a) 3/2.x - 1 = 1/2x - 3/5
b) 1/2x + 1/2 . (x-2) = 3/4 - 2x
c) ( x - 1/2 ) - 1/4 = 0
d) 4 mũ x-3 + 1 = 17
Cho mik xin lời giải chi tiết nx ạ
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`a, 3/4 - 5/4 :(x-1) =1/2`
`=> 5/4:(x-1)= 3/4 -1/2`
`=> 5/4:(x-1)= 3/4 - 2/4`
`=> 5/4:(x-1)= 1/4`
`=> x-1= 5/4 : 1/4`
`=> x-1=5`
`=>x=5+1`
`=>x=6`
__
`(1/2-x)^2 -2^2 =12`
`=> (1/2-x)^2 = 12+4`
`=> (1/2-x)^2= 16`
`=> (1/2-x)^2 =4^2`
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}-x=4\\\dfrac{1}{2}-x=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\)
__
`(1/2)^(2x-1) =1/16`
`=> (1/2)^(2x-1) = (1/2)^4`
`=> 2x-1=4`
`=> 2x=4+1`
`=>2x=5`
`=>x=5/2`
\(a,\dfrac{3}{4}-\dfrac{5}{4}:\left(x-1\right)=\dfrac{1}{2}\)
\(\dfrac{5}{4}:\left(x-1\right)=\dfrac{3}{4}-\dfrac{1}{2}\)
\(\dfrac{5}{4}:\left(x-1\right)=\dfrac{1}{4}\)
\(x-1=\dfrac{5}{4}:\dfrac{1}{4}\)
\(x-1=5\)
\(x=6\)
\(\left(\dfrac{1}{2}-x\right)^2-2^2=12\)
\(\left(\dfrac{1}{2}-x\right)^2-4=12\)
\(\left(\dfrac{1}{2}-x\right)^2=16\)
\(\left(\dfrac{1}{2}-x\right)^2=4^2hoặc\left(\dfrac{1}{2}-x\right)^2=\left(-4\right)^2\)
\(\dfrac{1}{2}-x=4hoặc\dfrac{1}{2}-x=-4\)
=>1/2 -x =4 1/2 -x= -4
=> x=1/2-4 x=1/2-(-4)
=>x=-7/2 x=9/2
vậy x∈{-7/2 ; 9/2}
\(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{16}\)
\(=>\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^4\)
\(=>2x-1=4\)
\(=>2x=5\)
\(=>x=\dfrac{5}{2}\)
c: \(=\dfrac{x^3+2x+2x^2+2x+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^2+2x+1}{x^2-x+1}\)
Ko cần đâu bn à mk mong bn đấy
a)\(\left(3x-1\right)\left(5-\frac{1}{2}x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5-\frac{1}{2}x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)
b)\(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
\(2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{8}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{29}{12}\\x=-\frac{13}{12}\end{cases}}\)
a)\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)
\(\Leftrightarrow\)3x - 1 = 0 hay \(\frac{-1}{2}\)x + 5 = 0
\(\Leftrightarrow\)3x = 1 I\(\Leftrightarrow\)\(\frac{-1}{2}\)x = -5
\(\Leftrightarrow\) x = \(\frac{1}{3}\) I\(\Leftrightarrow\) x = 10
b) 2 I \(\frac{1}{2}x-\frac{1}{3}\)I - \(\frac{3}{2}\)=\(\frac{1}{4}\)
\(\Leftrightarrow\) 2 I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{4}\)
\(\Leftrightarrow\) I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{7}{8}\) hay \(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{-7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x\) = \(\frac{29}{24}\) I\(\Leftrightarrow\)\(\frac{1}{2}x\) = \(\frac{-13}{24}\)
\(\Leftrightarrow\) x = \(\frac{29}{12}\) I\(\Leftrightarrow\) x = \(\frac{-13}{12}\)
c) (2x +\(\frac{3}{5}\))2 - \(\frac{9}{25}\)= 0
\(\Leftrightarrow\)(2x +\(\frac{3}{5}\))2 = \(\frac{9}{25}\)
\(\Leftrightarrow\) 2x +\(\frac{3}{5}\) = \(\frac{3}{5}\) hay 2x +\(\frac{3}{5}\)= \(\frac{-3}{5}\)
\(\Leftrightarrow\) 2x = 0 I \(\Leftrightarrow\)2x = \(\frac{-6}{5}\)
\(\Leftrightarrow\) x = 0 I \(\Leftrightarrow\) x = \(\frac{-3}{5}\)
d) 3(x -\(\frac{1}{2}\)) - 5(x +\(\frac{3}{5}\)) = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)3x - \(\frac{3}{2}\)- 5x - 3 = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)-2x + x - \(\frac{9}{2}\)- \(\frac{1}{5}\)= 0
\(\Leftrightarrow\)-x = \(\frac{-47}{10}\)
\(\Leftrightarrow\) x = \(\frac{47}{10}\)
a) \(2x-6=0\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=\dfrac{6}{2}=3\)
b) \(x^2-4x=0\)
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x=-\frac{13}{4}\)
\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)
\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)
\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)
\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)
\(\Leftrightarrow x=-\frac{6}{11}\)
d,e,f Tương tự
\(a,\dfrac{3}{2}\cdot x-1=\dfrac{1}{2}x-\dfrac{3}{5}\)
\(\Rightarrow\dfrac{3}{2}x-\dfrac{1}{2}x=-\dfrac{3}{5}+1\)
\(\Rightarrow\left(\dfrac{3}{2}-\dfrac{1}{2}\right)x=-\dfrac{3}{5}+\dfrac{5}{5}\)
\(\Rightarrow x=\dfrac{2}{5}\)
\(b,\dfrac{1}{2}x+\dfrac{1}{2}\left(x-2\right)=\dfrac{3}{4}-2x\)
\(\Rightarrow\dfrac{1}{2}x+\dfrac{1}{2}x+2x-1=\dfrac{3}{4}\)
\(\Rightarrow\left(\dfrac{1}{2}+\dfrac{1}{2}+2\right)x=\dfrac{3}{4}+1\)
\(\Rightarrow3x=\dfrac{7}{4}\)
\(\Rightarrow x=\dfrac{7}{4}:3\)
\(\Rightarrow x=\dfrac{7}{12}\)
\(c,\left(x-\dfrac{1}{2}\right)-\dfrac{1}{4}=0\)
\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{1}{4}+\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{4}+\dfrac{2}{4}\)
\(\Rightarrow x=\dfrac{3}{4}\)
\(d,4^{x-3}+1=17\)
\(\Rightarrow4^{x-3}=17-1\)
\(\Rightarrow4^{x-3}=16\)
\(\Rightarrow4^{x-3}=4^2\)
\(\Rightarrow x-3=2\)
\(\Rightarrow x=2+3\)
\(\Rightarrow x=5\)
#Toru
`3/2 x -1 =1/2x -3/5`
`=> 3/2x -1/2x = -3/5 +1`
`=> 2/2x= -3/5 + 5/5`
`=> x= 2/5`
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`1/2x +1/2(x-2) = 3/4 -2x`
`=> 1/2x + 1/2x - 2/2 = 3/4 -2x`
`=> 1/2x +1/2x +2x = 3/4 + 1`
`=> 1/2x +1/2x + 4/2x = 3/4 +4/4`
`=> 6/2x = 7/4`
`=> x= 7/4 : 3`
`=>x=7/12`
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`(x-1/2) -1/4=0`
`=> x-1/2=1/4`
`=> x=1/4 +1/2`
`=> x= 1/4 +2/4`
`=>x=3/4`
__
`4^(x-3) +1=17`
`=> 4^(x-3) =17-1`
`=> 4^(x-3)=16`
`=> 4^(x-3)=4^2`
`=> x-3=2`
`=>x=2+3`
`=>x=5`