x^2+1=3x
x+1/x=3
x^2+2+1/x^2=9 -> 1/x^2=7-x^2

(x^4+x^-2)/(x^2+5x+1)=
(x^4+7-x^2)/(3x+5x)=
(x^2(x^2-1)+7)/8x=
((3x-1)(3x-2)+7)/8x=
(9x^2-9x+9)/8x=
9/8•(x^2-x+1)/x=9/8•2x/x=
9/4

C

x^2+2x+4=0,
x^4=(-2x-4)^2=4x^2+16x+16
-> 4(-2x-4)+16x+16=8x

x+4/x=-2 -> 8/x=-4-2x
x^4/4+8/x-3=
x^4/4-4-2x-3=
8x/4-4-2x-3=-7

c

x+1/(16x)=3
x•4+1/(16x)•4=12
4x+1/(4x)=12
16x^2+1/16x^2=144-2=142

(16x^2+1/16x^2)(x+1/16x)=3•142
16x^3+x+1/16x+1/256x^3=426
=>
16x^3+1/(256x^3)=423

Inke question ke solution bhej do koi please bhut confusion h

ax^2+bx+c=0
x1=5+3sqrt(3)
-> x2=5-3sqrt(3)
x1+x2=5+5=10
x1•x2=5^2-9•3=-2

=> x^2-10x-2=0
a=1, b=-10, c=-2
(a^2+b^2+c^2)/(a+b+c)=
(1+100+4)/(-11)=-105/11
C

Isko bhi solve kardijiye

Ligiye sir समझ ligiye जब आपको olampiyad level के questions hi solve करने hain to

Plzzz solution...

Ap
b=a+c/(2)

Gp
c²=ad

Hp
d=ce/(c+e)

c²/a=ce/(c+e)
c/a=e/(c+e)
c²+ce=ea
c²+ce-ea=0

c=-e±√e²+4e/(2)
2c=-e+√e²+4e
(2c+e)²=(e²+4e)
4c²+e²+2ec=e²+4e
2c²+ec=2e
😐

Let x+y= a , y+z= b, z+x= c

Now,

Given, a⅓+b⅓+c⅓ = 0
If a+b+c=0 then a³+b³+c³= 3abc
So,
a+b+c = 3a⅓b⅓c⅓

a+b+c = 3(abc)⅓

2(x+y+z) = 3[(x+y)(y+z)(z+x)]⅓

8[x³+y³+z³+3(x+y)(y+z)(z+x)] = 27(x+y)(y+z)(z+x)

=> 8(x³+y³+z³) = 3(x+y)(y+z)(z+x)

=> (x³+y³+z³)= 3/8(x+y)(y+z)(z+x)

??

Put
bx + ay = ab/(a²+b²)+ ab/(a²+b²)

x = a/(a² + b²)
y = b/(a²+b²)

a² + b²/(a²+b²)² * a² + b²

=(a² + b²)/(a² + b²)

= 1

x^2+x=5
x^2+6x+9-5x-14=0
(x+3)^2-5(x+3)+1=0
(x+3)^2+1=5(x+3)

=>
x+3+1/(x+3)=5
x+3=a
a^3+1/a^3=(a+1/a)^3-3(a+1/a)=
5^3-3•5=110

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arctgx+arcsinx=pi/2
sin(arctgx+arcsinx)=sinpi/2

arcsinx=arctg(tgarcsinx)=
arctg(sinarcsinx/cosarcsinx)=
artg(x/sqrt(1-x^2))=
pi/2-arcctg(x/sqrt(1-x^2))

arctgx=pi/2-arcctgx

pi/2-arcctgx+pi/2-arcctg(x/sqrt(1-x^2))=pi/2
arctg(x/sqrt(1-x^2))+arcctgx=pi/2

Ikkala tarafini ctg lavorsak:

ctg(arctgx/sqrt(1-x^2)+arctgx)=ctgpi/2

(x/sqrt[1-x^2]•x-1)/(x/sqrt[1-x^2]+x)=0
x^2=sqrt(1-x^2)
Va x(1+sqrt[1-x^2]) teng emas 0

x^4=1-x^2
x^4+x^2-1=0
D=1+4=5
x^2=(-1+✔️5)/2
x=sqrt(-1+✔️5)/2