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AH
Akai Haruma
Giáo viên
1 tháng 5 2023

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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$\frac{2x}{3}=\frac{6}{x}$ ($x\neq 0$)

$\Rightarrow 2x^2=18$

$x^2=18:2=9=(-3)^2=3^2$

$\Rightarrow x=\pm 3$

---------------------

$\frac{x-1}{14}=\frac{4}{x}$ ($x\neq 0$)

$\Rightarrow x(x-1)=14.4=56$

$x^2-x-56=0$

$(x+7)(x-8)=0$

$\Rightarrow x+7=0$ hoặc $x-8=0$

$\Leftrightarrow x=-7$ hoặc $x=8$

---------------------------

$\frac{-x}{8}=\frac{-50}{x}$ ($x\neq 0$)

$\Rightarrow -x^2=8(-50)$

$x^2=400=20^2=(-20)^2$

$\Rightarrow x=\pm 20$

4 tháng 1 2022

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) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)

\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)

 

 

19 tháng 2 2022

\(\dfrac{5}{7}-x=\dfrac{9}{21}\\ \Rightarrow\dfrac{5}{7}-x=\dfrac{3}{7}\\ \Rightarrow x=\dfrac{5}{7}-\dfrac{3}{7}\\ \Rightarrow x=\dfrac{2}{7}\\ b,-x-\dfrac{1}{3}=\dfrac{2}{6}\\ \Rightarrow-x=\dfrac{2}{6}+\dfrac{1}{3}\\ \Rightarrow-x=\dfrac{2}{3}\\ \Rightarrow x=-\dfrac{2}{3}\\ \dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\\ \Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\\ \Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\\ \Rightarrow x=-\dfrac{13}{12}\)

26 tháng 4 2022

a. 5 - 3(x + 4) = -1

⇔ 5 - 3x - 12 = -1

⇔ 3x = -1 - 5 + 12

⇔ 3x = 6

⇔ x = 2

26 tháng 4 2022

\(d,2x^2-3=5\)

\(\Leftrightarrow2x^2=8\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x=\pm2\)

\(e,x\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)

19 tháng 5 2022

tách đi bạn

19 tháng 5 2022

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) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)

 

AH
Akai Haruma
Giáo viên
10 tháng 3 2023

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 \(x=\dfrac{2}{3}\)

20 tháng 5 2021

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 

22 tháng 1 2022

a) x= 2, x= 8.(6 : 3) = 16, x= 1. (27 : 9)= 3

b) x= 6 : (8 : 4) = 3, x= -1, x= -2 . -8 = x.x => 16 = x2 => 42 = x2 => x=4

        Tick cho mình đi ok

7 tháng 4 2022

a)\(x=\left(\dfrac{3}{56}\cdot\dfrac{28}{9}\right):\dfrac{-3}{7}=\dfrac{1}{6}:\dfrac{-3}{7}=-\dfrac{7}{18}\)

b)\(x=\left(\dfrac{7}{15}\cdot\dfrac{5}{3}\right)+\dfrac{3}{16}=\dfrac{7}{9}+\dfrac{3}{16}=\dfrac{139}{144}\)

7 tháng 4 2022

c)\(x=\left(\dfrac{5}{6}-\dfrac{2}{5}\right).5=\dfrac{13}{6}\)

d)\(=>x\left(\dfrac{3}{4}-\dfrac{2}{5}\right)=\dfrac{1}{6}\cdot\left(\dfrac{3}{7}+\dfrac{5}{7}\right)\)

\(x\cdot\dfrac{7}{20}=\dfrac{4}{21}=>x=\dfrac{4}{21}\cdot\dfrac{20}{7}=\dfrac{80}{147}\)