b, Ta có : \(\frac{2^7x9^2}{3^3x2^5}=\frac{2^52^2x3^23.3}{3.3^2x2^5}=\frac{2^2x3}{x}=12\)

ta có \(2^7.9^2:3^3.2^5\)

=\(2^7.3^4:3^2.2^5\)

=\(\left(2^7.2^5\right):\left(3^4.3^3\right)\)

=\(2^{12}:3^7\)

=\(4096:2187\)=\(1,872885231\approx2\)

\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}\)

\(=2\times\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}+\dfrac{1}{9\times11}\right)\)

\(=2\times\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)\)

\(=1-\dfrac{1}{11}\)

\(=\dfrac{11}{11}-\dfrac{1}{11}\)

\(=\dfrac{10}{11}\)

\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\\ =1-\dfrac{1}{11}\\ =\dfrac{10}{11}\)

\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{9\cdot11}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{2}{11}\)

\(=1-\dfrac{2}{11}\)

\(=\dfrac{9}{11}\)

\(3x\left(x-2\right)=3x.x-3x.2=3x^2-6x\)

=> Chọn A

CHọn A

\(\frac{3}{5×3}+\frac{3}{5×7}+\frac{3}{7×9}+\frac{3}{9×11}+\frac{3}{11×13}\)

\(=\frac{3}{2}×\left(\frac{2}{3×5}+\frac{2}{5×7}+\frac{2}{7×9}+\frac{2}{9×11}+\frac{2}{11×13}\right)\)

\(=\frac{3}{2}×\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(=\frac{3}{2}×\left(\frac{1}{3}-\frac{1}{13}\right)\)

\(=\frac{3}{2}×\left(\frac{13}{39}-\frac{3}{39}\right)\)

\(=\frac{3}{2}×\frac{10}{39}\)

\(=\frac{5}{13}\)

\(\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+\frac{3}{11.13}\)

\(=\frac{3}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{ 1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11} +\frac{1}{11}-\frac{1}{13}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{13}\right)\)

\(=\frac{3}{2}.\frac{10}{39}\)

\(=\frac{15}{39}\)

sửa đề câu a  và câu b  nhá  , mik nghĩ đề như này :

  \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)

 \(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)

= \(\frac{1}{1}-\frac{1}{215}\)

\(=\frac{214}{215}\)

b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)

    \(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)

\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)

\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)

\(A\cdot2=\frac{214}{215}\)

\(A=\frac{214}{215}:2\)

\(A=\frac{107}{215}\)

@ミ★Ŧɦươйǥ★彡 cảm ơn bạn nhiều

2/5 x 7 + 2/7 x9 + 2 /9 x 11 + ... + 2 /13 x 15

= 1 /5 - 1 /7 + 1/7 - 1/9 + 1/9 - 1/11 + ... + 1 /13 - 1/15

= 1/5 - 1/15

= 2/15

Vậy đáp án là B nhé

Đặt \(A=\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{13\cdot15}\)

Ta có : \(2A=\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+...+\frac{2}{13}-\frac{2}{15}\)

\(2A=\frac{2}{5}-\frac{2}{15}\)

\(2A=\frac{4}{15}\)

\(A=\frac{4}{15}:2\)

\(A=\frac{2}{15}\)

cám ơn mấy bn nhìu nha!!!

a: \(=\dfrac{\left(x^2+5\right)\left(x^2-5\right)+2x\left(x^2+5\right)}{x^2+5}=x^2+2x-5\)

b: \(=\dfrac{x^3-2x^2-x^2+2x+3x-6}{x-2}=x^2-x+3\)

\(\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+...+\frac{3}{129\cdot131}\)

\(=3\left(\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+....+\frac{1}{129\cdot131}\right)\)

\(=\frac{3}{2}\left(\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{129\cdot131}\right)\)

\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{129}-\frac{1}{131}\right)\)

\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{131}\right)\)

\(=\frac{3}{2}\cdot\frac{126}{655}\)

\(=\frac{378}{1310}=\frac{189}{655}\)