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Before going further, let us repeat an aspect of learning which is useful to keep in mind while formulating teaching strategies. A child who can add or subtract in the context of s

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Solve by factorization X 2 +(a/a+b + a+b/a)x+1 = 0 X 2 +(a/a+b + a+b/a)x+1 =>  X 2 +(a/a+b x a+b/ax + a/a+b .a+b/a) =>  X[x+a/a+b] +a+b/a[a+a*a+b]= 0 =>  X= -a

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Evaluate the given limit. Solution: In this question none of the earlier examples can help us. There's no factoring or simplifying to accomplish.  We can't rationalize &

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Show that the product of 3 consecutive positive integers is divisible by 6. Ans: n,n+1,n+2 be three consecutive positive integers We know that n is of the form 3q, 3q +1

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