}{4}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{86}+\dfrac{1}{85}+\dfrac{1}{84}+\dfrac{1}{83}+\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow x+100=0\).Do \(\dfrac{1}{86}+\dfrac{1}{85}+\dfrac{1}{84}+\dfrac{1}{83}+\dfrac{1}{4}\ne0\)
\(\Leftrightarrow x=-100\)
c)\(\dfrac{1}{\left(x^2+5\right)\left(x^2+4\right)}+\dfrac{1}{\left(x^2+4\right)\left(x^2+3\right)}+\dfrac{1}{\left(x^2+3\right)\left(x^2+2\right)}+\dfrac{1}{\left(x^2+2\right)\left(x^2+1\right)}=-1\)
\(\Leftrightarrow\dfrac{1}{\left(x^2+1\right)\left(x^2+2\right)}+\dfrac{1}{\left(x^2+2\right)\left(x^2+3\right)}+...+\dfrac{1}{\left(x^2+4\right)\left(x^2+5\right)}=-1\)
\(\Leftrightarrow\dfrac{1}{x^2+1}-\dfrac{1}{x^2+2}+\dfrac{1}{x^2+2}-\dfrac{1}{x^2+3}+...+\dfrac{1}{x^2+4}-\dfrac{1}{x^2+5}=-1\)
\(\Leftrightarrow\dfrac{1}{x^2+1}-\dfrac{1}{x^2+5}=-1\)\(\Leftrightarrow\dfrac{4}{x^4+6x^2+5}=-1\)
\(\Leftrightarrow\dfrac{x^4+6x^2+9}{x^4+6x^2+5}=0\Leftrightarrow x^4+6x^2+9=0\)
\(\Leftrightarrow\left(x^2+3\right)^2>0\forall x\) (vô nghiệm)
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 = \(\left\{50\right\}\)