[230-(15-5.x)].=390
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720 : [ 41 - (2\(x\) - 5)] = 22.5
41 - (2\(x\) - 5) = 720 :110
2\(x\) - 5 = 41 - \(\dfrac{72}{11}\)
2\(x\) - 5 = \(\dfrac{379}{11}\)
2\(x\) = \(\dfrac{379}{11}\) + 5
2\(x\) = \(\dfrac{434}{11}\)
\(x\) = \(\dfrac{434}{11}\) : 2
\(x\) = \(\dfrac{217}{11}\)
15 - \(x\) = 7 - (-2)
15 - \(x\) = 7 + 2
15 - \(x\) = 9
\(x\) = 15 - 9
\(x\) = 6
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(x+1)+(x+2)+(x+3)+(x+4)+(x+5)...+(x+15)=390
(x+x+x+x+....+x)+(1+2+3+4+5+.....+15)=390
Tổng của các số từ 1 đến 15 là:
(15+1)x15:2=120
Có 15 số x.
Vậy Xx15+120=390
Xx15=390-120
Xx15=270
X=270:15
X=18
X x 15 + ( 1+2+3+4+....+15) =390
vì 1+2+3+4+.....+15=120
nên 390 - 120 = 270
x = 270 : 15 = 18
x = 18
mình đang âm điểm
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\(=15:\left\{390:\left[125\cdot4-5\cdot32-210\right]\right\}\)
\(=15:\left\{390:130\right\}=15:3=5\)
=> 15:{390:[125.4-(5.32+210)]}
=> 15:{390:[500-(160+210)]}
=> 15:{390:[500-370]}
=> 15:{390:130}
=> 15:3=5
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=15 : { 390 : [ 125.4 - ( 5.32 + 35 . 6 ) ] }
=15 : { 390 : [ 500 - ( 160 + 35 . 6 ) ] }
=15 : { 390 : [ 500 - ( 160 + 210 ) ] }
=15 : { 390 : [ 500 - 370 ] }
=15 : { 390 : 130 }
= 15: 3
= 5
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\(15\times\left(32-24\div x\right)-230=70\)
\(15\times\left(32-24\div x\right)=300\)
\(32-24\div x=20\)
\(24\div x=12\)
\(x=2\)
\(15\times\left(32-24:x\right)-230=70\)
\(15\times\left(32-24:x\right)=70+230\)
\(15\times\left(32-24:x\right)=300\)
\(32-24:x=300:15\)
\(32-24:x=20\)
\(24:x=32-20\)
\(24:x=12\)
\(x=24:12\)
\(x=2\)
\(#WendyDang\)
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230 - 35 - 15 = ? x 2 = ? x ? = ?
= 230 - { 35 + 15 } = ? x 2 = ? x ?
= 230 - 50 = ? x 2 = ? x ?
= 180 x 2 = ? x ?
= 360 x 360
= ?
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a) \(\left[136+\left(18-x\right).5\right].11^6=11^7\)
\(\Rightarrow\left[136+\left(18-x\right).5\right]=11\)
\(\Rightarrow\left(18-x\right).5=11-136=-125\)
\(\Rightarrow18-x=-125:5=-25\)
\(\Rightarrow x=18-\left(-25\right)=18+25=43\)
b) \(x\in B\left(15\right);362< x\le390\)
\(\Rightarrow x\in\left\{375;390\right\}\)