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Kate123234
@Kate123234
July 2022
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Решите неравенство log2 (4^x + 81^x - 4*9^x + 3) > 2^x
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Verified answer
Log2_(4^x + 81^x - 4 * 9^x + 3) > 2^x;
log2_(4^x + 81^x - 4 * 9^x + 3) > log2_2^(2^x);
2 > 1; ⇒ 4^x + 81^x - 4* 9^x + 3 > 4^x;
81^x - 4* 9^x + 3 > 0;
9^x = t > 0;
t^2 - 4 * t + 3 > 0;
t1 = 1;
t2 = 3; (t - 1) * ( t - 3) > 0;
0 < t < 1 U t > 3;
9^x < 1 U 9^x > 3;
9^x < 9^0; U 9^x > 9^(1/2);
x < 0; U x > 1/2.
Ответ х ∈ ( - ∞ ; 0) U (0,5; +∞)
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Answers & Comments
Verified answer
Log2_(4^x + 81^x - 4 * 9^x + 3) > 2^x;log2_(4^x + 81^x - 4 * 9^x + 3) > log2_2^(2^x);
2 > 1; ⇒ 4^x + 81^x - 4* 9^x + 3 > 4^x;
81^x - 4* 9^x + 3 > 0;
9^x = t > 0;
t^2 - 4 * t + 3 > 0;
t1 = 1;
t2 = 3; (t - 1) * ( t - 3) > 0;
0 < t < 1 U t > 3;
9^x < 1 U 9^x > 3;
9^x < 9^0; U 9^x > 9^(1/2);
x < 0; U x > 1/2.
Ответ х ∈ ( - ∞ ; 0) U (0,5; +∞)