Vi bz = cy/a =cx-az/b =ay - bx/c

=>a(bz -cy )/a^2=b(cx-az)/b^2=c(ay-bx)/c^2

=>abz-acy/a^2=bcx =baz/b^2=cay-cbx/c^2

Theo tih chất của dãy tỉ số bằng nhau :

=>abz-acy/a^2=bcx=baz/b^2 =cay - cbx/c^2=a^2 + ....

=0/a^2 +b^2 +c^2 =0

Vi bz -cy /a=0=>bz=cy=y/b=z/c  (1)

Vi cx - az/b=0=>cx=a=>x/a =z/c  (2)

Từ (1) và (2) => x/a=y/b =z/c

GT : a//c 

        b// c

KL : a // b

\(A=4^{16}.5^{25}\)

\(A=\left(2^2\right)^{16}.5^{25}\)

\(A=2^{32}.5^{25}\)

\(A=2^{25+7}.5^{25}\)

\(A=\left(2^{25}.5^{25}\right).2^7\)

\(A=\left(2.5\right)^{25}.2^7\)

\(A=10^{25}.128=1280000....00\)( 28 chữ số )

Vậy A có 28 chữ số.

B= 1.2.3+2.3.4+ ... + (n - 1)n(n +1)

4B = 1.2.3.4 + 2.3.4.4 + ... + (n - 1)n(n + 1).4

4B = 1.2.3.4 - 0.1.2.3 + 2.3.4.5 - 1.2.3.4 + ... + (n - 1)n(n + 1)(n + 2) - [(n - 2)(n - 1)n(n + 1)]

4B = (n - 1)n(n + 1)(n + 2) - 0.1.2.3 = (n - 1)n(n + 1)(n + 2)

dùng dãy tỉ số bằng nhau

ta có:

\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+..+\frac{19}{9^2.10^2}\)

\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+..+\frac{10^2-9^2}{9^2.10^2}\)

\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+..+\frac{1}{9^2}-\frac{1}{10^2}=1-\frac{1}{10^2}

12-|3x+2015|-|-3|

=12-|3x+2015|+3

=B < 9

hay Bmax=9

<=>3x+2015=0

<=>....