Tìm x,biet :{[27x11-(x+12)x8]x 3-79}:5=100
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
{[27x11-(x+12)x8]x3-79}:5=100
[27x11-(x+12)x8]x3-79=100x5=500
[27x11-(x+12)x8]x3=500+79=579
27x11-(x+12)x8=579:3=193
27x11-(x+12)=193:8=24,125
x+12=27x11-24,125
x+12=297-24,125
x+12=272,875
x=272,875-12
x=260,875.
Vay x=260,875.
{[27x11-(x+12)x8]x3-79}:5=100
[297 -(x+12)x8]x3-79 =100x5
[297 -(x+12)x8]x3-79 =500
[297 -(x+12)x8]x3 =500+79
[297 -(x+12)x8]x3 =579
297 -(x+12)x8 =579:3
297 -(x+12)x8 =193
(x+12)x8 =297-193
(x+12)x8 =104
x+12 =104:8
x+12 =13
x =13-12
x =1
Thử lại: {[27x11-(1+12)x8]x3-79}:5=100
\(\left\{\left[27\times11-\left(x+12\right)\times8\right]\times3-79\right\}\div5=100\)
\(\left[27\times11-\left(x+12\right)\times8\right]\times3-79=100\times5\)
\(\left[27\times11-\left(x+12\right)\times8\right]\times3-79=500\)
\(\left[27\times11-\left(x+12\right)\times8\right]\times3=500+79\)
\(\left[27\times11-\left(x+12\right)\times8\right]\times3=579\)
\(27\times11-\left(x+12\right)\times8=579:3\)
\(27\times11-\left(x+12\right)\times8=193\)
\(297-\left(x+12\right)\times8=193\)
\(\left(x+12\right)\times8=297-193\)
\(\left(x+12\right)\times8=104\)
\(x+12=104:8\)
\(x+12=13\)
\(x=13-12\)
\(x=1\)
\(k\)\(dung\)\(cho\)\(minh\)\(nhe!\)
\(7\times11-\left(x+12\right)\times8\times3-79:5=100\)
\(\Rightarrow77-24x-288-15,8-100=0\)
\(\Rightarrow-24x=326,8\)
\(\Rightarrow x=-\dfrac{817}{60}\)
=>77-(x+12)*24-79:5=100
=>(x+12)*24=-38,8
=>x+12=-97/60
=>x=-817/60
{ [ 27 * 11 - ( x + 12 ) * 8 ] * 3 - 79 } : 5 = 100
[ 297 - ( x + 12 ) * 8 ] * 3 - 79 = 100 * 5
[ 297 - ( x + 12 ) * 8 ] * 3 - 79 = 500
[ 297 - ( x + 12 ) * 8 ] * 3 = 500 + 79
[ 297 - ( x + 12 ) * 8 ] * 3 = 579
297 - ( x + 12 ) * 8 = 579 : 3
297 - ( x + 12 ) * 8 = 193
( x + 12 ) * 8 = 297 - 193
( x + 12 ) * 8 = 104
x + 12 = 104 : 8
x + 12 = 13
x = 13 - 12
x = 1.^^
dấu " \(.\) " là dấu nhân
\(\left\{\left[27.11-\left(x+12\right).8\right].3-79\right\}:5=100\)
=>\(\left[297-8.x+12.8\right].3-79=100.5\)
=> \(\left[297-8.x+96\right].3-79=500\)
=> \(\left[297-8.x+96\right].3=500+79\)
=> \(\left[\left(297+96\right)-8.x\right].3=579\)
=> \(393-8.x=579:3\)
=> \(393-8.x=193\)
=> \(8.x=393-193\)
=> \(8.x=200=>x=200:8=25\)
Vậy \(x=25\)
Chúc bạn Hk tốt!!!!