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a) \(6x+15\times8=12\times\left(19-x\right)\)
\(6x+120=228-12x\)
\(6x+120-228+12x=0\)
\(18x-108=0\)
\(18x=108\)
\(x=6\)
b) \(160-\left(35\div x+3\right)\times15=15\)
\(160-\left(35\div x+3\right)=1\)
\(35\div x+3=159\)
\(35\div x=156\)
\(x=\dfrac{35}{156}\)
c) \(2x-\left(1309\div11-19\right)-2=0\)
\(2x-1309\div11-19=2\)
\(2x-119-19=2\)
\(2x-119=21\)
\(2x=140\)
\(x=70\)
d) \(\left(x-7\right)\times\left(2x-16\right)=0\)
\(x-7=0;2x-16=0\)
\(x=7;2x=16\)
\(x=7;x=8\)
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x*18 + x*2 = 160
=> 18x + 2x = 160
=> 20x = 160
=> x = 8
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\(x\times18 +x\times2=160\)
\(\Rightarrow20x=160\)
\(\Rightarrow x=8\)
X x 18 + X x 2 = 160 X x (18+2) = 160 X x 20 =160 X =160 : 20 X = 80.
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Ta có : 2x + 2x + 2 = 160
=> 2x(1 + 22) = 160
=> 2x . 5 = 160
=> 2x = 32
=> 2x = 25
=> x = 5
2x + 2x + 2 = 160
Ta thấy : 160 = 26 + 25
\(\Rightarrow\)2x = 25 nên x = 5
\(\Rightarrow\)2x+2 = 26 nên x = 4
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\(\left(x+2\right)+\left(x+4\right)+...+\left(x+20\right)=160\)
\(x+2+x+4+...+x+20=160\)
\(\left(x+x+x+...+x\right)+\left(2+4+...+20\right)=160\)
\(x\cdot20+\frac{\left(20+2\right)\cdot\left(\left(20-2\right):2+1\right)}{2}=160\)
\(x\cdot20+110=160\)
\(x\cdot20=160-110\)
\(x\cdot20=50\)
\(x=50:20\)
\(x=2,5\)
Vậy \(x=2,5\)
[ x + 2 ] + [ x + 4 ] + [ x + 6 ] + ... + [ x + 20 ] = 160
( 20 + 2 ) x 10 : 2 + 10x = 160
22 x 10 : 2 + 10x = 160
220 : 2 + 10x = 160
110 + 10x = 160
10x = 160 - 110
10x = 50 ; x = 5
[ 1/2 + 1/4 + 1/8 + ... + 1.64 ] /x = 3/8
=> [ ( 1/2 + 1/4 ) + ( 1/8 + 1/16 ) + ( 1/32 + 1/64 ) ] / x = 3/8
[ 3/4 + 3/16 + 3/64 ] / x = 3/8
[ 48/64 + 12/64 + 3/64 ] / x = 3/8
=> 63/64x = 3/8
=> 64x = 21 x 8
64x = 168
x = 2,625
k chắc nha
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2 x X x 3 - 160 = 710
=> 2 x 3 x X = 710 + 160 = 870
=> 6 x X = 870
=> X = 870 : 6
=> X = 145
Vậy ...
Muốn ủng hộ thì ti.ck đúng mk nha =)
2 . x . 3 - 160 = 710
=> x . 6 - 160 = 710
x . 6 = 710 + 160
x . 6 = 870
x = 870 : 6
x = 145
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a, 134: [x-3] = 35 + 160: 5
134 : [x-3] = 35 + 32
134 : [x-3] = 67
x - 3 = 134: 67
x-3 = 2
x = 2 + 3
x = 5
Bài b bạn ghi lại đề nha
a,134:(x-3)=35+160:5
=> 134:(x-3)=35+32
=> 134:(x-3)= 67
=> x - 3 = 134 : 67
=> x-3 = 2
=> x = 2 + 3 = 5
b,[(10-x).2]:3-2=3
=> [(10-x).2]:3 = 3+2
=> [(10-x).2]:3 = 5
=> (10 -x) . 2 = 5.3
=> (10 - x).2 = 15
=> 10-x = 15: 2
=> 10 - x =7,5
=> x = 10 - 7,5 = 2,5
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A = \(\dfrac{3}{6\cdot8}+\dfrac{3}{8\cdot10}+...+\dfrac{3}{158\cdot160}\)
A = \(\dfrac{3}{2}\cdot\left(\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}+...+\dfrac{2}{158\cdot160}\right)\)
A = \(\dfrac{3}{2}\cdot\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+...+\dfrac{1}{158}-\dfrac{1}{160}\right)\)
A = \(\dfrac{3}{2}\cdot\left(\dfrac{1}{6}-\dfrac{1}{160}\right)\)
A = \(\dfrac{3}{2}\cdot\dfrac{77}{480}\)
A = \(\dfrac{77}{320}\)
Vậy: A = \(\dfrac{77}{320}\)
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=>(x+x+...+x)+(5+7+9+...+15)=160 ( 6 số hạng x)
=>6*x+(15+5)*[(15-5):2+1]:2=160
=>6*x+20*6:2=160
=>6*x+60=160
=>6*x=160-60
=>6*x=100
=>x=100:6
=>x=50/3
( x + 2 ) . 16 . x = 160 . x
( x + 2 ) . 16 . x = 16 . 10 . x
x + 2 = 10
x = 8
( x + 2 ) . 16 . x = 160 . x
( x + 2 ) . 16 . x = 16 . 10 . x
x + 2 = 10
x = 8