(2.x-1)50=2.x-1
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NM
Nguyễn Minh Hằng
8 tháng 10 2021
x=1 nha bạn !
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Những câu hỏi liên quan
DP
7 tháng 7 2017
\(c,\)\(\left(x-1\right)+\left(x-2\right)+....+\left(x-100\right)=50\)
\(\left(x+x+...+x\right)-\left(1+2+...+100\right)=50\)
\(100x-5050=50\)
\(100x=50+5050\)
\(100x=5100\)
\(\Rightarrow x=\frac{5100}{100}=51\)
DP
7 tháng 7 2017
\(a,\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+....+\left(x+100\right)=5750\)
\(\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)
\(100x+5050=5750\)
\(100x=5750-5050\)
\(100x=700\)
\(\Rightarrow x=7\)
\(b,x+\left(1+2+3+...+50\right)=2000\)
\(x+\frac{\left[1+50\right]\cdot\left[\left(50-1\right)\div1+1\right]}{2}=2000\)
\(x+1275=2000\)
\(\Rightarrow x=2000-1275=725\)
VN
16 tháng 8 2016
a) \(\frac{5-x}{4x^2-8x}\) + \(\frac{7}{8x}\) = \(\frac{x-1}{2x\left(x-2\right)}\) +\(\frac{1}{8x-16}\) ĐKXĐ : x #0, x#2, x#-2
<=> \(\frac{5-x}{4x\left(x-2\right)}\) + \(\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}\) + \(\frac{1}{8\left(x-2\right)}\)
<=> \(\frac{2\left(5-x\right)}{8x\left(x-2\right)}+\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{4\left(x-1\right)}{8x\left(x-2\right)}+\frac{x}{8x\left(x-2\right)}\)
=> 10 - 2x + 7x - 14 = 4x - 4 + x
<=>-2x + 7x - 4x + x = -4 - 10 + 14
<=>x=-14
VN
3
8 tháng 3 2017
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)
DH
0
N
2
16 tháng 2 2021
Ta có: \(\dfrac{x+1}{99}+\dfrac{x+2}{98}+...+\dfrac{x+50}{50}+50=0\)
\(\Leftrightarrow\dfrac{x+1}{99}+1+\dfrac{x+2}{98}+1+...+\dfrac{x+50}{50}+1=0\)
\(\Leftrightarrow\dfrac{x+100}{99}+\dfrac{x+100}{98}+...+\dfrac{x+100}{50}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}+...+\dfrac{1}{50}\right)=0\)
mà \(\dfrac{1}{99}+\dfrac{1}{98}+...+\dfrac{1}{50}>0\)
nên x+100=0
hay x=-100
Vậy: S={-100}
16 tháng 2 2021
\(\dfrac{x+1}{99}+\dfrac{x+2}{98}+...+\dfrac{x+50}{50}+50=0\)
\(\Leftrightarrow\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)+...+\left(\dfrac{x+50}{50}+1\right)=0\)
\(\Leftrightarrow\dfrac{x+100}{99}+\dfrac{x+100}{98}+...+\dfrac{x+100}{50}=0\)
\(\Leftrightarrow\left(x+100\right).\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+...+\dfrac{1}{50}\right)=0\)
\(\Leftrightarrow x+100=0\) (vì \(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+...+\dfrac{1}{50}>0\) )
\(\Leftrightarrow x=-100\)
21 tháng 7 2022
a: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\cdot\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
=>x=3 hoặc x=-10/7
b: \(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow13\left(x+3\right)+x^2-9-12x-42=0\)
\(\Leftrightarrow x^2-12x-51+13x+39=0\)
\(\Leftrightarrow x^2+x-12=0\)
=>(x+4)(x-3)=0
=>x=-4
NK
1
24 tháng 1 2016
Ta có:x1+x2+...+x49+x50+x51=(x1+x2)+...+(x49+x50)+x51=0
= 1+1+...+1 +x51=0 (có 25 số 1)
=25 +x51=0
=>x51=0-25= -25
Ta có:x50+x51 =1
x50+(-25)=1
x50 =1-(-25)=26
Vậy x50= 26
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