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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
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) \(\dfrac{2}{5}\)-\(\left(\dfrac{1}{10}-x\right)\)=\(\left(\dfrac{-2}{5}-\dfrac{1}{2}\right)^2\)
\(\dfrac{2}{5}\)- \(\left(\dfrac{1}{10}-x\right)\)= \(\dfrac{1}{20}\)
\(\left(\dfrac{1}{10}-x\right)\)= \(\dfrac{2}{5}\)-\(\dfrac{1}{20}\)
\(\left(\dfrac{1}{10}-x\right)\)=\(\dfrac{7}{20}\)
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) \(x+\dfrac{3}{10}=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}-\dfrac{3}{10}\)
\(x=\dfrac{-7}{10}\)
b) \(x+\dfrac{5}{6}=\dfrac{2}{5}-\left(-\dfrac{2}{3}\right)\)
\(x+\dfrac{5}{6}=\dfrac{2}{5}+\dfrac{2}{3}\)
\(x+\dfrac{5}{6}=\dfrac{16}{15}\)
\(x=\dfrac{16}{15}-\dfrac{5}{6}\)
\(x=\dfrac{7}{30}\)
c) \(1\dfrac{2}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x=-\dfrac{4}{5}-\dfrac{3}{7}\)
\(\dfrac{7}{5}x=\dfrac{-43}{35}\)
\(\Rightarrow x=\dfrac{-43}{49}\)
d) \(\left[x+\dfrac{3}{4}\right]-\dfrac{1}{3}=0\)
\(\left[x+\dfrac{3}{4}\right]=0+\dfrac{1}{3}\)
\(\left[x+\dfrac{3}{4}\right]=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}-\dfrac{3}{4}\)
\(x=\dfrac{-5}{12}\)
e) \(\left[x+\dfrac{4}{5}\right]-\left(-3,75\right)=-\left(-2,15\right)\)
\(\left[x+\dfrac{4}{5}\right]+3,75=2,15\)
\(x+\dfrac{4}{5}=2,15-3,75\)
\(x+\dfrac{4}{5}=-\dfrac{8}{5}\)
\(x=\dfrac{-8}{5}-\dfrac{4}{5}\)
\(x=\dfrac{-12}{5}\)
f) \(\left(x-2\right)^2=1\)
\(\Rightarrow x=1\)
Sức chịu đựng có giới hạn -.-
- Mình tiếp tục cho Nguyễn Phương Trâm nhé.
g, \(\left(2x-1\right)^3=-27\)
\(\Rightarrow\left(2x-1\right)^3=\left(-3\right)^3\)
\(\Rightarrow2x-1=-3\)
\(\Rightarrow2x=-2\)
=> \(x=-1\)
- Vậy x = -1
h,\(\dfrac{x-1}{-15}=-\dfrac{60}{x-1}\)
\(\Rightarrow\left(x-1\right)^2=-60.\left(-15\right)\)
\(\Rightarrow\left(x-1\right)^2=900 \)
\(\Rightarrow\left(x-1\right)^2=30^2\Rightarrow x-1=30\)
=> x = 31
i,\(x:\left(\dfrac{-1}{2}\right)^3=\dfrac{-1}{2}\)
=> \(x:\left(-\dfrac{1}{8}\right)=-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{16}\)
- Vậy x=\(\dfrac{1}{16}\)
j, \(\left(\dfrac{3}{4}\right)^5.x=\left(\dfrac{3}{4}\right)^7\)
\(\Rightarrow \left(\dfrac{3}{4}\right).x=\left(\dfrac{3}{4}\right)^2\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^2:\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}\)
- Vạy x = \(\dfrac{3}{4}\)
k, \(8^x:2^x=4\Rightarrow\left(8:2\right)^x=4\)
=>\(4^x=4\)
=> x = 1
- Vậy x = 1
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,
\(-3\left(\dfrac{2}{5}x-\dfrac{1}{5}\right)-x\left(x-\dfrac{1}{2}\right)=\dfrac{1}{6}-x^2\)
\(\Rightarrow-\dfrac{6}{5}x+\dfrac{3}{5}-x^2+\dfrac{1}{2}x=\dfrac{1}{6}-x^2\)
\(\Rightarrow-\dfrac{7}{10}x=\dfrac{1}{6}-\dfrac{3}{5}-x^2+x^2\)
\(\Rightarrow-\dfrac{7}{10}x=-\dfrac{13}{30}\Leftrightarrow x=\dfrac{13}{21}\)
1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)
=>4x=18
hay x=9/2
2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)
=>4x=108
hay x=27
3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)
\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)
=>4x=12
hay x=3
b, \(\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{x+17-x+2}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow x=19\)
Chúc bạn học tốt!!!
a, \(\dfrac{x+1}{5}+\dfrac{x+3}{4}=\dfrac{x+5}{3}+\dfrac{x+7}{2}\)
\(\Rightarrow\dfrac{x+1}{5}+2+\dfrac{x+3}{4}+2=\dfrac{x+5}{3}+2+\dfrac{x+7}{2}+2\)
\(\Rightarrow\dfrac{x+11}{5}+\dfrac{x+11}{4}-\dfrac{x+11}{3}-\dfrac{x+11}{2}=0\)
\(\Rightarrow\left(x+11\right)\left(\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\right)=0\)
\(\Rightarrow x+11=0\Rightarrow x=-11\)
Vậy x = -11
b, \(\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{15}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow x=15\)
Vậy 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