Tìm x\(\in\)Z, biết:
a)3-5+(-x+3)=6
b)(-4-x)+(4-15)=-15
c)(11+x)-(-11-9)=32
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1.
a.
11 + x : 5 = 13
x : 5 = 13 - 11
x : 5 = 2
x = 2 . 5
x = 10
a: \(\dfrac{x}{0.9}=\dfrac{5}{6}\)
\(\Leftrightarrow x=\dfrac{3}{4}\)
b: \(\dfrac{-6}{x}=\dfrac{9}{-15}\)
\(\Leftrightarrow x=10\)
c: \(\dfrac{\dfrac{14}{15}}{\dfrac{9}{10}}=\dfrac{x}{\dfrac{3}{7}}\)
\(\Leftrightarrow x=\dfrac{3}{7}\cdot\dfrac{14}{15}:\dfrac{9}{10}=\dfrac{2}{5}\cdot\dfrac{10}{9}=\dfrac{20}{45}=\dfrac{4}{9}\)
b: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{3}=\dfrac{2x+3y-z}{2\cdot2+3\cdot5-3}=\dfrac{32}{16}=2\)
Do đó: x=4; y=10; z=6
a)
\(x+\left(x+2\right)+\left(x+4\right)+...+\left(x+98\right)=0\)
\(x+x+2+x+4+...+x+98=0\)
\(50x+\left(98+2\right).\left[\left(98-2\right):2+1\right]:2=0\)
\(50x+100.49:2=0\)
\(50x+49.50=0\)
\(50x=0-49.50\)
\(50x=-2450\)
\(x=-2450:50\)
\(x=-49\)
b)
\(\left(x-5\right)+\left(x-4\right)+\left(x-3\right)+...+\left(x+11\right)+\left(x+12\right)=99\)
\(x+x+x+...+x-5-4-3-...+11+12=99\)
\(18x+6+7\text{+ 8 + 9 + 10 + 11 + 12 = 99}\)
\(18x+63=99\)
\(18x=99-63\)
\(18x=36\)
\(x=36:18\)
\(x=2\)
a)
\(\begin{array}{l}x + \left( { - \frac{1}{5}} \right) = \frac{{ - 4}}{{15}}\\x = \frac{{ - 4}}{{15}} + \frac{1}{5}\\x = \frac{{ - 4}}{{15}} + \frac{3}{{15}}\\x = \frac{{ - 1}}{{15}}\end{array}\)
Vậy \(x = \frac{{ - 1}}{{15}}\).
b)
\(\begin{array}{l}3,7 - x = \frac{7}{{10}}\\x = 3,7 - \frac{7}{{10}}\\x = \frac{{37}}{{10}} - \frac{7}{{10}}\\x=\frac{30}{10}\\x = 3\end{array}\)
Vậy \(x = 3\).
c)
\(\begin{array}{l}x.\frac{3}{2} = 2,4\\x.\frac{3}{2} = \frac{{12}}{5}\\x = \frac{{12}}{5}:\frac{3}{2}\\x = \frac{{12}}{5}.\frac{2}{3}\\x = \frac{8}{5}\end{array}\)
Vậy \(x = \frac{8}{5}\)
d)
\(\begin{array}{l}3,2:x = - \frac{6}{{11}}\\\frac{{16}}{5}:x = - \frac{6}{{11}}\\x = \frac{{16}}{5}:\left( { - \frac{6}{{11}}} \right)\\x = \frac{{16}}{5}.\frac{{ - 11}}{6}\\x = \frac{{ - 88}}{{15}}\end{array}\)
Vậy \(x = \frac{{ - 88}}{{15}}\).
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) 3-5+(-x+3)=6`
`=>5+(-x+3)=3-6`
`=>5+(-x+3)=-3`
`=>-x+3=-3-5`
`=>-x+3=-8`
`=>-x=-8-3`
`=>-x=-11`
`=>x=11`
__
`b)(-4-x)+(4-15)=-15`
`=>(-4-x)+-11=-15`
`=>-4-x=-15-(-11)`
`=>-4-x=-15+11`
`=>-4-x=-4`
`=>x=-4-(-4)`
`=>x=-4+4`
`=>x=0`
`c)(11+x)-(-11-9)=32`
`=>(11+x)-(-20)=32`
`=>(11+x)+20=32`
`=>11+x=32-20`
`=>11+x=12`
`=>x=12-11`
`=>x=1`
`a)3-5+(-x+3)=6`
`5+(-x+3)=3-6`
`5+(-x+3)=-3`
`-x+3=-3-5`
`-x+3=-8`
`-x=-8-3`
`-x=-11`
`x=11`
`b,(-4-x)+(4-15)=-15`
`(-4-x)+(-11)=-15`
`-4-x=-15-(-11)`
`-4-x=-15+11`
`-4-x=-4`
`x=-4-(-4)`
`x=-4+4`
`x=0`
`c)(11+x)-(-11-9)=32`
`(11+x)-(-20)=32`
`(11+x)+20=32`
`11+x=32-20`
`11+x=12`
`x=12-11`
`x=1`