plz help me in this maths problem
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Answer:
in 6th question
Step-by-step explanation:
x=[tex]\sqrt{2}[/tex]+1
so 1/x=1/([tex]\sqrt{2}[/tex]+1)
now
[tex](x+1/x)^{2}[/tex]=[tex]x^{2}[/tex]+1/[tex]x^{2}[/tex]+2
now put value of x and 1/x
{([tex]\sqrt{2}[/tex]+1)+1/([tex]\sqrt{2}[/tex]+1)}whole square=[tex]x^{2}[/tex]+1/[tex]x^{2}[/tex]+2
{2+1+2[tex]\sqrt{2}[/tex]}/[tex]\sqrt{2}[/tex]+1}whole square -2 =[tex]x^{2}[/tex]+1/[tex]x^{2}[/tex]
{[tex](3+2root2)^{2}[/tex]/3+2root2}-2=[tex]x^{2}[/tex]+1/[tex]x^{2}[/tex]
3+2root2-2
1+2[tex]\sqrt{2}[/tex]=[tex]x^{2}[/tex]+1/[tex]x^{2}[/tex]
similarly you can solve by using identity
(a+b)whole cube idientiy equate and you will get it
please forgive me for not solving 7 question
if you understood and was able to solve 7 question
then please mark brainleist