1)x²-160; 4)x²-3≥0;. Created by Zhetpisbay. algebra-ru

1)x²-160; 4)x²-3≥0;...

October 2021 1 47 Report

1)x²-16<0;
2)x²-95≤0;
3)x²+1>0;
4)x²-3≥0;

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meripoppins60 1) x² - 16 < 0
х₁ = - 4 или х₂ = 4
-4 4
--------°\\\\\\\\\\\\\\\\\°-----> x
  x∈ (-4; 4)

2) x² - 95 ≤ 0
х₁ = - √95 или х₂ = √95
-√95   √95
--------*\\\\\\\\\\\\\\\\\\\\\\\\*-----> x
  x∈ [-√95 ; √95]

3) x² + 1 > 0
x² > -1 - НЕТ решения

4) x² - 3 ≥ 0
х₁ = - √3 или х₂ = √3
-√3    √3
\\\\\\\\\\\\*---------------------*\\\\\\\\\\> x
  x∈ (-∞; -√3] U [√3; +∞)

3 votes Thanks 4

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