Răspuns:
Explicație pas cu pas:
a) 2(x+√2)-√50<√2(2x-3)-4 <=>
2x + 2√2 - √(25·2) < 2√2 ·x -3√2 - 4 =>
√(25·2) = √(5²·2) = 5√2 =>
2x + 2√2 - 5√2 < 2√2 ·x -3√2 - 4 =>
2x-3√2 < 2√2 ·x -3√2-4 =>
2x-2√2 ·x < -4 <=>
2x·(1-√2) < -4 => x < -2/(1-√2)
Amplificam cu 1+√2 =>
x < (-2-2√2)/(1-2) =>
x < (2+2√2) => x ∈(-∞ ; 2+2√2)
b) 2(2-√2) ≤ 2√2(x-7) I:2 =>
2-√2 ≤ √2(x-7) =>
2-√2 ≤ √2 ·x -7√2 =>
2-√2+7√2 ≤ √2 ·x =>
2-6√2 ≤ √2 ·x =>
√2 ·x ≥ 2-6√2 I : √2 =>
x ≥ (2/√2) - 6 =>
x ≥ (2√2)/2 - 6 =>
x ≥ √2 - 6 => x ∈ (√2-6 ; +∞)
