a,\(A=1993^{1^{2\times3\times4\times...\times1994}}=1993^1=1993\)

b,\(B=1994^{\left(225-1^2\right)\times\left(225-2^2\right).....\left(225-50^2\right)}\)

     \(=1994^{\left(225-1^2\right)\times\left(225-2^2\right)...\left(225-15^2\right)...\left(225-50^2\right)}\)

     \(=1994^{\left(225-1^2\right)\times\left(225-2^2\right)...\left(225-225\right)...\left(225-50^2\right)}\)

     \(=1994^{\left(225-1^2\right)\times\left(225-2^2\right)...\times0\times...\left(225-50^2\right)}\)

     \(=1994^0=1\)

c, \(C=\frac{2^{10}\times3^{31}+2^{40}\times3^6}{2^{11}\times3^{31}+2^{41}\times3^6}\)

       \(=\frac{2^{10}\times3^6\times\left(1\times3^{25}+2^{30}\times1\right)}{2^{11}\times3^6\times\left(1\times3^{25}+2^{30}\times1\right)}\)

       \(=\frac{2^{10}}{2^{11}}=\frac{1}{2}\)

\(2017\cdot \left(225-1^2\right)\left(225-2^2\right)....\left(225-15^2\right).....\left(225-56^2\right)\)

\(=2017\cdot224\cdot221\cdot\cdot\cdot\cdot\cdot0\cdot\cdot\cdot\left(-2911\right)\)

\(=0\)

Ta có : D = (1 + 2 + 2² + 2³ + ....... + 2²⁰⁰⁴) - 2²⁰⁰⁵

Đặt A = 1 + 2 + 2² + 2³ + ....... + 2²⁰⁰⁴

=> 2A = 2 + 2² + 2³ + ....... + 2²⁰⁰⁵ 

=> 2A - A = 2²⁰⁰⁵ - 1

=> A = 2²⁰⁰⁵ - 1

Thay vào ta có : D = (1 + 2 + 2² + 2³ + ....... + 2²⁰⁰⁴) - 2²⁰⁰⁵

=> D = 2²⁰⁰⁵ - 1 - 2²⁰⁰⁵

=> D = -1

cậu làm còn thiếu bước kìa Nguyễn Việt Hoàng

thế còn các phần khác thì sao bạn

a) Ta có: \(-\dfrac{3}{2}\sqrt{9-4\sqrt{5}}+\sqrt{\left(-4\right)^2\cdot\left(1+\sqrt{5}\right)^2}\)

\(=\dfrac{-3}{2}\left(\sqrt{5}-2\right)+4\cdot\left(\sqrt{5}+1\right)\)

\(=\dfrac{-3}{2}\sqrt{5}+3+4\sqrt{5}+4\)

\(=\dfrac{5}{2}\sqrt{5}+7\)

b) Ta có: \(\left(1+\dfrac{1}{\tan^225^0}\right)\cdot\sin^225^0-\tan55^0\cdot\tan35^0\)

\(=\dfrac{\tan^225^0+1}{\tan^225^0}\cdot\sin25^0-1\)

\(=\left(\dfrac{\sin^225^0}{\cos^225^0}+1\right)\cdot\dfrac{\cos^225^0}{\sin^225^0}\cdot\sin25^0-1\)

\(=\dfrac{\sin^225^0+\cos^225^0}{\cos^225^0}\cdot\dfrac{\cos^225^0}{\sin25^0}-1\)

\(=\dfrac{1}{\sin25^0}-1\)

\(=\dfrac{1-\sin25^0}{\sin25^0}\)

a: \(\left(18\dfrac{1}{3}:\sqrt{225}+8\dfrac{2}{3}\cdot\sqrt{\dfrac{49}{4}}\right):\left[\left(12\dfrac{1}{3}+8\dfrac{6}{7}\right)-\dfrac{\left(\sqrt{7}\right)^2}{\left(3\sqrt{2}\right)^2}\right]:\dfrac{1704}{445}\)

\(=\left(\dfrac{55}{3}:15+\dfrac{26}{3}\cdot\dfrac{7}{4}\right):\left[\left(12+\dfrac{1}{3}+8+\dfrac{6}{7}\right)-\dfrac{7}{18}\right]\cdot\dfrac{445}{1704}\)

\(=\left(\dfrac{55}{45}+\dfrac{91}{6}\right):\left[20+\dfrac{101}{126}\right]\cdot\dfrac{445}{1704}\)

\(=\dfrac{295}{18}:\dfrac{2621}{126}\cdot\dfrac{445}{1704}\)

\(=\dfrac{295}{18}\cdot\dfrac{126}{2621}\cdot\dfrac{445}{1704}\simeq0,21\)

b: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

c: \(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{n+1}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{n}{n+1}\)

\(=\dfrac{1}{n+1}\)

d: \(-66\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\right)+124\cdot\left(-37\right)+63\cdot\left(-124\right)\)

\(=-66\cdot\dfrac{33-22+6}{66}+124\left(-37-63\right)\)

\(=-17-12400=-12417\)

e: \(\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)

\(=\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)

\(=\dfrac{7}{4}\cdot33\cdot\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)

\(=33\cdot\dfrac{7}{4}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)

\(=33\cdot\dfrac{7}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)

\(=33\cdot\dfrac{7}{4}\cdot\dfrac{4}{21}=\dfrac{33\cdot1}{3}=11\)