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Sudar
@Sudar
August 2022
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Помогите срочно! Решите уравнение методом разложения на множители :
a) 2^x * x - 4x - 4 +2^x =0
b) 2x^2 sin x - 8 sin x + 4 = x^2
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azamurat
А) 2^x(x+1)-4(x+1)=0
(2^x-4)(x+1)=0
2^x-4=0 x+1=0
2^x=4 x₂=-1
2^x=2^2
x₁=2
б) 2^x2sinx-8sinx+4-x^2=0
x^2(2sinx-1)-4(2sinx-1)=0
(x^2-4)(2sinx-1)=0
x^2-4=0 2sinx-1=0
x₁=2 x₂=-2 sinx=1/2 x=pi/4
2 votes
Thanks 3
azamurat
жми спс
azamurat
;)
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Answers & Comments
(2^x-4)(x+1)=0
2^x-4=0 x+1=0
2^x=4 x₂=-1
2^x=2^2
x₁=2
б) 2^x2sinx-8sinx+4-x^2=0
x^2(2sinx-1)-4(2sinx-1)=0
(x^2-4)(2sinx-1)=0
x^2-4=0 2sinx-1=0
x₁=2 x₂=-2 sinx=1/2 x=pi/4