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Inessa678
@Inessa678
July 2022
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решите плиз очень надо
3*9^x+11*3^x 4
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fsspo
3*9^x + 11*3^x = 4
3*3^2x + 11*3^x - 4 = 0
3^x = t ; t>0
3t^2 + 11t - 4 = 0
D = b^2 - 4*a*c
D = 11^2 + 4*3*4 = 121 + 48 = 169
t1,2 = (-b ± √D) / 2*a
t1 = (-11 + 13)/6 = 2/6 = 1/3
t2 = (-11 - 13)/6 = -24/6 = -4 - лишний корень, т.к. t>0
3^x = 1/3
x = -1
2 votes
Thanks 1
Inessa678
можешь еще такое решить 0 4^x-2 5^x+1> 1, 5
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Answers & Comments
3*3^2x + 11*3^x - 4 = 0
3^x = t ; t>0
3t^2 + 11t - 4 = 0
D = b^2 - 4*a*c
D = 11^2 + 4*3*4 = 121 + 48 = 169
t1,2 = (-b ± √D) / 2*a
t1 = (-11 + 13)/6 = 2/6 = 1/3
t2 = (-11 - 13)/6 = -24/6 = -4 - лишний корень, т.к. t>0
3^x = 1/3
x = -1