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\(22.\left(x+32\right)-5=55.179^0\\ 22.\left(x+32\right)-5=55.1\\ 22.\left(x+32\right)-5=55\\ 22.\left(x+32\right)=55+5\\ 22.\left(x+32\right)=60\\ x+32=60:22\\ x+32=\dfrac{30}{11}\\ x=\dfrac{30}{11}-32\\ x=\dfrac{-322}{11}\)
\(22.\left(x+32\right)-5=55.179^0\)
=>22.(x+32)-5=55.1
=>22.(x+32)-5=55
=>22.(x+32)=55+5
=>22.(x+32)=60
=>x+32=60:22
=>x+32=\(\dfrac{30}{11}\)
=>x=\(\dfrac{30}{11}-32\)
\(=>x=-\dfrac{322}{11}\)
Vậy............
2^2.(x+3^2)-5=55
4.(x+9)=55+5=60
(x+9)=60:4=15
x=15-9=6
vậy x=6
a) x+32=5
x=5-32
x=-27
b)x-4=-7
x=-7+4
x=-3
c) 10-(x+12)=5-(22-30)
10-x-12=5-22+30
-2-x=13
x=-2-13
=-15
#H
a; \(\dfrac{93}{17}\): \(x\) + (- \(\dfrac{21}{17}\)) : \(x\) + \(\dfrac{22}{7}\): \(\dfrac{22}{3}\) = \(\dfrac{5}{14}\)
\(\dfrac{94}{17}\) \(\times\) \(\dfrac{1}{x}\) - \(\dfrac{21}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)
\(\dfrac{72}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)
\(\dfrac{72}{17x}\) = \(\dfrac{5}{14}\) - \(\dfrac{3}{7}\)
\(\dfrac{72}{17x}\) = - \(\dfrac{1}{14}\)
17\(x\) = 72.(-14)
17\(x\) = - 1008
\(x\) = - 1008 : 17
\(x\) = - \(\dfrac{1008}{17}\)
Vậy \(x\) \(=-\dfrac{1008}{17}\)
b; - \(\dfrac{32}{27}\) - (3\(x\) - \(\dfrac{7}{9}\))3 = - \(\dfrac{24}{27}\)
- \(\dfrac{32}{27}\) + \(\dfrac{24}{27}\) = (3\(x\) - \(\dfrac{7}{9}\))3
(3\(x-\dfrac{7}{9}\))3 = - \(\dfrac{8}{27}\)
(3\(x-\dfrac{7}{9}\))3 = (- \(\dfrac{2}{3}\))3
3\(x-\dfrac{7}{9}\) = - \(\dfrac{2}{3}\)
3\(x\) = - \(\dfrac{2}{3}\) + \(\dfrac{7}{9}\)
3\(x\) = \(\dfrac{1}{9}\)
\(x\) = \(\dfrac{1}{9}\) : 3
\(x\) = \(\dfrac{1}{27}\)
Vậy \(x=\dfrac{1}{27}\)
`22(x+32)-5=55. 179^0`
`=>22(x+32)-5=55.1`
`=>22(x+32)-5=55`
`=> 22(x+32)=55+5`
`=>22(x+32)=60`
`=>x+32=60:22`
`=>x+32=30/11`
`=>x=30/11-32`
`=>x=-322/11`
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