giải phương trình \(\frac{100}{x}-\frac{100}{x+20}=\frac{1}{3}\)
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1 tháng 2 2022
a: \(\Leftrightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{9}-\dfrac{1}{10}\right)\cdot\left(x-1\right)+\dfrac{1}{10}x-x=-\dfrac{9}{10}\)
\(\Leftrightarrow\dfrac{9}{10}x-\dfrac{9}{10}-\dfrac{9}{10}x=-\dfrac{9}{10}\)
=>-9/10=-9/10(luôn đúng)
b: \(\Leftrightarrow\dfrac{195x+195+130x+195+117x+195+100x+195}{195}=\dfrac{22\cdot39+4\cdot65+6\cdot39+40\cdot5}{195}\)
=>347x+780=1552
=>347x=772
hay x=772/347
DN
25 tháng 12 2017
a) \(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
\(\Leftrightarrow\left(\frac{x-5}{100}-1\right)+\left(\frac{x-4}{101}-1\right)+\left(\frac{x-3}{102}-1\right)=\left(\frac{x-100}{5}-1\right)+\left(\frac{x-101}{4}-1\right)+\left(\frac{x-102}{3}-1\right)\)
\(\Leftrightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)
\(\Leftrightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
\(\Leftrightarrow x=105\)
b) \(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5\)
\(\Leftrightarrow\left(\frac{29-x}{21}+1\right)+\left(\frac{27-x}{23}+1\right)+\left(\frac{25-x}{25}+1\right)+\left(\frac{23-x}{27}+1\right)+\left(\frac{21-x}{29}+1\right)=0\)
\(\Leftrightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)
\(\Leftrightarrow x=50\)
NC
1
LM
1
4 tháng 4 2020
\(\frac{100}{\frac{6}{5}x}+\frac{100}{x}=\frac{11}{3}\)
ĐK: \(x\ne0\)
\(\Leftrightarrow100+\frac{100}{x}.\frac{6}{5}x=\frac{11}{3}.\frac{6}{5}x\)
\(\Leftrightarrow100+120=\frac{22}{5}x\Leftrightarrow\frac{22}{5}x=220\Leftrightarrow x=50\left(tm\right)\)
H9
HT.Phong (9A5)
CTVHS
26 tháng 8 2023
\(0,1^{2x-1}=100\)
\(\Leftrightarrow0,1^{2x-1}=0,1^{log_{0,1}100}\)
\(\Leftrightarrow2x-1=log_{0,1}100\)
\(\Leftrightarrow2x-1=-2\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Chọn A
K
0
25 tháng 2 2019
\(\frac{x+5}{95}+\frac{x+3}{97}+\frac{x+1}{99}=\frac{x+15}{85}+\frac{x+20}{80}+\frac{x+25}{75}.\)
\(\frac{x+5}{95}+1+\frac{x+3}{97}+1+\frac{x+1}{99}+1-\frac{x+15}{85}-1-\frac{x+20}{80}-1-\frac{x+25}{75}-1=0\)
\(\frac{x+100}{95}+\frac{x+100}{97}+\frac{x+100}{99}-\frac{x+100}{85}-\frac{x+100}{80}-\frac{x+100}{75}=0\)
\(\left(x+100\right).\left(\frac{1}{95}+\frac{1}{97}+\frac{1}{99}-\frac{1}{85}-\frac{1}{80}-\frac{1}{75}\right)=0\)
\(\Rightarrow x+100=0\Rightarrow x=-100\)
\(\frac{1}{95}+\frac{1}{97}+\frac{1}{99}-\frac{1}{85}-\frac{1}{80}-\frac{1}{75}\ne0\)
\(\frac{100\left(x+20\right)}{x\left(x+20\right)}-\frac{100x}{x\left(x+20\right)}=\frac{1}{3}\)
\(\frac{100x+2000-100x}{x\left(x+20\right)}=\frac{1}{3}\)
\(\frac{2000}{x\left(x+20\right)}=\frac{1}{3}\)
\(\Rightarrow x^2+20x=3.2000\)
\(\Rightarrow x^2+20x-6000=0\)
ĐKXĐ: \(x\ne0;x\ne-2\)
Ta có: \(\frac{100x+2000-100x}{x\left(x+20\right)}=\frac{1}{3}\)
\(\Leftrightarrow\frac{2000}{x^2+20x}=\frac{1}{3}\)
\(\Leftrightarrow x^2+20x=6000\)
\(\Leftrightarrow x^2+2.10x+100=6100\)
\(\Leftrightarrow\left(x+10\right)^2=6100\)
\(\Leftrightarrow\orbr{\begin{cases}x=10\sqrt{61}-10\left(TM\right)\\x=-10\sqrt{61}-10\left(TM\right)\end{cases}}\)
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