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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
10 tháng 3 2022
1: =>x=3/5-1/5=2/5
b: =>x/3=5/8+1/8=3/4
=>x=9/4
3: =>10/3x=3+1/4+6+3/4=10
=>x=10:10/3=3
AH
Akai Haruma
Giáo viên
1 tháng 5 2023
Lời giải:
$\frac{x}{3}-\frac{1}{4}=\frac{-5}{6}$
$\frac{x}{3}=\frac{1}{4}-\frac{5}{6}=\frac{-7}{12}$
$x=\frac{-7}{12}.3=\frac{-7}{4}$
-------------------
$\frac{2x}{3}=\frac{6}{x}$ ($x\neq 0$)
$\Rightarrow 2x^2=18$
$x^2=18:2=9=(-3)^2=3^2$
$\Rightarrow x=\pm 3$
---------------------
$\frac{x-1}{14}=\frac{4}{x}$ ($x\neq 0$)
$\Rightarrow x(x-1)=14.4=56$
$x^2-x-56=0$
$(x+7)(x-8)=0$
$\Rightarrow x+7=0$ hoặc $x-8=0$
$\Leftrightarrow x=-7$ hoặc $x=8$
---------------------------
$\frac{-x}{8}=\frac{-50}{x}$ ($x\neq 0$)
$\Rightarrow -x^2=8(-50)$
$x^2=400=20^2=(-20)^2$
$\Rightarrow x=\pm 20$
OY
1 tháng 8 2021
1/ \(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
=\(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).0\)
=\(0\)
29 tháng 6 2023
a: =>6/x=x/24
=>x^2=144
=>x=12 hoặc x=-12
b: =>x(1-7/12+3/8)=5/24
=>x*19/24=5/24
=>x=5/24:19/24=5/19
c: =>(x-1/3)^2=1+3/4+1/2=9/4
=>x-1/3=3/2 hoặc x-1/3=-3/2
=>x=11/6 hoặc x=-7/6
d: =>(x-3)^2=16
=>x-3=4 hoặc x-3=-4
=>x=-1 hoặc x=7
e: =>9/x=-1/3
=>x=-27
f: =>x-1/2=0 hoặc -x/2-3=0
=>x=1/2 hoặc x=-6
HP
8 tháng 3 2022
\(a.x+\dfrac{1}{6}=-\dfrac{3}{8}\)
\(\Leftrightarrow x=-\dfrac{13}{24}\)
\(b.2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)
\(\Leftrightarrow2-\dfrac{3}{4}+x=\dfrac{7}{12}\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)
\(c.\dfrac{1}{2}x+\dfrac{1}{8}x=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{5}{8}x=\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{6}{5}\)
\(d.75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}-\left(\dfrac{-1}{2}\right)^2\)
\(=\dfrac{75}{100}-\dfrac{3}{2}+\dfrac{1}{2}:\dfrac{5}{12}-\dfrac{1}{4}\)
\(=-\dfrac{3}{4}+\dfrac{6}{5}-\dfrac{1}{4}\)
\(=\dfrac{1}{5}\)
DC
8 tháng 3 2022
a) \(x+\dfrac{1}{6}=\dfrac{-3}{8}\)
\(x=\dfrac{-3}{8}-\dfrac{1}{6}\)
\(x=\dfrac{-13}{24}\)
vậy x =....
b) \(2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)
\(\dfrac{3}{4}-x=2-\dfrac{7}{12}\)
\(\dfrac{3}{4}-x=\dfrac{17}{12}\)
\(x=\dfrac{3}{4}-\dfrac{17}{12}\)
\(x=\dfrac{-2}{3}\)
vậy x =....
4 tháng 9 2021
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 \(x=\dfrac{2}{3}\)
O
18 tháng 4 2022
a) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)
Th1 : \(x-\dfrac{1}{2}=0\)
\(x=0+\dfrac{1}{2}\)
\(x=\dfrac{1}{2}\)
Th2 : \(-3-\dfrac{x}{2}=0\)
\(\dfrac{x}{2}=-3\)
\(x=\left(-3\right)\cdot2\)
\(x=-6\)
Vậy \(x\) = \(\left(\dfrac{1}{2};-6\right)\)
b) \(x-\dfrac{1}{8}=\dfrac{5}{8}\)
\(x=\dfrac{5}{8}+\dfrac{1}{8}\)
\(x=\dfrac{3}{4}\)
c) \(-\dfrac{1}{2}-\left(\dfrac{3}{2}+x\right)=-2\)
\(\dfrac{3}{2}+x=-\dfrac{1}{2}-\left(-2\right)\)
\(\dfrac{3}{2}+x=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}-\dfrac{3}{2}\)
\(x=0\)
d) \(x+\dfrac{1}{3}=\dfrac{-12}{5}\cdot\dfrac{10}{6}\)
\(x+\dfrac{1}{3}=-4\)
\(x=-4-\dfrac{1}{3}\)
\(x=-\dfrac{13}{3}\)
O
5 tháng 3 2022
\(x-\dfrac{1}{2}=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}+\dfrac{1}{2}\)
\(x=\dfrac{5}{4}\)
\(x+\dfrac{7}{8}=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}-\dfrac{7}{8}\)
\(x=\dfrac{-1}{8}\)
\(\dfrac{1}{2}\cdot x-\dfrac{1}{4}=\dfrac{-1}{2}\)
\(\dfrac{1}{2}\cdot x=\dfrac{-1}{2}+\dfrac{1}{4}\)
\(\dfrac{1}{2}\cdot x=\dfrac{-1}{4}\)
\(x=\dfrac{-1}{4}\div\dfrac{1}{2}\)
\(x=\dfrac{-1}{2}\)
Câu D ko bt
T
27 tháng 9 2023
\(a,x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{\left(-1\right)}{3}\)
\(x+\dfrac{1}{3}=\dfrac{2}{5}+\dfrac{1}{3}\)
\(x=\dfrac{2}{5}+\dfrac{1}{3}-\dfrac{1}{3}\)
\(x=\dfrac{2}{5}\)
\(---\)
\(b,\dfrac{5}{8}-x=\dfrac{-3}{20}-\dfrac{\left(-1\right)}{6}\)
\(\dfrac{5}{8}-x=\dfrac{-3}{20}+\dfrac{1}{6}\)
\(\dfrac{5}{8}-x=\dfrac{-9}{60}+\dfrac{10}{60}\)
\(\dfrac{5}{8}-x=\dfrac{1}{60}\)
\(x=\dfrac{5}{8}-\dfrac{1}{60}\)
\(x=\dfrac{75}{120}-\dfrac{2}{120}\)
\(x=\dfrac{73}{120}\)
#\(Toru\)
`@` `\text {Ans}`
`\downarrow`
\(x+\dfrac{1}{6}=\dfrac{-3}{8}\)
`=>`\(x=\dfrac{-3}{8}-\dfrac{1}{6}\)
`=>`\(x=-\dfrac{13}{24}\)
Vậy, `x =`\(-\dfrac{13}{24}\)
\(x+\dfrac{1}{6}=-\dfrac{3}{8}\)
\(\Rightarrow x=\dfrac{-3}{8}-\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{13}{24}\)