No.1 : (9^x + 9^-x) = 98, maka nilai (3^x + 3^-x) = ... no.2 : x+y=1 dan x^3 + y^3 = 49, maka x^2 + .... Question from @gwenlauren36

izharulhaq78 $\displaystyle (3^x+3^{-x})^2=(3^x)^2+2\times3^x\times3^{-x}+(3^{-x})^2\\(3^x+3^{-x})^2=3^{x\times2}+2\times3^{x-x}+3^{-x\times2}\\(3^x+3^{-x})^2=3^{2x}+2\times3^{0}+3^{-2x}\\(3^x+3^{-x})^2=(3^2)^{x}+2\times1+(3^2)^{-x}\\(3^x+3^{-x})^2=9^{x}+2+9^{-x}\\(3^x+3^{-x})^2=9^{x}+9^{-x}+2\\(3^x+3^{-x})^2=98+2\\(3^x+3^{-x})^2=100\\3^x+3^{-x}=\sqrt{100}\\\boxed{\boxed{3^x+3^{-x}=10}}$

$\displaystyle (x+y)^3=x^3+3x^2y+3xy^2+y^3\\(x+y)^3=x^3+y^3+3(x^2y+xy^2)\\(x+y)^3=x^3+y^3+3xy(x+y)\\1=49+3xy(1)\\1-49=3xy\\-48=3xy\\-16=xy\\\\(x+y)^2=x^2+2xy+y^2\\(1)^2=x^2+2(-16)+y^2\\1=x^2-32+y^2\\1+32=x^2+y^2\\\boxed{\boxed{33=x^2+y^2}}$

imamjhe 1. 9^x + 9^-x = 98
3^2x + 3^-2x = 98
(3^x + 3^-x)^2 - 2 = 98
(3^x + 3^-x)^2 = 100
3^x + 3^-x = 10

2. x^3 + y^3 = 49
(x+y)^3 - 3xy(x+y) = 49
(1)^3 - 3xy(1) = 49
1 - 3xy = 49
-3xy = 48
xy = -16

maka x^2 + y^2 adalah
(x+y)^2 = 1^2
x^2 + 2xy + y^2 = 1
x^2 + y^2 = 1 - 2xy

subtitusi xy ke persamaan

X^2 + y^2 = 1 -2(-16)
x^2 + y^2 = 33

Bantuin yg nomor 11 dan 12 dongg pleasee makasi banyakk

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