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\(\dfrac{x-130}{20}\)+\(\dfrac{x-100}{25}\)+\(\dfrac{x-60}{30}\)+\(\dfrac{x-10}{35}\)=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
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 ...
Pt\(\Leftrightarrow\dfrac{x+98}{2}+1+\dfrac{x+96}{4}+1+\dfrac{x+65}{35}+1=\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+49}{51}+1\)
\(\Leftrightarrow\dfrac{x+100}{2}+\dfrac{x+100}{4}+\dfrac{x+100}{35}-\dfrac{x+100}{97}-\dfrac{x+100}{95}-\dfrac{x+100}{51}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{35}-\dfrac{1}{97}-\dfrac{1}{35}-\dfrac{1}{51}\right)=0\)
\(\Leftrightarrow x+100=0\Leftrightarrow x=-100\)
Vậy...
\(PT\Leftrightarrow\left(\dfrac{x-70}{130}-1\right)+\left(\dfrac{x-25}{175}-1\right)+\left(\dfrac{x-50}{150}-1\right)+\left(\dfrac{x-275}{25}+3\right)=0\)
\(\Leftrightarrow\left(x-200\right)\left(\dfrac{1}{130}+\dfrac{1}{175}+\dfrac{1}{150}+\dfrac{1}{25}\right)=0\Leftrightarrow x=200\).
Vậy...
\(\dfrac{x}{30}=\dfrac{x}{40}+\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{4x}{120}=\dfrac{3x}{120}+\dfrac{90}{120}\)
\(\Leftrightarrow4x=3x+90\)
\(\Leftrightarrow4x-3x-90=0\)
\(\Leftrightarrow x-90=0\)
\(\Leftrightarrow x=90\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{90\right\}\)
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}
\(\Leftrightarrow3x\left(x-10\right)=60x-60\left(x-10\right)\)
\(\Leftrightarrow3x\left(x-10\right)=600\)
\(\Leftrightarrow x^2-10x-200=0\)
=>(x-20)(x+10)=0
=>x=20 hoặc x=-10
\(\dfrac{60}{x-10}-\dfrac{60}{x}=\dfrac{3}{10}\)đk : x khác 10 ; 0
\(\Leftrightarrow600x-600\left(x-10\right)=3x\left(x-10\right)\)
\(\Leftrightarrow3x^2-30x-6000=0\Leftrightarrow x=50;x=-40\left(tm\right)\)
a) \(\dfrac{x+43}{57}+\dfrac{x+46}{54}=\dfrac{x+49}{51}+\dfrac{x+52}{48}\)
\(\left(\dfrac{x+43}{57}+1\right)+\left(\dfrac{x+46}{54}+1\right)=\left(\dfrac{x+49}{51}+1\right)+\left(\dfrac{x+52}{48}\right)\)
\(\dfrac{x+43+57}{57}+\dfrac{x+46+54}{54}-\dfrac{x+49+51}{51}-\dfrac{x+52+48}{48}=0\)
\(\dfrac{x+100}{57}+\dfrac{x+100}{54}-\dfrac{x+100}{51}-\dfrac{x+100}{48}=0\)
\(\left(x+100\right)\left(\dfrac{1}{57}+\dfrac{1}{54}-\dfrac{1}{51}-\dfrac{1}{48}\right)=0\)
Mà \(\dfrac{1}{57}+\dfrac{1}{54}-\dfrac{1}{51}-\dfrac{1}{48}\ne0\)
Nên: \(x+100=0\)
\(x=-100\)
a, đk : x khác 5;-6
\(x^2+12x+36+x^2-10x+25=2x^2+23x+61\)
\(\Leftrightarrow2x+61=23x+61\Leftrightarrow21x=0\Leftrightarrow x=0\)(tm)
b, đk : x khác 1;3
\(x^2+2x-15=x^2-1-8\Leftrightarrow2x-15=-9\Leftrightarrow x=3\left(ktmđk\right)\)
pt vô nghiệm
a, đk : x khác 5;-6
x2+12x+36+x2−10x+25=2x2+23x+61x2+12x+36+x2−10x+25=2x2+23x+61
⇔2x+61=23x+61⇔21x=0⇔x=0⇔2x+61=23x+61⇔21x=0⇔x=0(tm)
b, đk : x khác 1;3
x2+2x−15=x2−1−8⇔2x−15=−9⇔x=3(ktmđk)x2+2x−15=x2−1−8⇔2x−15=−9⇔x=3(ktmđk)
pt vô nghiệm
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)