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BINH SAI DIEU KIEN c
12/25/2008 1 Nguyễn Quang Minh 0 . 0 . . . /1000 0 00 00/10 000/1 . . . 222 111 2 1 321 21 21 = + = + =+ =+ − r b a r b a r b a r b a nnnn n n K K K p rr p br p ar p br p bb p ab p ar p ab p aa K K K rba rba rba p p p rrr bbb aaa ω ω ω ω ω ω 0WKBBP 0WNK T1 12/25/2008 2 =+ ++ + =+ ++ + =+ ++ + =+ =+ − 0 . 0 . 0 . rrba brba arba K p rr K p br K p ar K p br K p bb K p ab K p ar K p ab K p aa ω ω ω 0WKBBP 0WNK T1 [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] ( ) acba ccba bcba acba rrba brba arba KtacKtab taa K KtccKtbcKtaa KtbcKtbbKtaa KtacKtabKtaa KtrrKtbrKtaa KtbrKtbbKtaa KtarKtabKtaa ω ω ω ω ω ω ω ++−= =+++ =+++ =+++ =++++ =++++ =++++ =+ =+ − 1 0 0 0 0 . 0 . 0 . 0WKBBP 0WNK T1 12/25/2008 3 [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] ( ) [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] ( ) [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] 2. 02. 01. 1. 1. 1. 1. 1. 1. 1. 1 01.1. 01.1. 0 0 1 0 0 0 2. 2. 1. 1. 1. 1. 1. tcc KKtcc tbc tbb Ktbc tbb tbc tcc Ktbc tbb K KtccKtbc KtbcKtbb tac taa Ktac taa tab tccKtac taa tab tbc tab taa Ktab taa tac tbcKtab taa tab tbb KtacKtab taa K KtccKtbcKtac KtbcKtbbKtab KtacKtabKtaa c ccc b cc bcb ccb bcb a ccb a bcb acba ccba bcba acba ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω −=⇒=+ = −+ − +−= =++ =++ = −+ −+ − = −+ −+ − ++−= =+++ =+++ =+++ Ký hiệu dòng a K a b K a c K c a E 1 ( -1 K a b E 1b ×a b.1 E 2 ( -1 K b c E 1c ×a E 2c ×b.1 c.2 E 3 -1 K c K a K b K c 12/25/2008 4 Ký hiệu dòng a K a b K a c K c a E 1 ( -1 K a b [ ] [ ] [ ] ( ) acba KtacKtab taa K ω ++−= 1 Ký hiệu dòng a K a b K a c K c a E 1 ( -1 K a b E 1b ×a b.1 [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] ( ) [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] 01.1. 0 1 0 0 0 1. =++ = −+ −+ − ++−= =+++ =+++ =+++ bcb a bcb acba ccba bcba acba KtbcKtbb tab taa Ktab taa tac tbcKtab taa tab tbb KtacKtab taa K KtccKtbcKtac KtbcKtbbKtab KtacKtabKtaa ω ω ω ω ω ω ω 12/25/2008 5 Ký hiệu dòng a K a b K a c K c a E 1 ( -1 K a b E 1b ×a b.1 E 2 ( -1 K b c E 1c ×a E 2c ×b.1 c.2 E 3 -1 K c K a K b K c [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] ( ) [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] ( ) [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] 2. 02. 01. 1. 1. 1. 1. 1. 1. 1. 1 01.1. 01.1. 0 0 1 0 0 0 2. 2. 1. 1. 1. 1. 1. tcc KKtcc tbc tbb Ktbc tbb tbc tcc Ktbc tbb K KtccKtbc KtbcKtbb tac taa Ktac taa tab tccKtac taa tab tbc tab taa Ktab taa tac tbcKtab taa tab tbb KtacKtab taa K KtccKtbcKtac KtbcKtbbKtab KtacKtabKtaa c ccc b cc bcb ccb bcb a ccb a bcb acba ccba bcba acba ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω −=⇒=+ = −+ − +−= =++ =++ = −+ −+ − = −+ −+ − ++−= =+++ =+++ =+++ 12/25/2008 6 [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] ( ) [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] ( ) [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] 2. 02. 01. 1. 1. 1. 1. 1. 1. 1. 1 01.1. 01.1. 0 0 1 0 0 0 2. 2. 1. 1. 1. 1. 1. tcc KKtcc tbc tbb Ktbc tbb tbc tcc Ktbc tbb K KtccKtbc KtbcKtbb tac taa Ktac taa tab tccKtac taa tab tbc tab taa Ktab taa tac tbcKtab taa tab tbb KtacKtab taa K KtccKtbcKtac KtbcKtbbKtab KtacKtabKtaa c ccc b cc bcb ccb bcb a ccb a bcb acba ccba bcba acba ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω −=⇒=+ = −+ − +−= =++ =++ = −+ −+ − = −+ −+ − ++−= =+++ =+++ =+++ [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] ( ) [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] ( ) [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] 2. 02. 01. 1. 1. 1. 1. 1. 1. 1. 1 01.1. 01.1. 0 0 1 0 0 0 2. 2. 1. 1. 1. 1. 1. tcc KKtcc tbc tbb Ktbc tbb tbc tcc Ktbc tbb K KtccKtbc KtbcKtbb tac taa Ktac taa tab tccKtac taa tab tbc tab taa Ktab taa tac tbcKtab taa tab tbb KtacKtab taa K KtccKtbcKtac KtbcKtbbKtab KtacKtabKtaa c ccc b cc bcb ccb bcb a ccb a bcb acba ccba bcba acba ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω −=⇒=+ = −+ − +−= =++ =++ = −+ −+ − = −+ −+ − ++−= =+++ =+++ =+++ 12/25/2008 7 [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] ( ) [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] ( ) [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] 2. 02. 01. 1. 1. 1. 1. 1. 1. 1. 1 01.1. 01.1. 0 0 1 0 0 0 2. 2. 1. 1. 1. 1. 1. tcc KKtcc tbc tbb Ktbc tbb tbc tcc Ktbc tbb K KtccKtbc KtbcKtbb tac taa Ktac taa tab tccKtac taa tab tbc tab taa Ktab taa tac tbcKtab taa tab tbb KtacKtab taa K KtccKtbcKtac KtbcKtbbKtab KtacKtabKtaa c ccc b cc bcb ccb bcb a ccb a bcb acba ccba bcba acba ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω −=⇒=+ = −+ − +−= =++ =++ = −+ −+ − = −+ −+ − ++−= =+++ =+++ =+++ [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] ( ) [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] ( ) [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] 2. 02. 01. 1. 1. 1. 1. 1. 1. 1. 1 01.1. 01.1. 0 0 1 0 0 0 2. 2. 1. 1. 1. 1. 1. tcc KKtcc tbc tbb Ktbc tbb tbc tcc Ktbc tbb K KtccKtbc KtbcKtbb tac taa Ktac taa tab tccKtac taa tab tbc tab taa Ktab taa tac tbcKtab taa tab tbb KtacKtab taa K KtccKtbcKtac KtbcKtbbKtab KtacKtabKtaa c ccc b cc bcb ccb bcb a ccb a bcb acba ccba bcba acba ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω −=⇒=+ = −+ − +−= =++ =++ = −+ −+ − = −+ −+ − ++−= =+++ =+++ =+++ 12/25/2008 8 [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] ( ) [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] ( ) [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] 2. 02. 01. 1. 1. 1. 1. 1. 1. 1. 1 01.1. 01.1. 0 0 1 0 0 0 2. 2. 1. 1. 1. 1. 1. tcc KKtcc tbc tbb Ktbc tbb tbc tcc Ktbc tbb K KtccKtbc KtbcKtbb tac taa Ktac taa tab tccKtac taa tab tbc tab taa Ktab taa tac tbcKtab taa tab tbb KtacKtab taa K KtccKtbcKtac KtbcKtbbKtab KtacKtabKtaa c ccc b cc bcb ccb bcb a ccb a bcb acba ccba bcba acba ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω −=⇒=+ = −+ − +−= =++ =++ = −+ −+ − = −+ −+ − ++−= =+++ =+++ =+++ =−++ =+++ =+++ 01422 0023 05.023 cba cba cba KKK KKK KKK Giải hệ phương trình chuẩn số liên hệ như sau 12/25/2008 9 Ký hiệu dòng a K a b K a c K c a 3 1 2 0.5 E 1 ( -1 -0.33333 -0.66667 -0.16667 K a b 3 2 0 E 1b ×a -0.33333 -0.66667 -0.16667 b.1 2.666667 1.333333 -0.16667 E 2 ( -1 -0.5 0.0625 K b c 4 -1 E 1c ×a -1.33333 -0.33333 E 2c ×b.1 -0.66667 0.083333 c.2 2 -1.25 E 3 -1 0.625 K c K a =-0.5 K b =-0.25 K c =0.625 ðánh giá ñộ chính xác: - Xác ñịnh sai số trung phương - Sai số trung phương trọng số ñơn vị P – Trọng số của hàm ñại lượng cần ñánh giá ñộ chính xác x x x x P mm m m P 1 0 2 2 0 = = 0 m 12/25/2008 10 ðánh giá ñộ chính xác: - Xác ñịnh sai số trung phương 2 3 1 A C E B D α α 12 α S S S ?, = YX mm ?= S m ðánh giá ñộ chính xác: - Xác ñịnh sai số trung phương - Sai số trung phương trọng số ñơn vị P – Trọng số của hàm ñại lượng cần ñánh giá ñộ chính xác x x x x P mm m m P 1 0 2 2 0 = = 0 m . tbcKtab taa tab tbb KtacKtab taa K KtccKtbcKtac KtbcKtbbKtab KtacKtabKtaa c ccc b cc bcb ccb bcb a ccb a bcb acba ccba bcba acba ω ω ω ω ω ω ω ω ω ω ω ω ω ω. tbcKtab taa tab tbb KtacKtab taa K KtccKtbcKtac KtbcKtbbKtab KtacKtabKtaa c ccc b cc bcb ccb bcb a ccb a bcb acba ccba bcba acba ω ω ω ω ω ω ω ω ω ω ω ω ω ω