(2x-15)5=(2x-15)3
=>(2x-15)5-(2x-15)3=0
=>(2x-15)3.(2x-15)2-(2x-15)3.1=0
=>(2x-15)3.((2x-15)2-1)=0
=>(2x-15)3=0=>2x-15=0=>2x=15=>x=15/2
hoặc (2x-15)2-1=0=>(2x-15)2=0=>2x-15=1,-1=>2x=16,14=>x=8,7
Vậy x=15/2,8,7.
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ta co ( 2x-15)5= (2x-15)3
=> (2x-15)5-(2x-15)3=0
=> (2x-15)3 .{(2x-15)2-1}=0
=> (2x-15)3=0 hoac (2x-15)2-1=0
=> 2x=15 hoac (2x-15)2=1
=> x=15/2 hoac 2x-15=-1;1
=> x=15/2 hoac 2x= 14;16 => x=7;8
vay x=15/2;7;8
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