\(\dfrac{9+x}{13-x}=\dfrac{5}{6}\) ( Với X < 13)
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`[-5]/13 . 2/11+[-5]/11 . 9/13+1`
`=[-5]/13 . 2/11+[-5]/13 . 9/11+1`
`=[-5]/13 . (2/11+9/11)+1`
`=[-5]/13 . 11/11+1`
`=[-5]/13+1=[-5]/13+13/13=8/13`
a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)
⇔\(7\left(x-3\right)=5\left(x+5\right)\)
⇔\(7x-21=5x+25\)
⇔\(7x-21-5x-25=0\)
⇔\(2x-46=0\)
⇔\(2x=46\)
⇔\(x=23\)
a) 4/9 x ( 3/5 + 8/5 - 2/10 )
= 4/9 x 1
=4/9
b) ( 312 + 325 - 247 ) : 13
= 390 : 13
= 30
a/\(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)
\(x=1+\dfrac{5}{7}+\dfrac{13}{14}\)
\(x=\dfrac{14}{14}+\dfrac{10}{14}+\dfrac{13}{14}\)
\(x=\dfrac{37}{14}\)
Vậy \(x=\dfrac{37}{14}\)
b/\(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+\dfrac{3}{5}+\dfrac{6}{5}=\dfrac{11}{3}\)
\(x+\dfrac{9}{5}=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}-\dfrac{9}{5}\)
\(x=\dfrac{55}{15}-\dfrac{27}{15}\)
\(x=\dfrac{28}{15}\)
Vậy \(x=\dfrac{28}{15}\)
#kễnh
a) \(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)
\(x-\dfrac{23}{14}=1\)
\(x=1+\dfrac{23}{14}\)
\(x=\dfrac{37}{14}\)
b) \(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+1+\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+\dfrac{9}{5}=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}-\dfrac{9}{5}\)
\(x=\dfrac{28}{15}\)
a: \(=\dfrac{37}{4}+\dfrac{117}{16}+\dfrac{1}{4}=\dfrac{19}{2}+\dfrac{117}{16}=\dfrac{269}{16}\)
b: \(=1+\left(\dfrac{9}{10}+\dfrac{8}{10}\right):\dfrac{19}{6}=1+\dfrac{17}{10}\cdot\dfrac{6}{19}=\dfrac{146}{95}\)
c: \(=\dfrac{1}{4}-\dfrac{6}{4}+\dfrac{6}{5}=\dfrac{-5}{4}+\dfrac{6}{5}=\dfrac{-1}{20}\)
\(\dfrac{x-2}{5}=\dfrac{1-x}{6}\\ =>\left(x-2\right)\cdot6=\left(1-x\right)\cdot5\\ =>6x-12=5-5x\\ =>6x+5x=5+12\\ =>11x=17\\ x=\dfrac{17}{11}\)
\(=\dfrac{-5}{9}\left(\dfrac{-5}{13}+\dfrac{6}{13}\right)=\dfrac{-5}{9}\cdot\dfrac{1}{13}=\dfrac{-5}{117}\)
\(\left(-\dfrac{5}{9}\times\dfrac{-5}{13}\right)+\left(-\dfrac{5}{9}\times\dfrac{6}{13}\right)\)
\(=-\dfrac{5}{9}\times\dfrac{-5}{13}+-\dfrac{5}{9}\times\dfrac{6}{13}\)
\(=-\dfrac{5}{9}\times\left(\dfrac{-5}{13}+\dfrac{6}{13}\right)\)
\(=-\dfrac{5}{9}\times\dfrac{1}{13}\)
\(=\dfrac{-5}{117}\)
a) \(\dfrac{x}{y}=\dfrac{9}{7}\)⇒\(\dfrac{x}{9}=\dfrac{y}{7}\)
\(\dfrac{y}{z}=\dfrac{7}{3}\)⇒\(\dfrac{y}{7}=\dfrac{z}{3}\)
⇒\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau,ta có:
\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=-\dfrac{15}{5}=-3\)
⇒\(\left\{{}\begin{matrix}x=-3.9=-27\\y=-3.7=-21\\z=-3.3=-9\end{matrix}\right.\)
c: Ta có: 5x=8y=20z
nên \(\dfrac{x}{\dfrac{1}{5}}=\dfrac{y}{\dfrac{1}{8}}=\dfrac{z}{\dfrac{1}{20}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{\dfrac{1}{5}}=\dfrac{y}{\dfrac{1}{8}}=\dfrac{z}{\dfrac{1}{20}}=\dfrac{x-y-z}{\dfrac{1}{5}-\dfrac{1}{8}-\dfrac{1}{20}}=\dfrac{3}{\dfrac{1}{40}}=120\)
Do đó: x=24; y=15; z=6
a) Áp dụng t/c dtsbn:
\(\dfrac{x}{7}=\dfrac{y}{13}=\dfrac{x+y}{7+13}=\dfrac{40}{20}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.7=14\\y=2.13=26\end{matrix}\right.\)
b) \(\dfrac{3}{x}=\dfrac{7}{y}\Rightarrow\dfrac{x}{3}=\dfrac{y}{7}\)
Và \(x+16=y\Rightarrow y-x=16\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{y-x}{7-3}=\dfrac{16}{4}=4\)
\(\Rightarrow\left\{{}\begin{matrix}x=4.3=12\\y=4.7=28\end{matrix}\right.\)
\(\dfrac{9+x}{13-x}=\dfrac{5}{6}\left(x< 13\right)\\ 6\times\left(x+9\right)=5\times\left(13-x\right)\\ 6\times x+54=65-5\times x\\ 6\times x+5\times x=65-54\\ 11\times x=11\\ x=1\left(TM\right)\)
`[9+x]/[13-x]=5/6` `(x < 13)`
`=>6xx(9+x)=5xx(13-x)`
`=>6xx9+6xx x=5xx13+5xx (-x)`
`=>54+6xx x=65-5xx x`
`=>6xx x+5xx x=65-54`
`=>(6+5)xx x=11`
`=>11xx x=11`
`=>x=11:11=1` (t/m `x < 13`)