Mình đã tự giải được bài này rồi. Các bạn không cần giải nữa đâu nhé.

=154183446 b nhé !

\(234678x657=154183446\)

\(=20^2\)\(+18^2\)\(+16^2\)\(+...+4^2\)\(+2^2\)\(-19^2\)\(-17^2\)\(-15^2\)\(-3^2\)\(-1^2\)

\(=\left(20^2-19^2\right)\)\(+\left(18^2-17^2\right)\)\(+...+\left(2^2-1^2\right)\)

\(=39+35+31+..+7+3\)

Tổng trên bằng dãy số:3;7;11;...;31;35;39

Số số hạng là

(39-3):4+1=10

Tổng dãy số trên là

(39+3)x10/2=210

\(=>39+35+31+...+7+3=210\)

\(\left(20^2+18^2+16^2+...+4^2+2^2\right)-\left(19^2+17^2+15^2+...+3^2+1^2\right)\)

\(=20^2-19^2+18^2-17^2+...+4^2-3^2+2^2-1^2\)

=(20−19)(20+19)+(18−17)(18+17)+(16−15)(16+15)+....+(4−3)(4+3)+(2−1)(2+1)

=20+19+18+17+16+15+...+4+3+2+1

=\(\frac{20\cdot21}{2}=\frac{4202}{2}=210\)

a=(x+3)

b=(x+1/2)

c=(xy^2+1)

Good luck!

\(a,\left(x+3\right)^2\)

\(b,\left(x+\frac{1}{2}\right)^2\)

\(c,\left(xy^2+1\right)^2\)

\(\frac{5}{3\times6}+\frac{5}{6\times9}+\frac{5}{9\times12}+....+\frac{5}{99\times102}.\)

Đặt :

 \(A=\frac{5}{3\times6}+\frac{5}{6\times9}+\frac{5}{9\times12}+....+\frac{5}{99\times102}.\)

\(\frac{3}{5}\times A=\frac{3}{5}\times\left(\frac{5}{3\times6}+\frac{5}{6\times9}+...+\frac{5}{99\times102}\right)\)

\(\frac{3}{5}A=\frac{3}{3\times6}+\frac{3}{6\times9}+...+\frac{3}{96\times99}\)

\(\frac{3}{5}A=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+....+\frac{1}{96}-\frac{1}{99}\)

\(\frac{3}{5}A=\frac{1}{3}-\frac{1}{99}\)

\(\frac{3}{5}A=\frac{32}{99}\)

\(A=\frac{32}{99}\div\frac{3}{5}=\frac{160}{297}\)

\(A=\frac{5}{3\times6}+\frac{5}{6\times9}+.....+\frac{5}{99\times102}\)

\(A=\frac{5}{3}\left[\left(\frac{1}{3}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{9}\right)+.....+\left(\frac{1}{99}-\frac{1}{102}\right)\right]\)

\(A=\frac{5}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+....+\frac{1}{99}-\frac{1}{102}\right)\)

\(A=\frac{5}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)

\(A=\frac{5}{3}.\frac{11}{34}\)

\(A=\frac{55}{102}\)