\(\dfrac{35}{-30}-\dfrac{78}{30}\)
K
Khách
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NA
21 tháng 1 2022
\(a,\dfrac{20}{30}=\dfrac{30}{45}v\text{ì}20.45=30.30=900\\ b,\dfrac{-25}{35}=\dfrac{-55}{77}v\text{ì}:\left(-25\right).77=35.\left(-55\right)=-1925\)
11 tháng 2 2022
\(E=\dfrac{-1}{3}-\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{3}{5}+\dfrac{1}{5}=-1\)
31 tháng 8 2021
a: \(\dfrac{20}{30}=\dfrac{2}{3}\)
\(\dfrac{30}{45}=\dfrac{2}{3}\)
Do đó: \(\dfrac{20}{30}=\dfrac{30}{45}\)
b: \(\dfrac{-25}{35}=\dfrac{-5}{7}\)
\(\dfrac{-55}{77}=\dfrac{-5}{7}\)
Do đó: \(-\dfrac{25}{35}=-\dfrac{55}{77}\)
31 tháng 8 2021
a,20/30=2/3
30/45=2/3
->20/30=30/45
b -25/35=-5/7
-55/77=-5/7
-> -25/35=-55/77
26 tháng 1 2023
a: 20/30=2/3
30/45=2/3
=>20/30=30/45
b: -25/35=-5/7
-55/77=-5/7
=>-25/35=-55/77
24 tháng 2 2021
\(\dfrac{3}{6}=\dfrac{3:3}{6:3}=\dfrac{1}{2}\\ \dfrac{18}{24}=\dfrac{18:6}{24:6}=\dfrac{3}{4}\\ \dfrac{5}{35}=\dfrac{5:5}{35:5}=\dfrac{1}{7}\\ \dfrac{40}{90}=\dfrac{40:10}{90:10}=\dfrac{4}{9}\\ \dfrac{75}{30}=\dfrac{75:15}{30:15}=\dfrac{5}{2}\)
24 tháng 2 2021
\(\dfrac{3}{6}=\dfrac{3:3}{6:3}=\dfrac{1}{2}\\ \dfrac{18}{24}=\dfrac{18:6}{24:6}=\dfrac{3}{4}\\ \dfrac{5}{35}=\dfrac{5:5}{35:5}=\dfrac{1}{7}\\ \dfrac{40}{90}=\dfrac{40:10}{90:10}=\dfrac{4}{9}\\ \dfrac{75}{30}=\dfrac{75:15}{30:15}=\dfrac{5}{2}\)
3 tháng 2 2021
\(\dfrac{x-130}{20}\)+\(\dfrac{x-100}{25}\)+\(\dfrac{x-60}{30}\)+\(\dfrac{x-10}{35}\)=10
⇔\(\dfrac{2625\left(x-130\right)}{52500}\)+\(\dfrac{2100\left(x-100\right)}{52500}\)+\(\dfrac{1750\left(x-60\right)}{52500}\)+\(\dfrac{1500\left(x-10\right)}{52500}\)=\(\dfrac{525000}{52500}\)
⇔2625\(x\)-341250+2100\(x\)-210000+1750\(x\)-105000+1500\(x\)-15000=525000
⇔ 7975\(x\) = 1196250
⇔ \(x\) = \(\dfrac{1196250}{7975}\)
⇔\(x \) = 150
AH
Akai Haruma
Giáo viên
23 tháng 8 2023
Lời giải:
PT $\Leftrightarrow \frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+35}{65}+1+\frac{x+40}{60}+1$
$\Leftrightarrow \frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}$
$\Leftrightarrow (x+100)(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60})=0$
Dễ thấy $\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}<0$
$\Rightarrow x+100=0$
$\Leftrightarrow x=-100$ (tm)
G
3
D
8 tháng 2 2023
`[x+35]/1984-[x+30]/1989+[x+19]/2000+[x+23]/[1996=-2`
`<=>[x+35]/1984+1-[x+30]/1989-1+[x+19]/2000+1+[x+23]/1996+1=0`
`<=>[x+2019]/1984-[x+2019]/1989+[x+2019]/2000+[x+2019]/1996=0`
`<=>(x+2019)(1/1984-1/1989+1/2000+1/1996)=0`
`=>x+2019=0`
`<=>x=-2019`
NT
Nguyễn thành Đạt
CTVHS
8 tháng 2 2023
\(\dfrac{x+35}{1984}-\dfrac{x+30}{1989}+\dfrac{x+19}{2000}+\dfrac{x+23}{1996}\text{=}-2\)
\(\Leftrightarrow\dfrac{x+35}{1984}-\dfrac{x+30}{1989}+\dfrac{x+19}{2000}+\dfrac{x+23}{1996}+3-1\text{=}0\)
\(\Leftrightarrow\left(\dfrac{x+35}{1984}+1\right)-\left(\dfrac{x+30}{1989}+1\right)+\left(\dfrac{x+19}{2000}+1\right)+\left(\dfrac{x+23}{1996}+1\right)\text{=}0\)
\(\Leftrightarrow\dfrac{x+2019}{1984}-\dfrac{x+2019}{1989}+\dfrac{x+2019}{2000}+\dfrac{x+2019}{1996}\text{=}0\)
\(\Leftrightarrow\left(x+2019\right)\left(\dfrac{1}{1984}-\dfrac{1}{1989}+\dfrac{1}{2000}+\dfrac{1}{1996}\right)\text{=}0\)
\(\Leftrightarrow\left(x+2019\right)\text{=}0\)
\(\Leftrightarrow x\text{=}-2019\)
\(\dfrac{35}{-30}-\dfrac{78}{30}\)
\(=\dfrac{-35}{30}-\dfrac{78}{30}\)
\(=\dfrac{-35-78}{30}=\dfrac{-113}{30}\)
\(\dfrac{35}{-30}-\dfrac{78}{30}=-\dfrac{113}{30}\)