b2c2>=2(abbcca)Chứng minh HOC24 Lớp học Lớp học Tất cả Lớp 12 Lớp 11 Lớp 10 Lớp 9 Lớp 8 Lớp 7 Lớp 6 Lớp 5 Lớp 4 Lớp 3 Lớp 2 Lớp 1 Hỏi đáp Đề thi Video bài giảng Khóa học TinVerified by Toppr Consider, a 2b 2c 2–ab–bc–ca=0 Multiply both sides with 2, we get 2(a 2b 2c 2–ab–bc–ca)=0 ⇒ 2a 22b 22c 2–2ab–2bc–2ca=0 ⇒ (a 2–2abb 2)(b 2–2bcc 2)(c Multiplying and dividing the expression by 2, = 2 (a2 b2 c2 − ab − bc – ca) / 2 = (2a2 2b2 2c2 − 2ab − 2bc − 2ca) / 2 = (a2 − 2ab b2 b2 − 2bc c2 c2− 2ca a2) / 2
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Factorise a2+b2+c2+2(ab+bc+ca)-Q = a2 b2 c2 − ab − bc − ca = 1 2a b c⊤ 2 − 1 − 1 − 1 2 − 1 − 1 − 1 2 ⏟ = L a b c where matrix L is the Laplacian of the cycle graph with 3 vertices, whose (signed) incidence matrix is CThe number of diagonals, k that can be drawn from one vertex of an x sided rectangle to all other vertices is equal to the number of sides of the rectangle minus 4 Frame the formula for the given
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If abc=0, then write the value of a2/bcb2/cac2/ab rohitsingh4524 rohitsingh4524 Math Secondary School answered • expert verified ProcedureTo prove this equation nonnegative, you will have to convert the equation in terms of perfect square form containing a,b and c Now, a²b²c²abbcca = ½ • ( 2a²2b²2c²2ab2bc 2ca )If a, b, c are rational numbers such that a2 b2 c2 − ab − bc − ca = 0, prove that a = b = c Q If a 2 , b 2 , c 2 are in AP, prove that a b c , b c a , c a b are in AP
= a 2 b 2 2ab 2bc 2ca Using identity a2 b2 2ab = (a b) 2 We get, = (a b) 2 2bc 2ca = (a b) 2 2c (b a) or (a b) 2 2c (a b) Taking (a b) common = (a b)(a b 2c )To prove this equation nonnegative, you will have to convert the equation in terms of perfect square form containing a,b and c Now, a²b²c²abbcca = ½ • ( 2a²2b²2c²2ab2bc 2ca )Factories by Taking Out Common Factors Ab( B2 C2) (C2 B2) Ca( B2 C2) CISCE ICSE Class 9 Question Papers 10 Textbook Solutions Important Solutions 15
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