rozwiaz nierówności
1.
√2x + 5 ≤ - √2x - 1
2.
2x - 1 ≥ 3x
2x > - 10
√2x + √2x ≤ -1 -5
2√2x ≤ -6 /:2√2
x ≤ -6/2√2
x ≤ -3/√2 * √2/√2
x ≤ -3√2/2
x∈ ( - ∞ , -3√2/2 >
2x -3x ≥ 1
-x ≥ 1 /:(-1)
x ≤ -1
x∈( - ∞ , -1 >
2x > - 10 /:2
x > -5
x∈ ( -5 , + ∞ )
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1.
√2x + 5 ≤ - √2x - 1
√2x + √2x ≤ -1 -5
2√2x ≤ -6 /:2√2
x ≤ -6/2√2
x ≤ -3/√2 * √2/√2
x ≤ -3√2/2
x∈ ( - ∞ , -3√2/2 >
2.
2x - 1 ≥ 3x
2x -3x ≥ 1
-x ≥ 1 /:(-1)
x ≤ -1
x∈( - ∞ , -1 >
2x > - 10 /:2
x > -5
x∈ ( -5 , + ∞ )