Tìm x, biết:
a) x - 4 15 = - 3 10
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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 \(x = \frac{{ - 88}}{{15}}\).
Bài 1:
a: BCNN(10;12)=60
b: BCNN(24;10)=120
c: BCNN(4;14;26)=364
d: BCNN(6;8;10)=120
Lời giải:
a.
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x}{2}=\frac{y}{\frac{3}{2}}=\frac{z}{\frac{4}{3}}=\frac{x-y}{2-\frac{3}{2}}=\frac{15}{\frac{1}{2}}=30\)
\(\Rightarrow \left\{\begin{matrix} x=60\\ y=45\\ z=40\end{matrix}\right.\)
b)
Từ đkđb suy ra \(\frac{10x}{1}=\frac{5y}{\frac{1}{3}}=\frac{z}{\frac{1}{6}}=\frac{10x-5y+z}{1-\frac{1}{3}+\frac{1}{6}}=\frac{25}{\frac{5}{6}}=30\)
\(\Rightarrow \left\{\begin{matrix} x=3\\ y=2\\ z=5\end{matrix}\right.\)
`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`
`a,`\(2^x -15= 2^4+1\)
`-> 2^x-15=17`
`-> 2^x=17+15`
`-> 2^x=32`
`-> 2^x=2^5`
`-> x=5`
`b,` Có phải đề là \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\) ?
`=>`\(\dfrac{x+1}{65}+1+\dfrac{x+2}{64}+1=\dfrac{x+3}{63}+1+\dfrac{x+4}{62}+1\)
`=>`\(\dfrac{x+1+65}{65}+\dfrac{x+2+64}{64}-\dfrac{x+3+63}{63}-\dfrac{x+4+62}{62}=0\)
`=>`\(\dfrac{x+66}{65}+\dfrac{x+66}{64}-\dfrac{x+66}{63}-\dfrac{x+66}{62}=0\)
`=>`\(\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}\right)=0\)
Mà `1/65+1/64-1/63-1/62 \ne 0`
`-> x+66=0`
`-> x=-66`
a: =>2^x=2^4+16=32
=>x=5
b: Sửa đề: \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\)
=>\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+2}{64}+1\right)=\left(\dfrac{x+3}{63}+1\right)+\left(\dfrac{x+4}{62}+1\right)\)
=>x+66=0
=>x=-66
a) x+7=-12
x=(-12)-7
x=-19
b)x-15=-21
x=(-21)+15
x=-6
c)13-x=20
x=13-20
x=-7
d)17-(2+x)=3
x=17-3
x=14
x=14-2
x=12
Bài 1:
a)-54
b)-8
Bài 2:
a)(x-14):5=415:413
⇔(x-14):5=42
⇔(x-14):5=16
⇔x-14=80
⇔x=94
b)7x-15x=15-175
⇔-8x=-160
⇔x=20