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27
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27 ⎧ ⎧ ⎪ F ' x1 ( ) (1) = 2 * 1 2 − 1 = 0 ⎪0 < E ⎪ ⎪ зад ⎪ ( ) (1) ⎨F ' x2 = − 1 2 ⎪ 1 ⇒ ⎨ − = E зад ⎪ ⎪ 2 ⎪ ( ) (1) ⎪ F ' x3 = 2 1 ⎪1 ⎪ = E зад ⎩ ⎩2 2. По x1: ( F ' x1 ( 2 +1) ) = 2 x ( ) − 1 = 2⎛⎜ 12 − α 1 2 ( 2) ⎞ * 0 ⎟ − 1 = 1 − 0 *α ( 2) x − 1 = 0 ⎝ x1 ⎠ 1 ( ) F ' x1 = 0 2 α ( 2) x = 0 1 2 +1 1 1 x1 = − 0 *1 = 2 2 ( ) F ' x1 = 0 2 α ( 2) x = 0 1 2 +1 1 1 x1 = − 0 *1 = 2 2 1. По x2: ( F ' x2 ( 2 +1) ) = ⎡⎢2 12 − α ( 2) ⎛ 1 ⎞⎤ 3 3 * ⎜ − ⎟⎥ − = 1 + α ( 2 ) x − = α ( 2 ) x − 1 ⎣ x2 ⎝ 2 ⎠⎦ 2 2 2 2 2 ⎛ 1 ⎞ ⎛ ( 2) 1 ⎞ 1 ( 2) 1 ⎜ − ⎟ * ⎜ α x2 − ⎟ = α x2 − ⎝ 2⎠ ⎝ 2⎠ 2 4 1 2 1 α ( 2) x = * = 2 4 1 2 ( 2 +1) 1 1⎛ 1⎞ 1 1 3 x2 = − ⎜− ⎟ = + = 2 2⎝ 2⎠ 2 4 4 2. По x3 ( ( ) ) = 2 * 32 − 2 − 12 = 12 F ' x3 2