8/x + 3 = 56/91
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H9
HT.Phong (9A5)
CTVHS
8 tháng 7 2023
Bài 1:
\(101\cdot125+101\cdot25-101\cdot50\)
\(=101\cdot\left(125+25-50\right)\)
\(=101\cdot100\)
\(=10100\)
Bài 2:
\(76\cdot115+56\cdot24+59\cdot24\)
\(=76\cdot115+24\cdot\left(56+59\right)\)
\(=76\cdot115+24\cdot115\)
\(=115\cdot\left(76+24\right)\)
\(=115\cdot100\)
\(=11500\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
30 tháng 1
2 < 3
4 < 8
9 < 34
56 < 78
91 < 9999
9991 < 999999999
LD
2
26 tháng 7 2023
a,91-5(5+x)=61
=>25+5x=91-61=30
=>5x=30-25=5
=>x=1
b,\([\left(x+34\right)-50]\)x2=56
=>(x+34)-50=56:2=28
=>x+34=28+50=78
=>x=78-34=44.
c,1045-\([2015-\left(3x-24\right)]\)=5
=>2015-(3x-24)=1045-5=1040
=>3x-24=2015-1040=975
=>3x=975+24=999
=>x=999:3=333
d,\([195-\left(3x-27\right)]\)x39=4212
=>195-(3x-27)=4212:39=108
=>3x-27=195-108=87
=>3x=87+27=117
=>x=39
e,30-3(x-2)=18
=>30-3x+6=18
=>30-3x=18-6=12
=>3x=30-12=18
=>x=18:3=6
26 tháng 7 2023
a) \(...\Rightarrow5\left(5+x\right)=91-61=30\)
\(\Rightarrow\left(5+x\right)=30:5=6\Rightarrow x=6-5=1\)
b) \(...\Rightarrow\left(x+34\right)-50=56:2=28\)
\(\Rightarrow\left(x+34\right)=28+50=78\Rightarrow x=78-34=44\)
c) \(...\Rightarrow2015-\left(3x-24\right)=1045-5=1040\)
\(\Rightarrow\left(3x-24\right)=2015-1040=975\)
\(\Rightarrow3x=975+24=999\Rightarrow x=999:3=333\)
d) \(...\Rightarrow195-\left(3x-27\right)=4212:39=108\)
\(\Rightarrow\left(3x-27\right)=195-108=87\)
\(\Rightarrow3x=87+27=114\Rightarrow x=114:3=38\)
e) \(...\Rightarrow3\left(x-2\right)=30-18=12\Rightarrow x-2=12:3=4\)
\(\Rightarrow x=4+2=6\)
GL
2
TT
8 tháng 2 2021
\(\dfrac{x+3}{97}+\dfrac{x+5}{95}+\dfrac{x+9}{91}=\dfrac{x+91}{9}+\dfrac{x+92}{8}+\dfrac{x+61}{39}\)
=> \(\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+9}{91}+1=\dfrac{x+91}{9}+1+\dfrac{x+92}{8}+1+\dfrac{x+61}{39}+1\)
=> \(\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}=\dfrac{x+100}{9}+\dfrac{x+100}{8}+\dfrac{x+100}{39}\)
=> \(\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}-\dfrac{x+100}{9}-\dfrac{x+100}{8}-\dfrac{x+100}{39}=0\)
=> \(\left(x+100\right).\left(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\right)=0\)
=> x = - 100 (do \(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\ne0\)
8 tháng 2 2021
Ta có: \(\dfrac{x+3}{97}+\dfrac{x+5}{95}+\dfrac{x+9}{91}=\dfrac{x+91}{9}+\dfrac{x+92}{8}+\dfrac{x+61}{39}\)
\(\Leftrightarrow\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+9}{91}+1=\dfrac{x+91}{9}+1+\dfrac{x+92}{8}+1+\dfrac{x+61}{39}+1\)
\(\Leftrightarrow\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}=\dfrac{x+100}{9}+\dfrac{x+100}{8}+\dfrac{x+100}{39}\)
\(\Leftrightarrow\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}-\dfrac{x+100}{9}-\dfrac{x+100}{8}-\dfrac{x+100}{39}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\right)=0\)
mà \(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\ne0\)
nên x+100=0
hay x=-100
Vậy: S={-100}
KM
6
NK
0
\(\dfrac{8}{x}+3=\dfrac{56}{91}\)
\(\dfrac{8}{x}=\dfrac{56}{91}-3\)
\(\dfrac{8}{x}=-\dfrac{31}{13}\)
\(x=-\dfrac{8\times13}{31}=-\dfrac{104}{31}\)
`8/x + 3 = 56/91`
`=> 8/x= 56/91-3`
`=>8/x=-31/13`
`=> 8 xx 13 =-31x`
`=> -31x=104`
`=>x=104:(-31)`
`=>x=-104/31`