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a) \(-2\left(x-\dfrac{1}{3}\right)-5\left(x+\dfrac{1}{3}\right)=\dfrac{1}{2}x\)
\(\Leftrightarrow-2x+\dfrac{2}{3}-5x-\dfrac{5}{3}=\dfrac{1}{2}x\)
\(\Leftrightarrow-7x-1=\dfrac{1}{2}x\)
\(\Leftrightarrow-7x-\dfrac{1}{2}x=1\)
\(\Leftrightarrow-\dfrac{15}{2}x=1\)
\(\Leftrightarrow x=1:\left(-\dfrac{15}{2}\right)=-\dfrac{2}{15}\)
b) \(-\left(\dfrac{1}{2}x-\dfrac{3}{4}\right)-\left(-x+1\right)=\dfrac{3}{2}\)
\(\Leftrightarrow-\dfrac{1}{2}x+\dfrac{3}{4}+x-1=\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{1}{2}x-\dfrac{1}{4}=\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{1}{2}x=\dfrac{3}{2}+\dfrac{1}{4}=\dfrac{7}{4}\)
\(\Leftrightarrow x=\dfrac{7}{4}:\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{7}{2}\)
a, -5/7+ 1+ 30/-7< x < -1/6+ 1/3 +5/6
<=> -4< x <1
<=> x = -3; -2; -1; 0
a, \(\dfrac{-5}{7}+1+\dfrac{30}{-7}\le x\le\dfrac{-1}{6}+\dfrac{1}{3}+\dfrac{5}{6}\)
<=> -4 \(\le x\le1\)
Do x \(\in Z\Rightarrow x=-4;-3;-2;-1;0;1\)
b, \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)
<=> -\(\dfrac{1}{12}< x< \dfrac{1}{8}\)
Do x \(\in Z\Rightarrow x=0;1\)
@Mai Tran
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{38.39}=\dfrac{5}{3}-x\)
=> \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+......+\dfrac{1}{38}-\dfrac{1}{39}=\dfrac{5}{3}-x\)
=> \(1-\dfrac{1}{39}=\dfrac{5}{3}-x\)
=> \(\dfrac{39}{39}-\dfrac{1}{39}=\dfrac{65}{39}-\dfrac{39x}{39}\)
=> 39-1=65-39x
=> 38=65-39x
=> 39x=-38+65
=> 39x=27
=> x=\(\dfrac{9}{13}\)
vậy x= \(\dfrac{9}{13}\)
a: \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
nên \(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{6}{10}=\dfrac{3}{5}\)
hay \(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b: \(\Leftrightarrow5-\dfrac{4}{7}x=13\)
nên 4/7x=-8
hay x=-12
c: \(\left(x+\dfrac{1}{2}\right)\cdot\left(\dfrac{2}{3}-2x\right)=0\)
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
d: \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
nên 1/6x=5/12
hay x=5/2
a) \(\left(\dfrac{1}{2}x-3\right)\left(-\dfrac{1}{3}+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=0\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=0+3\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3:\dfrac{1}{2}\\x=0-\left(-\dfrac{1}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{1}{3}\end{matrix}\right.\)
d) \(9x^2=1\)
\(\Leftrightarrow x^2=1:9\)
\(\Leftrightarrow x^2=\dfrac{1}{9}\)
\(\Leftrightarrow x^2=\left(\dfrac{1}{3}\right)^2\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
b1:
\(-5.\left(x-\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\Leftrightarrow-5x+1-\frac{1}{2}x+\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(\Leftrightarrow-\frac{11}{2}x+\frac{4}{3}=\frac{3}{2}x-\frac{5}{6}\Leftrightarrow-\frac{11}{2}x-\frac{3}{2}x=-\frac{5}{6}-\frac{4}{3}\Leftrightarrow-7x=-\frac{13}{6}\)
\(\Leftrightarrow x=\frac{-13}{6}:\left(-7\right)=\frac{13}{42}\)
b2:
a)\(3^{3x-1}=9^{x-2}\Leftrightarrow3^{3x-1}=\left(3^2\right)^{x-2}\Leftrightarrow3^{3x-1}=3^{2x-4}\Leftrightarrow3x-1=2x-4\)
<=>3x-2x=-4+1<=>x=-3
b)\(\left(x-1\right)^4=\left(x-1\right)^6\Leftrightarrow\left(x-1\right)^4-\left(x-1\right)^6=0\Leftrightarrow\left(x-1\right)^4\left[1-\left(x-1\right)^2\right]=0\)
\(\Leftrightarrow\left(x-1\right)^4\left[1-\left(x^2-2x+1\right)\right]=0\Leftrightarrow\left(x-1\right)^4\left(1-x^2+2x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^4\left(-x^2+2x\right)=0\Leftrightarrow\left(x-1\right)^4.\left(-x\right).\left(x-2\right)=0\)
<=>(x-1)4=0 hoặc -x=0 hoặc x-2=0 <=> x=1 hoặc x=0 hoặc x=2