【ベストコレクション】 (ab bc ca)^2 964063-(ab+bc+ac)^2
Let us take abc common from (abbcca) then we getabc(c1 a1 b1)⇒abc(a1 b1 c1)Hence, (abbcca) can be expressed as abc(a1 b1 c1)= 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 )A 2 b 2 c 2 = (a b c) 2 2(ab bc ca) = (2 5 3) 2 2(2×5 5×3 3×2) = 100 62 = 38 Answer (2 2 5 2 3 2) = 38 How To Use the (a 2 b 2 c 2) Formula Give Steps?
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(ab+bc+ac)^2
(ab+bc+ac)^2-The (a b c)2 formula is used to calculate the squares of three numbers with different operations a plus b minus c Whole Square Formula is one of the major algebraic identities and can be P dilip_k ⇒ v = 2ab 2ca 2bc ⇒ 2ab 2ca = v −2bc ⇒ 2a(b c) = v − 2bc ⇒ a = v −2bc (2(b c)) Answer link
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AB BM > AM (i) Similarly, in ΔACM, AC CM > AM (ii) Adding equation (i) and (ii), AB BM MC AC > AM AM AB BC AC > 2AM Yes, the given expression is true(i) (ab – bc) (bc – ca) (ca ab) = ab – bc bc – ca ca – ab = ab – ab – bc bc – ca ca = 0 Thus, the sum of the given expressions is 0 (ii) (a – b ab) (b – c bc) (c – a ac) = a – bA 2 b 2 c 2 2 × 59 = 625 Given, ab bc ca = 59 a 2 b 2 c 2 118 = 625 a 2 b 2 c 2 118 – 118 = 625 – 118 subtracting 118 from both the sides Therefore, a 2 b 2 c 2 = 507
Notice $2 ( a^2 b^2 b^2 ab bc ac ) = (ab)^2 (ac)^2 (bc)^2 \geq 0 $ Hence, $$ a^2 b^2 c^2 \geq ab bc ac $$ Now, use this and the fact that $$ (ab c)^2 =If abc=0 and a^2b^2c^2=abbcac, then it follows that 0=(abc)^2=a^2b^2c^22(abbcac), or a^2b^2c^2=2(abbcac) Put this together andWe have to show that $$(abc)^2((ab)^2(bc)^2(ca)^2)\le (abbcca)(a^2b^2c^2)^2$$ multiplying this out we get $${a}^{5}b{a}^{5}c{a}^{4}{b}^{2}{a}^{4}bc{a}^{4}{c}^{2}
(abbcca)² = (ab) ²(bc)²(ca)²2(ab)(bc)2(bc)(ca)2(ca)(ab) hence, (ab)²(bc)²(ca)²2ab²c2abc²2a²bc (ab)²(bc)²(ca)²2abc(bca)∴ a 2 b 2 c 2 = 1 − 4 = − 3 and a 2 b 2 b 2 c 2 c 2 a 2 = (a b b c c a) 2 − 2 a b c (a b c) a 2 b 2 b 2 c 2 c 2 a 2 = 4 − 6 = − 2 Now, a 4 b 4 c 4 = (a 2 b 2 c 2) 2 − 2 (a 2 b 2Divide each term in 2(abbc ca) = v 2 ( a b b c c a) = v by 2 2 and simplify Tap for more steps abbc ca = v 2 a b b c c a = v 2 Subtract bc b c from both sides of the equation
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