a) (2x+1)^2 – (x -1)^2 c) 1/4 ( a + 1)^2 – 4/9 ( a – 2)^2

2 bình luận về “a) (2x+1)^2 – (x -1)^2 c) 1/4 ( a + 1)^2 – 4/9 ( a – 2)^2”

1. a)

   (2x+1)^2-(x-1)^2

   = [2x+1+(x-1)].[2x+1-(x-1)]

   = (2x+1+x-1)(2x+1-x+1)

   = 3x(x+2)

   c)

   1/4(a+1)^2-4/9(a-2)^2

   = [1/2(a+1)]^2-[2/3(a-2)]^2

   = (1/2a+1/2)^2-(2/3a-4/3)^2

   = [1/2a+1/2-(2/3a-4/3)].[1/2a+1/2+(2/3a-4/3)]

   = (1/2a+1/2-2/3a+4/3)(1/2a+1/2+2/3a-4/3)

   = (-1/6a+11/6)(7/6a-5/6)

   $#<33$

    

   Trả lời
2. Giải đáp:

    a)

   (2x+1)^{2}-(x-1)^{2}

   =[(2x+1)-(x-1)].[(2x+1)+(x-1)]

   =(2x+1-x+1).(2x+1+x-1)

   =3x.(x+2)

   b)

   \frac{1}{4}.(a+1)^{2}-\frac{4}{9}.(a-2)^{2}

   =[\frac{1}{2}.(a+1)]^{2}-[\frac{2}{3}.(a-2)]^{2}

   =(\frac{1}{2}a+\frac{1}{2})^{2}-(\frac{2}{3}a-\frac{4}{3})^{2}

   =[(\frac{1}{2}a+\frac{1}{2})-(\frac{2}{3}a-\frac{4}{3})].[(\frac{1}{2}a+\frac{1}{2})+(\frac{2}{3}a-\frac{4}{3})]

   =(\frac{1}{2}a+\frac{1}{2}-\frac{2}{3}a+\frac{4}{3}).(\frac{1}{2}a+\frac{1}{2}+\frac{2}{3}a-\frac{4}{3})

   =(-\frac{1}{6}a+\frac{11}{6}).(\frac{7}{6}a-\frac{5}{6})

   #DYNA

   Trả lời