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a) \(\dfrac{2}{5}\)-\(\left(\dfrac{1}{10}-x\right)\)=\(\left(\dfrac{-2}{5}-\dfrac{1}{2}\right)^2\)
\(\dfrac{2}{5}\)- \(\left(\dfrac{1}{10}-x\right)\)= \(\dfrac{1}{20}\)
\(\left(\dfrac{1}{10}-x\right)\)= \(\dfrac{2}{5}\)-\(\dfrac{1}{20}\)
\(\left(\dfrac{1}{10}-x\right)\)=\(\dfrac{7}{20}\)
x = \(\dfrac{1}{10}\)-\(\dfrac{7}{20}\)
x = \(\dfrac{-1}{4}\)
Chúc bn học tốt
a: =>4x-6-9=5-3x-3
=>4x-15=-3x+2
=>7x=17
hay x=17/7
b: \(\Leftrightarrow\dfrac{2}{3x}-\dfrac{1}{4}=\dfrac{4}{5}-\dfrac{7}{x}+2\)
=>2/3x+21/3x=4/5+2+1/4=61/20
=>23/3x=61/20
=>3x=23:61/20=460/61
hay x=460/183
a: TH1: x>=0
=>x+x=1/3
=>x=1/6(nhận)
TH2: x<0
Pt sẽ là -x+x=1/3
=>0=1/3(loại)
b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x^2-x-2=0\end{matrix}\right.\Leftrightarrow x=2\)
c: \(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{2}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{x-20-2x+2}{\left(x-1\right)\left(x-20\right)}=\dfrac{-3}{4}\)
\(\Leftrightarrow-3\left(x^2-21x+20\right)=4\left(-x-18\right)\)
\(\Leftrightarrow3x^2-63x+60=4x+72\)
=>3x^2-67x-12=0
hay \(x\in\left\{22.51;-0.18\right\}\)
\(\Leftrightarrow\dfrac{2}{x-3}-\dfrac{2}{x-2}+\dfrac{1}{x-8}-\dfrac{1}{x-3}+\dfrac{1}{x-20}-\dfrac{1}{x-8}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{x-3}-\dfrac{2}{x-2}=\dfrac{-3}{4}\)
\(\Leftrightarrow4\left(x-2\right)-8\left(x-3\right)=-3\left(x-3\right)\left(x-2\right)\)
\(\Leftrightarrow4x-8-8x+24+3\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow3x^2-15x+18-4x+16=0\)
\(\Leftrightarrow3x^2-19x+34=0\)
\(\text{Δ}=\left(-19\right)^2-4\cdot3\cdot34=-47< 0\)
Do đó: Phương trình vô nghiệm
\(\left|x+\dfrac{1}{1.5}\right|+\left|x+\dfrac{1}{5.9}\right|+\left|x+\dfrac{1}{9.14}\right|+...+\left|x+\dfrac{1}{397.401}\right|\ge0\)
\(\Rightarrow101x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow x+\dfrac{1}{1.5}+x+\dfrac{1}{5.9}+...+x+\dfrac{1}{397.401}=101x\)
\(\Rightarrow101x+\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{397.401}\right)=x\)
\(\Rightarrow\dfrac{1}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{397.401}\right)=x\)
\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+....+\dfrac{1}{397}-\dfrac{1}{401}\right)\)
\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{401}\right)\)
\(\Rightarrow x=\dfrac{1}{4}.\dfrac{400}{401}\)
\(\Rightarrow x=\dfrac{100}{401}\)
a)
\(\left(3x+\dfrac{1}{3}\right)\left(x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x+\dfrac{1}{3}=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{9}\\x=\dfrac{1}{2}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{3}{2}\right)\left(2x+1\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{3}{2}>0\\2x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{3}{2}< 0\\2x+1< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\)
a,
\(\dfrac{1}{4}x-1+\dfrac{1}{3}\left(\dfrac{5}{2}x-7\right)-\left(\dfrac{5}{8}x-2\right)=\dfrac{7}{2}\)
\(\Rightarrow\dfrac{1}{4}x-1+\dfrac{5}{6}x-\dfrac{7}{3}-\dfrac{5}{8}x+2=\dfrac{7}{2}\)
\(\Rightarrow\dfrac{1}{4}x+\dfrac{5}{6}x-\dfrac{5}{8}x=\dfrac{7}{2}+1+\dfrac{7}{3}-2\)
\(\Rightarrow\dfrac{11}{24}x=\dfrac{29}{6}\)
\(\Rightarrow x=\dfrac{116}{11}\)
b,
\(\left|2-\dfrac{3}{2}x\right|-4=x+2\)
\(\Rightarrow\left|2-\dfrac{3}{2}x\right|=x-2\)
\(\Rightarrow\left[{}\begin{matrix}2-\dfrac{3}{2}x=x+2\\2-\dfrac{3}{2}x=-\left(x+2\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2-\dfrac{3}{2}x=x+2\\2-\dfrac{3}{2}x=-x-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2-2=x+\dfrac{3}{2}x\\2+2=-x+\dfrac{3}{2}x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{5}{2}x=0\\\dfrac{1}{2}x=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)
c,
\(-3\left(\dfrac{2}{5}x-\dfrac{1}{5}\right)-x\left(x-\dfrac{1}{2}\right)=\dfrac{1}{6}-x^2\)
\(\Rightarrow-\dfrac{6}{5}x+\dfrac{3}{5}-x^2+\dfrac{1}{2}x=\dfrac{1}{6}-x^2\)
\(\Rightarrow-\dfrac{7}{10}x=\dfrac{1}{6}-\dfrac{3}{5}-x^2+x^2\)
\(\Rightarrow-\dfrac{7}{10}x=-\dfrac{13}{30}\Leftrightarrow x=\dfrac{13}{21}\)
a: \(\dfrac{x+1}{5}+\dfrac{x+1}{6}=\dfrac{x+1}{7}+\dfrac{x+1}{8}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}-\dfrac{1}{8}\right)=0\)
=>x+1=0
hay x=-1
b: \(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)
=>x-2010=0
hay x=2010
c: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+17\right)}=\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}\)
=>x=15
`|x+1/3|+|x+2/3|+|x+2/5|+|x+3/2|=33x`
`@TH1: x >= -1/3`
`=>x+1/3+x+2/3+x+2/5+x+3/2=33x`
`=>29x=29/10`
`=>x=1/10` (t/m)
`@TH2: -2/3 <= x < -1/3`
`=>-x-1/3+x+2/3+x+2/5+x+3/2=33x`
`=>31x=67/30`
`=>x=67/930` (ko t/m)
`@TH3:-2/5 <= x < -2/3`
`=>-x-1/3-x-2/3+x+2/5+x+3/2=33x`
`=>33x=9/10`
`=>x=3/110` (ko t/m)
`@TH4:-3/2 <= x < -2/5`
`=>-x-1/3-x-2/3-x-2/5+x+3/2=33x`
`=>35x=1/10`
`=>x=1/350` (ko t/m)
`@TH5: x < -3/2`
`=>-x-1/3-x-2/3-x-2/5-x-3/2=33x`
`=>37x=-29/10`
`=>x=-29/370` (ko t/m)
có VT \(\ge\) 0 với mọi x
=>VP:33x\(\ge\) 0 \(\Rightarrow\) x\(\ge\)0
\(\Rightarrow\) |x+1/3|\(\ge\)0;|x+2/3|\(\ge\) 0;|x+2/5|\(\ge\) 0;|x+3/2|\(\ge\) 0
=> (x+1/3)+(x+2/3)+(x+2/5)+(x+3/2)=33x
=>(x+x+x+x)+(1/3+2/3+2/5+3/2)=33x
=>4x+29/10=33x
=> 29/10=33x-4x
=>29/10=29x
=>x=29/10:29
=>x=1/10