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2) $\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}$
$=>\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1$
$=>\dfrac{x+4}{2000}+\dfrac{2000}{2000}+\dfrac{x+3}{2001}+\dfrac{2001}{2001}=\dfrac{x+2}{2002}+\dfrac{2002}{2002}+\dfrac{x+1}{2003}+\dfrac{2003}{2003}$
$=>\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}$
$=>\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0$
$=>(x+2004)(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}=0$
$=>x+2004=0$
$=>x=-2004$
3) Ta có : $A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}$
$=>A=\dfrac{1}{2}+\dfrac{1}{12}+...+\dfrac{1}{99.100}>\dfrac{1}{2}+\dfrac{1}{12}=\dfrac{7}{12}$
$=>A>\dfrac{7}{12}(1)$
Ta lại có : $A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$
$=>A=(1-\dfrac{1}{2}+\dfrac{1}{3})-(\dfrac{1}{4}-\dfrac{1}{5})-...-(\dfrac{1}{98}-\dfrac{1}{99})-\dfrac{1}{100}<(1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}$
$=>A<\dfrac{5}{6}(2)$
Từ (1)(2) => đpcm.
`@` `\text {Ans}`
`\downarrow`
`1,`
`3/16 - (x - 5/4) - (3/4 + (-7)/8 - 1) = 2 1/2`
`=> 3/16 - x + 5/4 - (-1/8 - 1) = 2 1/2`
`=> 3/16 - x + 5/4 - (-9/8) = 2 1/2`
`=> 3/16 - x + 19/8 = 2 1/2`
`=> 3/16 - x = 2 1/2 - 19/8`
`=> 3/16 - x =1/8`
`=> x = 3/16 - 1/8`
`=> x = 1/16`
Vậy, `x = 1/16`
`2,`
`1/2* (1/6 - 9/10) = 1/5 - x + (1/15 - (-1)/5)`
`=> 1/2 * (-11/15) = 1/5 - x + 4/15`
`=> -11/30 = x + 1/5 - 4/15`
`=> x + (-1/15) = -11/30`
`=> x = -11/30 + 1/15`
`=> x = -3/10`
Vậy, `x = -3/10.`
a)
\(\left|x-2\right|-\dfrac{3}{5}=\dfrac{1}{2}\\ \left|x-2\right|=\dfrac{1}{2}+\dfrac{3}{5}\\ \left|x-2\right|=\dfrac{11}{10}\\ =>\left[{}\begin{matrix}x-2=\dfrac{11}{10}\\x-2=-\dfrac{11}{10}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{31}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{7}{3}\right):\dfrac{-1}{3}=0,4\\ x-\dfrac{7}{3}=0,4\cdot\dfrac{-1}{3}\\ x-\dfrac{7}{3}=-\dfrac{2}{15}\\ x=-\dfrac{2}{15}+\dfrac{7}{3}\\ x=\dfrac{11}{5}\)
c)
\(\left|x-3\right|=5\\ =>\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\left[{}\begin{matrix}x=5+3\\x=-5+3\end{matrix}\right.\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
d)
\(\left(2x+3\right)^2=25\\ =>\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
e)
\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\dfrac{1}{4}:x=-\dfrac{7}{20}\)
\(x=\dfrac{1}{4}:\dfrac{-7}{20}\\ x=-\dfrac{5}{7}\)
f)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\\ =>x-\dfrac{1}{2}=\dfrac{1}{3}\\ x=\dfrac{1}{3}+\dfrac{1}{2}\\ x=\dfrac{5}{6}\)
\(a,\)\(\frac{x^7}{81}=27\)
\(\Rightarrow x^7=3^3.3^4=3^7\)
\(\Rightarrow x=3\)
\(b,\left(x^2\right)^4=\frac{x^{18}}{x^{10}}\)
\(\Rightarrow x^{18}=x^{10}.x^6\)
\(\Rightarrow x^{18}-x^{16}=0\)
\(\Rightarrow x^{16}\left(x^2-1\right)=0\)
\(\Rightarrow x^{16}\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x=\pm1\end{cases}}\)
\(\frac{x^7}{81}=27\Rightarrow x^7=27\cdot81=2187\)
\(x^7=2187\Leftrightarrow x^7=3^7\Rightarrow x=3\)
Vậy x=3
\(\dfrac{-3}{5}-x=\dfrac{21}{10}\)
\(x=\dfrac{-3}{5}-\dfrac{21}{10}\)
\(x=\)-\(\dfrac{27}{10}\)
\(x:\dfrac{2}{9}=\dfrac{9}{2}\)
\(x.\dfrac{9}{2}=\dfrac{9}{2}\)
\(x=\dfrac{9}{2}:\dfrac{9}{2}\)
\(x=1\)
\(\dfrac{x}{9}=\dfrac{5}{3}\)
\(x.3=5.9\)
\(x.3=45\)
\(x=45:3=15\)
\(x:\left(\dfrac{2}{5}\right)^3=\left(\dfrac{5}{2}\right)^3\)
\(x:\dfrac{8}{125}=\dfrac{125}{8}\)
\(x.\dfrac{125}{8}=\dfrac{125}{8}\)
\(x=\dfrac{125}{8}:\dfrac{125}{8}=1\)
\(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right)+\dfrac{4}{5}\\ =-\dfrac{5}{21}:\dfrac{4}{5}+\dfrac{5}{21}\\ =\left(-\dfrac{5}{21}+\dfrac{5}{21}\right):\dfrac{4}{5}\\ =0:\dfrac{4}{5}\\ =0.\)
Sửa cho mk dòng đầu là :4/5 và dòng tiếp theo mk thiếu :4/5
c) \(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)
⇔\(\left(x+4\right)\left(x+4\right)=100\)
⇔\(\left(x+4\right)^2=10^2\)
⇔\(\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=6\\x=-14\end{matrix}\right.\)
\(c,ĐK:x\ne-4\\ PT\Leftrightarrow\left(x+4\right)^2=100\\ \Leftrightarrow\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(tm\right)\\x=-14\left(tm\right)\end{matrix}\right.\\ d,ĐK:x\ne-2;x\ne-3\\ PT\Leftrightarrow\left(x-1\right)\left(x+3\right)=\left(x-2\right)\left(x+2\right)\\ \Leftrightarrow x^2+2x-3=x^2-4\\ \Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\)
\(\dfrac{3}{2}x-\dfrac{7}{3}=-\dfrac{1}{4}\)
\(\dfrac{3}{2}x\) \(=\dfrac{1}{4}+\dfrac{7}{3}\)
\(\dfrac{3}{2}x\) \(=\dfrac{3+28}{12}=\dfrac{31}{12}\)
\(x\) \(=\dfrac{31}{12}.\dfrac{2}{3}=\dfrac{31.1}{6.3}=\dfrac{31}{18}\)
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