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CPhill thank you/I unerstood everuthing thants to you,bu i have another question?is that necessary to add brackets at the denominator ?
thank you but thats not what i asked
if it 5*2+27x+10 then it will be:10+27x+10 -> 20+27x
if its 5*x*2+27x+10 the it will be 10x+27x+10 -> 37x+10
OHHH i've found my mistake.
what i need to do is: 1/x^2-1>=1/x-1 ,
anyway,thanks to everyone who helped me:)
Alan that was a problem:for what x values the graph of function f(x)=1/x^2 wont be under the graph of function g(x)=1/x-1
so what i need to do is: 1/x^2-1>1/x-1
ok so its ok I thought i was wrong:)thank you very much you helped me a lot .
i think i am ok with the graph but i am a bit confused with the brackets/
after this [(x-1)(x-6)]/[(x-6)(x-2)]>0 can i simlify it so i can get (x-1)(x-2)>0
or i need to do [(x-1)(x-6)]*[(x-6)(x-2)]>0?
and the next step after (x-1)(x-6)/(x-6)(x-2)>0 is (x-1)/(x-2)>0 ?
so if i get it right even though x cant be some number i need to put this number on the graph?
but when i did (x-1)(x-6)/(x-6)(x-2)>0 we can see immediately that x cant be 1 and 6 cause the nemirator cant be 0
and i dont understand why i need more brackets
so i need to do (x-1)(x-6)(x-6)(x-2)>0 ?