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Ta có : \(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)
\(\Rightarrow\frac{3}{2x+1}+\frac{5.2}{2\left(2x+1\right)}-\frac{3.2}{3\left(2x+1\right)}=\frac{6}{13}\)
=> \(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)
=> \(\frac{3+5-2}{2x+1}=\frac{6}{13}\)
=> \(\frac{6}{2x+1}=\frac{6}{13}\)
=> 2x + 1 = 13
=> 2x = 12
=> x = 6
Vậy x = 6
\(\frac{3}{2x+1}+\frac{10}{2\left(2x+1\right)}-\frac{6}{3\left(2x+1\right)}=\frac{6}{13}\)
\(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)
\(\frac{6}{2x+1}=\frac{6}{13}\)
\(\Rightarrow2x+1=13\left(6=6\right)\)
\(2x=12\)
\(x=6\)
\(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)
\(\frac{3}{2x+1}+\frac{2.5}{2\left(2x+1\right)}-\frac{2.3}{3\left(2x+1\right)}=\frac{6}{13}\)
\(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)
\(\frac{3+5-2}{2x+1}=\frac{6}{13}\)
\(\frac{6}{2x+1}=\frac{6}{13}\)
\(2x+1=13\)
\(\Rightarrow x=6\)
\(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)
\(\frac{3}{2x+1}+\frac{2.5}{2\left(2x+1\right)}-\frac{3.2}{3\left(2x+1\right)}=\frac{6}{13}\)
\(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)
\(\frac{6}{2x+1}=\frac{6}{13}\)
\(\Rightarrow2x+1=13\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\)
\(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{6}{13}\)
\(\Rightarrow\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)
\(\Rightarrow\frac{6}{2x+1}=\frac{6}{13}\Rightarrow2x+1=13\Rightarrow x=6\)
mình giải hơi gọn có gì ko hiểu thì hỏi nha !
d,
\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)
e,
\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)
\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)
\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)
Vậy không tồn tại $x$ thỏa mãn đề bài.
f,
\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)
\(\Leftrightarrow 6x-3=10+6x\)
\(\Leftrightarrow 13=0\) (vô lý)
Vậy không tồn tại $x$ thỏa mãn đề bài.
a,
$0-|x+1|=5$
$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)
Do đó không tồn tại $x$ thỏa mãn điều kiện đề.
b,
\(2-|\frac{3}{4}-x|=\frac{7}{12}\)
\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)
c,
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)
\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)
a) \(\frac{-x}{2}+\frac{2x}{3}+x+\frac{1}{4}+2x+\frac{1}{6}=\frac{3}{8}.\)
\(\frac{-x}{2}+\frac{2x}{3}+3x+\frac{5}{12}=\frac{3}{8}\)
\(x.\left(-\frac{1}{2}+\frac{2}{3}+3\right)+\frac{5}{12}=\frac{3}{8}\)
\(x\cdot\frac{19}{6}=-\frac{1}{24}\)
x = -1/76
b) \(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)
\(\frac{3}{2x+1}+\frac{2.5}{2.\left(2x+1\right)}-\frac{2.3}{3.\left(2x+1\right)}=\frac{6}{13}\)
\(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)
\(\frac{3+5-2}{2x+1}=\frac{6}{13}\)
\(\frac{6}{2x+1}=\frac{6}{13}\)
=> 2x + 1 = 13
2x = 12
x = 6
\(\frac{3}{2x+1}\)+ \(\frac{10}{4x+2}\) - \(\frac{6}{6x+3}\) = \(\frac{12}{26}\)
\(\dfrac{3}{2x+1}\) + \(\dfrac{2.5}{2\left(2x+1\right)}\) - \(\dfrac{2.3}{3\left(2x+1\right)}\) = \(\dfrac{6}{13}\)
\(\dfrac{3}{2x+1}\) + \(\dfrac{5}{2x+1}\) - \(\dfrac{2}{2x+1}\) = \(\dfrac{6}{13}\)
\(\dfrac{3}{2x+1}\) + \(\dfrac{5}{2x+1}\) + \(\dfrac{-2}{2x+1}\) = \(\dfrac{6}{13}\)
\(\dfrac{6}{2x+1}\) = \(\dfrac{6}{13}\)
\(\Rightarrow\) (2x+1).6 = 6.13 2x + 1 = \(\dfrac{6.13}{6}\) 2x + 1 = 13 2x = 13 - 1 2x = 12 x = 12:2 x = 6 Vậy x = 6Ta có: \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}\)= \(\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}\)
= \(\dfrac{3+5-2}{2x+1}=\dfrac{6}{2x+1}=\dfrac{12}{26}\) \(\Rightarrow156=24x+12\Rightarrow24x=144\Rightarrow x=6\)
Vậy x=6.
Học tốt nha