\(\dfrac{25}{75};\dfrac{17}{34};\dfrac{121}{132}\)
rút gọn và quy đồng giùm tui nhé thank nhiều
\(\dfrac{38}{75}\)÷\(\dfrac{19}{25}\)+\(\dfrac{81}{25}\)+ 4.\(\dfrac{2}{7}\)
`35/75 : 19/25 + 81/25 + 4.2/7`
=`38/75 xx 25/18 + 81/25 + 8/7`
=` 2/3 + 81/25 + 8/7`
= `293/75 + 8/7`
= `2651/525`
\(\dfrac{75}{100}\)-\(\dfrac{17}{25}\)+\(\dfrac{13}{37}\)+\(\dfrac{1}{4}\)-\(\dfrac{8}{25}\)+\(\dfrac{24}{37}\)
\(=\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(-\dfrac{17}{25}-\dfrac{8}{25}\right)+\left(\dfrac{13}{37}+\dfrac{24}{37}\right)=1\)
6. ÉT O ÉT
\(\dfrac{125}{90}=\dfrac{25}{...}\) / \(\dfrac{84}{91}=\dfrac{...}{13}\) / \(\dfrac{75}{...}=\dfrac{25}{15}=\dfrac{...}{3}\)
\(\dfrac{6}{7}=\dfrac{...}{21}=\dfrac{54}{...}=\dfrac{...}{126}\)
125/90=25/18
84/91=12/13
75/45=25/15=5/3
125/90=25/18
84/91=12/13
75/45=25/15=5/3
6/7=18/21=54/63=104/126
125/90 = 25/18
84/91 = 12/13
75/45 = 25/15 = 5/3
6/7 = 18/21 = 54/63 = 108/126
1.Tính
\(\dfrac{38}{75}:\dfrac{19}{25}\)
\(\dfrac{38}{75}:\dfrac{19}{25}=\dfrac{38}{75}\times\dfrac{25}{19}=\dfrac{2}{3}\)
\(\left|\dfrac{-25}{10}\right|\) x 0,2 + 75% : \((\dfrac{5}{2^{ }})^2\)
`|-25/10|×0,2+75%:(5/2)^2`
`=25/10×1/5+3/4:25/4`
`=5/10+3/4×4/25`
`=1/2+3/25`
`=25/50+6/50`
`=31/50`
giải phương trình sau:
\(\dfrac{x+25}{75}+\dfrac{x+30}{70}=\dfrac{x+35}{65}+\dfrac{x+40}{60}\)
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)
1/ (\(\left(-\dfrac{2}{3}\right)\)\(^2\) x \(\dfrac{-9}{8}\) - 25% x \(\dfrac{-16}{5}\)
2/ -1\(\dfrac{2}{5}\) x 75% + \(\dfrac{-7}{5}\) x 25%
3/ -2\(\dfrac{3}{7}\) x (-125%) + \(\dfrac{-17}{7}\) x 25%
4/ (-2)\(^3\) x (\(\dfrac{3}{4}\) x 0.25) : (2\(\dfrac{1}{4}\) - 1\(\dfrac{1}{6}\))
1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)
\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)
\(=\dfrac{-1}{2}+\dfrac{4}{5}\)
\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)
2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)
\(=\dfrac{-7}{5}\cdot\dfrac{3}{4}+\dfrac{-7}{5}\cdot\dfrac{1}{4}\)
\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)
3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)
\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)
\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)
\(=\dfrac{17}{7}\)
4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)
\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)
\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)
K = \(\left(3\dfrac{1}{3}.1,9+9,5:4\dfrac{1}{3}\right).\left(\dfrac{62}{75}-\dfrac{4}{25}\right)\)
\(K=\left(\dfrac{10}{3}\cdot\dfrac{19}{10}+\dfrac{19}{2}:\dfrac{13}{3}\right)\cdot\dfrac{2}{3}\)
\(=\dfrac{665}{117}\)
\(K=\left(\dfrac{10}{3}\cdot\dfrac{19}{10}+\dfrac{19}{2}:\dfrac{13}{3}\right)\cdot\left(\dfrac{62}{75}-\dfrac{12}{25}\right)=\left(\dfrac{19}{3}+\dfrac{57}{26}\right)\cdot\dfrac{2}{3}=\dfrac{665}{78}\cdot\dfrac{2}{3}=\dfrac{665}{117}\)
giải phương trình sau
\(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=0\)
Sửa đề: \(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=-5\)Ta có: \(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=-5\)
\(\Leftrightarrow\dfrac{74-x}{26}+1+\dfrac{75-x}{25}+1+\dfrac{76-x}{24}+1+\dfrac{77-x}{23}+1+\dfrac{78-x}{22}+1=0\)
\(\Leftrightarrow\dfrac{100-x}{26}+\dfrac{100-x}{25}+\dfrac{100-x}{24}+\dfrac{100-x}{23}+\dfrac{100-x}{22}=0\)
\(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}\right)=0\)
mà \(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}>0\)
nên 100-x=0
hay x=100
Vậy: S={100}
Ta có : \(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=-5\)
\(\Leftrightarrow\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}+5=0\)
\(\Leftrightarrow\dfrac{74-x}{26}+1+\dfrac{75-x}{25}+1+\dfrac{76-x}{24}+1+\dfrac{77-x}{23}+1+\dfrac{78-x}{22}+1=0\)
\(\Leftrightarrow\dfrac{100-x}{26}+\dfrac{100-x}{25}+\dfrac{100-x}{24}+\dfrac{100-x}{23}+\dfrac{100-x}{22}=0\)
\(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}\right)=0\)
Thấy : \(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}\ne0\)
\(\Rightarrow100-x=0\)
\(\Leftrightarrow x=100\)
Vậy ...
a,\(\dfrac{x}{-3}\text{=}\dfrac{-5}{15}\) b,\(\dfrac{1173}{x}\text{=}\dfrac{3}{5}\)
c,\(\dfrac{2}{x}\text{=}\dfrac{y}{15}\text{=}\dfrac{-25}{75}\)
a) \(x=\dfrac{-3.\left(-5\right)}{15}=1\)
b) \(x=\dfrac{5.1173}{3}=1955\)
c) Tách làm 2 phép tính
(1) \(\dfrac{y}{15}=\dfrac{-25}{75}\rightarrow y=\dfrac{-25.15}{75}=-5\)
(2) \(\dfrac{2}{x}=\dfrac{-25}{75}\rightarrow x=\dfrac{75.2}{-25}=-6\)