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Lời giải:
$\frac{x}{3}-\frac{1}{4}=\frac{-5}{6}$
$\frac{x}{3}=\frac{1}{4}-\frac{5}{6}=\frac{-7}{12}$
$x=\frac{-7}{12}.3=\frac{-7}{4}$
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$\frac{2x}{3}=\frac{6}{x}$ ($x\neq 0$)
$\Rightarrow 2x^2=18$
$x^2=18:2=9=(-3)^2=3^2$
$\Rightarrow x=\pm 3$
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$\frac{x-1}{14}=\frac{4}{x}$ ($x\neq 0$)
$\Rightarrow x(x-1)=14.4=56$
$x^2-x-56=0$
$(x+7)(x-8)=0$
$\Rightarrow x+7=0$ hoặc $x-8=0$
$\Leftrightarrow x=-7$ hoặc $x=8$
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$\frac{-x}{8}=\frac{-50}{x}$ ($x\neq 0$)
$\Rightarrow -x^2=8(-50)$
$x^2=400=20^2=(-20)^2$
$\Rightarrow x=\pm 20$
a) \(\dfrac{5}{x}=\dfrac{-10}{12}.\Rightarrow x=-6.\)
b) \(\dfrac{4}{-6}=\dfrac{x+3}{9}.\Rightarrow x+3=-6.\Leftrightarrow x=-9.\)
c) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)
\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)
\(\dfrac{5}{7}-x=\dfrac{9}{21}\\ \Rightarrow\dfrac{5}{7}-x=\dfrac{3}{7}\\ \Rightarrow x=\dfrac{5}{7}-\dfrac{3}{7}\\ \Rightarrow x=\dfrac{2}{7}\\ b,-x-\dfrac{1}{3}=\dfrac{2}{6}\\ \Rightarrow-x=\dfrac{2}{6}+\dfrac{1}{3}\\ \Rightarrow-x=\dfrac{2}{3}\\ \Rightarrow x=-\dfrac{2}{3}\\ \dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\\ \Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\\ \Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\\ \Rightarrow x=-\dfrac{13}{12}\)
a. 5 - 3(x + 4) = -1
⇔ 5 - 3x - 12 = -1
⇔ 3x = -1 - 5 + 12
⇔ 3x = 6
⇔ x = 2
\(d,2x^2-3=5\)
\(\Leftrightarrow2x^2=8\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow x=\pm2\)
\(e,x\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)
a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
Lời giải:
a.
$\frac{2}{3}x-\frac{7}{6}=\frac{12}{7}-\frac{1}{2}=\frac{17}{14}$
$\frac{2}{3}x=\frac{17}{14}+\frac{7}{6}=\frac{50}{21}$
$x=\frac{50}{21}: \frac{2}{3}=\frac{25}{7}$
b.
$(1\frac{1}{2}+\frac{5}{3}-\frac{1}{6}):x=\frac{3}{4}-\frac{1}{2}$
$3:x=\frac{1}{4}$
$x=3: \frac{1}{4}=12$
c: Ta có: \(\dfrac{1}{3}-\dfrac{7}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow x\cdot\dfrac{7}{8}=\dfrac{1}{12}\)
\(\Leftrightarrow x=\dfrac{1}{12}\cdot\dfrac{8}{7}=\dfrac{2}{21}\)
d: Ta có: \(\dfrac{3}{2}x+\dfrac{1}{7}=\dfrac{7}{8}\cdot\dfrac{64}{49}\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}=1\)
hay \(x=\dfrac{2}{3}\)
a,2/5 = 2/5 ; 3/8=6/16 ; 1/9=3/27
b, 4/3=8/6 ; -1=-1 ; -4/-2=-8/4
tick cho mik nhé
a)\(x=\left(\dfrac{3}{56}\cdot\dfrac{28}{9}\right):\dfrac{-3}{7}=\dfrac{1}{6}:\dfrac{-3}{7}=-\dfrac{7}{18}\)
b)\(x=\left(\dfrac{7}{15}\cdot\dfrac{5}{3}\right)+\dfrac{3}{16}=\dfrac{7}{9}+\dfrac{3}{16}=\dfrac{139}{144}\)