(-2). x - (-21) = 15
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a) 35-15+x=70
=> 20+x=70
=> x=50
b) 568-(2x+14)=15
=> 2x+14=553
=> 2x=539
=> x=269,5
c) x-105:21=15
=> x-5=15
=> x=20
d) (x-105)*21=15
=> x-105=5/7
=> x=5/7+105
=> x=740/7
a, 35 - 15 + x = 70 b, 568 - (2.x + 14 )= 15
x = 70 - 35 + 15 2.x + 14 = 568 - 15
x = 50 2.x + 14 =553 => 2.x =553-14=539 =>x=539 : 2 =269.5
c, x - 105 : 21 =15 d, (x - 105) . 21 = 15
x - 105 = 15 x 21 x - 105 = 15 : 21
x - 105 = 315 x - 105 = \(\frac{5}{7}\)
x = 315+105 x = \(\frac{5}{7}\)+ 105
x = 420 x = \(\frac{740}{7}\)
a) Ta có: \(\dfrac{-11}{15}< \dfrac{x}{15}< \dfrac{-8}{15}\)
nên -11<x<-8
hay \(x\in\left\{-10;-9\right\}\)
b) Ta có: \(\dfrac{3}{7}< \dfrac{x}{21}< \dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{9}{21}< \dfrac{x}{21}< \dfrac{14}{21}\)
Suy ra: 9<x<14
hay \(x\in\left\{10;11;12;13\right\}\)
c) Ta có: \(\dfrac{-67}{21}< \dfrac{x}{168}< \dfrac{-3}{8}\)
nên \(\dfrac{-536}{168}< \dfrac{x}{168}< \dfrac{-63}{168}\)
Suy ra: -536<x<-63
hay \(x\in\left\{-535;-534;...;-64\right\}\)
3 x 15 + 21 x 15 + 85 x 5
= 45 + 315 + 425
= 785
15 - 30 + 40
= 25
21 + 19 - 50 + 10
= 0
\(\dfrac{1}{5}-\dfrac{1}{4}+2\)
\(=-\dfrac{1}{20}+2\)
\(=\dfrac{39}{20}\)
\(\left(\dfrac{1}{4}+\dfrac{1}{6}\right)\times\left(\dfrac{1}{2}-\dfrac{1}{4}\right)\)
\(=\dfrac{5}{12}\times\dfrac{1}{4}\)
\(=\dfrac{5}{12}\times\dfrac{3}{12}\)
\(=\dfrac{5}{48}\)
\(\dfrac{1}{10}+\dfrac{1}{5}-\dfrac{3}{4}\)
\(=-\dfrac{9}{20}\)
\(3\times15+21\times15+85\times5\\ =15\times\left(3+21\right)+425\\ =15\times24+425\\ =360+425\\ =785\)
\(15-30+40\\ =\left(15+40\right)-30\\ =55-30\\ =25\)
\(21+19-50+10\\ =\left(21+19\right)-\left(50-10\right)\\ =40-40\\ =0\)
\(\dfrac{1}{5}-\dfrac{1}{4}+2\)
\(=\dfrac{4}{20}-\dfrac{5}{20}+\dfrac{40}{20}\)
\(=\dfrac{\left(4+40\right)}{20}-\dfrac{5}{20}\)
\(=\dfrac{44}{20}-\dfrac{5}{20}\)
\(=\dfrac{39}{20}\)
\(\left(\dfrac{1}{4}+\dfrac{1}{6}\right)\times\left(\dfrac{1}{2}-\dfrac{1}{4}\right)\)
\(=\dfrac{5}{12}\times\dfrac{1}{4}\)
\(=\dfrac{5}{48}\)
\(\dfrac{1}{10}+\dfrac{1}{5}-\dfrac{3}{4}\)
\(=\dfrac{2}{20}+\dfrac{4}{20}-\dfrac{15}{20}\)
\(=\dfrac{6}{20}-\dfrac{15}{20}\)
\(=-\dfrac{9}{20}\)
1.
b) \(3^x+3^{x+2}=2430\)
\(\Rightarrow3^x.1+3^x.3^2=2430\)
\(\Rightarrow3^x.\left(1+3^2\right)=2430\)
\(\Rightarrow3^x.10=2430\)
\(\Rightarrow3^x=2430:10\)
\(\Rightarrow3^x=243\)
\(\Rightarrow3^x=3^5\)
\(\Rightarrow x=5\)
Vậy \(x=5.\)
c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=15\\2x-15=\pm1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=15:2\\2x-15=1\\2x-15=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\2x=16\\2x=14\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{15}{2};8;7\right\}.\)
Chúc bạn học tốt!
1) \(\left(x-273\right)\times38=0\)
\(\Rightarrow x-273=0:38\)
\(\Rightarrow x-273=0\)
\(\Rightarrow x=0+273\)
\(\Rightarrow x=273\)
Vậy x = 273
2) \(\left(179-x\right)-12=57\)
\(\Rightarrow179-x=57+12\)
\(\Rightarrow179-x=69\)
\(\Rightarrow x=179-69\)
\(\Rightarrow x=110\)
Vậy x = 110
3) \(x-105:21=15\)
\(\Rightarrow x-5=15\)
\(\Rightarrow x=15+5\)
\(\Rightarrow x=20\)
Vậy x = 20
4) \(\left(x-105\right):21=15\)
\(\Rightarrow x-105=15\times21\)
\(\Rightarrow x-105=315\)
\(\Rightarrow x=315+105\)
\(\Rightarrow x=420\)
Vậy x = 420
_Chúc bạn học tốt_
(-2).x+21=15
(-2).x=15-21
(-2).x=-6
x=(-6):(-2)
x=3
Vậy x=3