absolutní
hodnota – rovnice
zpět na 1.N, 1.Mn
určete všechna řešení daných rovnic
s absolutní hodnotou
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řešení, včetně postupů
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zadání |
podmínky, výpočet |
řešení |
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1) |
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x – 2 ≥
0 x ≥ 2 x – 2 = 9 x = 11 |
x – 2 ≤
0 x ≤ 2 -(x – 2) = 9 -x + 2 = 9 -x = 7 x = -7 |
x1 = 11 x2 = -7 |
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2) |
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6 – x ≥
0 6 ≥ x 2x = 6 – x 3x = 6 x = 2 |
6 – x ≤
0 6 ≤ x 2x = -(6 – x) 2x = -6 + x x = -6 nesplňuje podmínku 6 ≤ x |
x = 2 |
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3) |
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4 – x ≥
0 -x ≥ -4 x ≤ 4 10 = 4 – x x = 4 – 10 x = -6 |
4 – x ≤
0 -x ≤ -4 x ≥ 4 10 = -(4 – x) 10 = -4 + x x = 14 |
x1 = -6 x2 = 14 |
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4) |
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x + 7 ≥
0 x ≥ -7 x + 7 = x – 5 7 = -5 NŘ |
x + 7 ≤
0 x ≤ -7 -(x + 7) = x – 5 -x – 7 = x – 5 -2x = 2 x = -1 nesplňuje podmínku x < -7 |
NŘ |
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5) |
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2x + 6 ≥
0 2x ≥ -6 x ≥ -3 2x + 6 = x x = -6 nesplňuje podmínku x ≥-3 |
2x + 6 ≤
0 2x ≤ -6 x ≤ -3 -(2x + 6) = x -2x – 6 = x -3x = 6 x = -2 nesplňuje podmínku x ≤ -3 |
NŘ |
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6) |
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x – 3 ≥
0 x ≥ 3 3 – x = x – 3 -2x = -6 x = 3 |
3 – x ≤
0 x ≤ 3 3 – x = -(x – 3) 3 – x = -x + 3 x = x
současně ale musí splňovat podmínku x ≤ 3 |
x
≤ 3 resp. |
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7) |
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4 + x ≥
0 x ≥ -4 5x = 4 + x 4x = 4 x = 1 |
4 + x ≤
0 x ≤ -4 5x = -(4 + x) 5x = -4 - x 6x = -4 x = nesplňuje podmínku x ≤ -4 |
x = 1 |
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8) |
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x – 7 ≥
0 x ≥ 7 x – 7 = 2x -x = 7 x = -7 nesplňuje podmínku x ≥ 7 |
x – 7 ≤
0 x ≤ 7 -(x – 7) = 2x -x + 7 = 2x -3x = -7 x = |
x
= |
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9) |
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1 – 2x ≥
0 -2x ≥ -1 x ≤
1 – 2x = 50x 1 = 52x x = |
1 – 2x ≤
0 -2x ≤ -1 x ≥
-1 + 2x = 50x -1 = 48x x = nesplňuje podmínku x ≥ |
x
= |
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10) |
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4 – x ≥
0 -x ≥ -4 x ≤ 4 10 = 4 – x x = 4 – 10 x = -6 |
4 – x ≤
0 -x ≤ -4 x ≥ 4 10 = -(4 – x) 10 = -4 + x x = 14 |
x1 = -6 x2 = 14 |
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11) |
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3x + 5 ≥
0 3x ≥ -5 x ≥ 3x + 5 = 2x – 1 x = -6 nesplňuje podmínku x≥ |
3x + 5 ≤
0 3x ≤ -5 x ≤ -(3x + 5) = 2x – 1 -3x – 5 = 2x – 1 -5x = 4 x = nesplňuje podmínku x ≤ |
NŘ |
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12) |
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x – 2 ≥
0 x ≥ 2 x + 2 = x – 2 2 = -2 NŘ |
x – 2 ≤
0 x ≤ 2 x + 2 = -(x – 2) x + 2 = -x + 2 2x = 0 x = 0 |
x = 0 |
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2 – x ≥
0 -x ≥ -2 x ≤ 2 2x – 7= 2 – x 3x = 9 x = 3 nesplňuje podmínku x ≤ 2 |
2 – x ≤
0 -x ≤ -2 x ≥ 2 2x – 7 = -(2 – x) 2x – 7 = -2 + x x = 5 |
x = 5 |
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14) |
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x ≥ 0 8 – x = 4 – x 8 = 4 NŘ |
x ≤ 0 8 – x = 4 – (-x) 8 – x = 4 + x -2x = -4 x = 2 nesplňuje podmínku x ≤ 0 |
NŘ |
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15) |
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2 – x ≥
0 -x ≥ -2 x ≤ 2 3 – (2 – x) = x – 1 3 – 2 + x = x – 1 1 + x = x – 1 1 = -1 NŘ |
2 – x ≤
0 -x ≤ -2 x ≥ 2 3 – (-(2 – x)) = x – 1 3 + 2 – x = x - 1 5 – x = x – 1 -2x = -6 x = 3 |
x = 3 |
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16) |
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4 – x ≥
0 -x ≥ -4 x ≤ 4
3 = 4x – x2 x2 – 4x + 3 = 0 (x – 3)(x – 1) = 0 x1 = 3, x2 = 1 |
4 – x ≤
0 -x ≤ -4 x ≥ 4
3 = -4x + x2 X2 – 4x – 3 = 0 X3,4 =
X3 = 2 + x4 = 2 – x3 podmínku x
≥ 4 splňuje, x4 tuto podmínku nesplňuje |
x1 = 3 x2 = 1 x3 = 2 + |
