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\(\frac{1}{3}x+\frac{2}{5}=\frac{6}{5}+\frac{1}{2}x\)
\(\frac{1}{3}x-\frac{1}{2}x=\frac{6}{5}-\frac{2}{5}\)
\(-\frac{1}{6}x=\frac{4}{5}\)
\(x=\frac{4}{5}:\left(-\frac{1}{6}\right)\)
\(x=-\frac{24}{5}\)
Ta có: \(\frac{1}{3}x+\frac{2}{5}=\frac{6}{5}+\frac{1}{2}x\)
\(\Leftrightarrow\frac{1}{2}x-\frac{1}{3}x=\frac{2}{5}-\frac{6}{5}\)
\(\Leftrightarrow\frac{1}{6}x=-\frac{4}{5}\)
\(\Rightarrow x=-\frac{24}{5}\)
\(\frac{2}{5}+\frac{3}{5}.\left(3x-3\right)=\frac{-5}{10}\)
\(\frac{2}{5}+\frac{3}{5}.3.\left(x-1\right)=\frac{-1}{2}\)
\(\frac{9}{5}.\left(x-1\right)=\frac{-1}{2}-\frac{2}{5}\)
\(\frac{9}{5}.\left(x-1\right)=\frac{-9}{10}\)
\(x-1=\frac{-9}{10}:\frac{9}{5}\)
\(x-1=\frac{-1}{2}\)
\(x=\frac{-1}{2}+1\)
\(x=\frac{1}{2}\)
\(\frac{2}{5}+\frac{3}{5}.\left(3x-3\right)=-\frac{5}{10}\)
\(\frac{3}{5}\left(3x-3\right)=-\frac{5}{10}-\frac{2}{5}\)
\(\frac{3}{5}\left(3x-3\right)=-\frac{9}{10}\)
\(\left(3x-3\right)=-\frac{9}{10}:\frac{3}{5}\)
\(3\left(x-1\right)=-\frac{3}{2}\)
\(x-1=-\frac{1}{2}\)
\(x=\frac{1}{2}\)
Vậy...
=6/15x(-9/6-20/6)
=6/15x[-(9/6+20/6)]
=6/15x(-29/6)
=-174/90
=-29/15
=>|3x-2/5|=1/35+90/35=91/35
=>3x-2/5=91/35 hoặc 3x-2/5=-91/35
=>3x-2/5=13/5 hoặc 3x-2/5=-13/5
=>3x=15/5=3 hoặc 3x=-11/5
=>x=-11/5 hoặc x=1
Lời giải:
$|3x-\frac{2}{5}|=\frac{1}{35}+\frac{18}{7}=\frac{13}{5}$
$\Rightarrow 3x-\frac{2}{5}=\frac{13}{5}$ hoặc $3x-\frac{2}{5}=\frac{-13}{5}$
$\Rightarrow 3x=3$ hoặc $3x=\frac{-11}{5}$
$\Rightarrow x=1$ hoặc $x=\frac{-11}{15}$
a, \(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)
\(\Leftrightarrow\left(x+1\right)\left(3x-5-3x+1\right)=x-4\Leftrightarrow-4\left(x+1\right)=x-4\)
\(\Leftrightarrow-4x-4=x-4\Leftrightarrow-4x-x=0\Leftrightarrow x=0\)
b, \(\left(x-2\right)\left(x+3\right)-\left(x+4\right)\left(x-7\right)=5-x\)
\(\Leftrightarrow x^2+x-6-x^2-3x+28=5-x\Leftrightarrow-2x+22=5-x\Leftrightarrow x=17\)
c, thiếu đề
d, \(3\left(x-7\right)\left(x+7\right)-\left(x-1\right)\left(3x+2\right)=13\)
\(\Leftrightarrow3x^2-147-3x^2+x+2=13\Leftrightarrow x=11+147=158\)
a.\(3x^2-2x-5-\left(3x^2+2x-1\right)=x-4\)
\(\Leftrightarrow-5x=0\Leftrightarrow x=0\)
b.\(x^2+x-6-\left(x^2-3x-28\right)=5-x\)
\(\Leftrightarrow5x=-17\Leftrightarrow x=-\frac{17}{5}\)
c.\(5\left(x^2-10x+21\right)-\left(5x^2-9x-2\right)=0\)
\(\Leftrightarrow-41x+107=0\Leftrightarrow x=\frac{107}{41}\)
d.\(3\left(x^2-49\right)-\left(3x^2-x-2\right)=13\Leftrightarrow x=158\)
à à mình nhân nhầm đấy, do công thức bị lỗi
dòng thứ 2 phải là : \(\left(3x-5\right)^2=-36\)( vô lí )
vì \(\left(3x-5\right)^2\ge0\forall x;-36< 0\)
\(\frac{3x-5}{18}=-\frac{2}{3x-5}\)
\(\Rightarrow\left(3x-5\right)^2=-16\)( vô lí )
vì \(\left(3x-5\right)^2\ge0\forall x;-16< 0\)