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\(165-\left\{3\left[416-\left(5^2.3-71\right)^2\right]+10^2\right\}:20\)
\(=165-\left\{3\left[416-\left(25.3-71\right)^2\right]+100\right\}:20\)
\(=165-\left\{3\left[416-\left(4\right)^2\right]+100\right\}:20\)
\(=165-\left\{3\left[416-16\right]+100\right\}:20\)
\(=165-\left\{3.400+100\right\}:2=165-1300:20\)
\(=165-65=100\)
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\(a,71-3(12-x)=125\\3(12-x)=71-125\\3(12-x)=-54\\12-x=-54:3\\12-x=-18\\x=12-(-18)\\x=30\\---\)
\(b,3(12+x)-41=-50\\3(12+x)=-50+41\\3(12+x)=-9\\12+x=-9:3\\12+x=-3\\x=-3-12\\x=-15\\---\)
\((14+2x)-34=-28\\14+2x=-28+34\\14+2x=6\\2x=6-14\\2x=-8\\x=-8:2\\x=-4\\Toru\)
3(12 - x) = 71 - 125
3(12 - x) = -54
12 - x = -54 : 3
12 -x = -18
x = 12 + 18
x = 30
b) 3(12+x)- 41= -50
3(12 + x) = -50 + 41
3(12 - x) = -9
12 - x = -9 : 3
12 - x = -3
x = 12 + 3
x = 15
c)(14+2x)-34=-28
( 14 + 2x) = -28 + 34
(14 + 2x) = 6
2x = 6 - 14
2x = -8
x = -8 : 2
x = -4
--thonezz--
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a,
2x-2018-x=1
=> x-2018=1
=> x=2019
b, (x+1)+(x+2)+(x+3)+...+(x+10)=165
Dãy trên có số số hạng là :
(10-1):1+1=10 ( số )
=> (x+1)+(x+2)+(x+3)+...+(x+10)=165
=> x+1+x+2+...+x+10=165
=> 10x.(1+...+10)=165
=> 10x.((10+1).10:2)=165
=> 10x.55=165
=> x=3
c, x+3.3-(20+49.2:7-18)=0
tính phần (20+49.2:7-18) ra là đc mà
![](https://rs.olm.vn/images/avt/0.png?1311)
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Theo đề bài ta có:
x+2x+3x+4x+5x+6x+7x+8x+9x+10x =-165
x(1+2+3+4+5+6+7+8+9+10) =-165
x.55 =-165
x =-165:55
x =-3
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$-165:x=3$
$x=(-165):3=-55$
-----------------
$2x-(-71)=-129$
$2x+71=-129$
$2x=-129-71=-200$
$x=-200:2=-100$
-165:x=3
x=-165:3
x=-55
Giải
2x-(-71)=-129
2x =-129+(-71)
2x. =-200
x. =-200:2
x. =-100