852. Поставьте вместо знака ∗ такой одночлен, чтобы трёхчлен но было представить в виде квадрата двучлена:а) ∗+56a+49;б) 36−12x+∗;в) 25a2+∗+41b2;г) 0,01b2+∗+100c2.
Q. 852. Поставьте вместо знака ∗ такой одночлен, чтобы трёхчлен но было представить в виде квадрата двучлена:а) ∗+56a+49;б) 36−12x+∗;в) 25a2+∗+41b2;г) 0,01b2+∗+100c2.
Recognize Perfect Square Trinomial: To represent a trinomial as the square of a binomial, the trinomial must be a perfect square trinomial. The general form of a perfect square trinomial is (a+b)2=a2+2ab+b2. We need to find the missing term that will complete the square for each case.
Complete the Square: 49: а) For the trinomial x2+56a+49, we recognize that 49 is a perfect square, 72. The term 56a suggests that the binomial form could be (x+7)2, where x is the variable part of the first term. To complete the square, we need the first term to be (7x)2, so x should be a. Therefore, the missing term is 7a×7a=49a2.
Complete the Square: 36: б) For the trinomial 36−12x+∗∗, we recognize that 36 is a perfect square, 62. The term −12x suggests that the binomial form could be (6−x)2. To complete the square, we need the last term to be (−x)2, so the missing term is x2.
Complete the Square: 25a2: в) For the trinomial 25a2+∗∗+(41)b2, we recognize that 25a2 is a perfect square, (5a)2, and (41)b2 is also a perfect square, (21b)2. The middle term should be 2⋅(5a)⋅(21b)=5ab. Therefore, the missing term is 5ab.
Complete the Square: 0.01b2: г) For the trinomial 0.01b2+∗∗+100c2, we recognize that 0.01b2 is a perfect square, (0.1b)2, and 100c2 is also a perfect square, (10c)2. The middle term should be 2×(0.1b)×(10c)=2bc. Therefore, the missing term is 2bc.
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