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\(\frac{7}{x}=\frac{x}{28}\) => \(x^2=7\cdot28=196=14^2\) => x = 14
Vậy x = 14
Ta có :
\(\frac{7}{x}=\frac{x}{28}\Rightarrow x.x=7.28\Rightarrow x^2=14.14\Rightarrow x=14\&-14\)
\(\text{ a, 5-(10x)=-7}\)
\(\Rightarrow\) 10x=5-(-7)
\(\Rightarrow\) 10x=12
\(\Rightarrow\) x=12:10
\(\Rightarrow\) x=1,2
b, (x-5).(2x+8)=0
\(\Rightarrow\) x-5=0 hoặc 2x+8=0
\(\Rightarrow\) x =0+5 \(\Rightarrow\) 2x =0+8
\(\Rightarrow\) x =5 \(\Rightarrow\) 2x =8
\(\Rightarrow\) x =8:2
\(\Rightarrow\) x =4
vậy x\(\in\){5;4}
c, 2x-9=-8+9
\(\Rightarrow\) 2x-9=1
\(\Rightarrow\) 2x =1+9
\(\Rightarrow\) 2x =10
\(\Rightarrow\) x =10:2
\(\Rightarrow\) x =5
d, |x-9|.(-8)=-16
\(\Rightarrow\)|x-9| =-16:(-8)
\(\Rightarrow\)|x-9| =2
\(\Rightarrow\) x-9 =\(\hept{\begin{cases}2\\-2\end{cases}}\)
trường hợp 1: x-9=2
\(\Rightarrow\) x =2+9
\(\Rightarrow\) x =11
trường hợp 2: x-9=-2
\(\Rightarrow\) x =-2+9
\(\Rightarrow\) x =7
vậy x \(\in\){11;7}
# học tốt #
\(x\in\left(\infty;-\infty\right)\)
\(\frac{19x+50}{14}=\frac{9}{1}\Rightarrow\left(19x+50\right)1=14.9\)
\(\frac{\left(19x+50\right)1}{19x}=\frac{14.9}{19x}\)
\(\frac{19x+50}{19x}=\frac{14.19}{19x}\)
\(\Rightarrow x=4\)
a, \(\left(19.x+2.5^2\right)\div14=\left(13-8\right)^2-4^2\)
\(\left(19.x+2.25\right)\div14=5^2-4^2\)
\(\left(19.x+2.25\right)\div14=25-16\)
\(\left(19.x+50\right)\div14=9\)
\(\left(19.x+50\right)=9.14\)
\(19.x+50=126\)
\(19.x=126-50\)
\(19.x=76\)
\(\Rightarrow x=76\div19\)
\(\Rightarrow x=4\)
Vậy x = 4
b, \(2.3^x=10.3^{12}+8.27^4\)
\(2.3^x=10.3^{12}+8.\left(3^3\right)^4\)
\(2.3^x=10.3^{12}+8.3^{12}\)
\(2.3^x=\left(10+8\right).3^{12}\)
\(2.3^x=18.3^{12}\)
\(2.3^x=2.3^3.3^{12}\)
\(2.3^x=2.3^{15}\)
\(\Rightarrow x=15\)
Vậy x = 15
1: =>5(2x+6)=40
=>2x+6=8
=>2x=2
=>x=1
2: =>12-(x+3)=256:64=4
=>(x+3)=8
=>x=5
3: =>2x-1=3 hoặc 2x-1=-3
=>x=2 hoặc x=-1
4: \(\Leftrightarrow3^{x+2017}=3^{2015}\)
=>x+2017=2015
=>x=-2
$x-28=8\cdot(-5)$
$x-28=-40$
$x=-40+28$
$x=-12$
`x-28 =8.(-5)`
`=> x-28= -40`
`=>x=-40+28`
`=>x=-12`
Vậy `x=-12`