Mọi người ơi, giúp em với ạ

a) \(\left(9^4.8+9^4.5\right):\left(9^2.\left(10-1\right)\right)\)

=\(9^4.13:9^3=13.9=117\)

b) 100-(75-25)=100-50=50

\(a,\left(2^3\cdot9^4+9^3\cdot45\right):\left(9^2\cdot10-9^2\right)\)

\(< =>\left(8\cdot6561+729\cdot45\right):\left(81\cdot10-81\right)\)

\(< =>\left(52488+32805\right):\left(810-81\right)\)

\(< =>85293:729\)

\(< =>117\)

\(b,100-\left[75-\left(7-2\right)^2\right]\)

\(< =>100-\left[75-\left(5\right)^2\right]\)

\(< =>100-\left[75-25\right]\)

\(< =>100-50\)

\(< =>50\)

\(c,\left(20\cdot2^4+12\cdot2^4\right)-\left(48\cdot2^2\right):8^2\)

\(< =>\left[2^4\cdot\left(20+12\right)\right]-\cdot\left(48\cdot4\right):8^2\)

\(< =>\left[2^4\cdot32\right]-192:8^2\)

\(< =>\left[16\cdot32\right]-192:64\)

\(< =>512-3\)

\(< =>509\)

Đặt tổng trên = A

\(A=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{98}+\frac{2}{192}\)

\(A.2=\frac{4}{3}+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{98}\)

\(A.2-A=\left(\frac{4}{3}+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{98}\right)-\left(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{98}+\frac{2}{192}\right)\)

\(A=\frac{4}{3}-\frac{2}{192}\)

\(A=\frac{127}{96}\)

\(A=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{98}+\frac{2}{192}\)

\(2A=\frac{4}{3}+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{98}\)

\(2A-A=\frac{4}{3}-\frac{2}{192}\)

\(A=\frac{4}{3}-\frac{2}{192}=\frac{127}{96}\)

\(a,=\left(2\sqrt{6}-4\sqrt{3}\right)\sqrt{6}+12\sqrt{2}=12-12\sqrt{2}+12\sqrt{2}=12\\ b,=\dfrac{6\left(3-\sqrt{3}\right)}{6}+\sqrt{3}=3-\sqrt{3}+\sqrt{3}=3\)

1 + 6 + 11 + 16 + ... + 45 + 51

số số hạng là : ( 51 - 1 ) : 5 + 1 = 11 ( số )

tổng là : ( 51 + 1 ) x 11 : 2 = 286

làm tương tự nhé. nhiều quá làm ko nổi

1 + 2 + 3 + ... + 98 + 99 + 100

số số hạng là : ( 100 - 1 ) : 1 + 1 = 100 ( số )

tổng là : ( 100 + 1 ) x 100 : 2 = 5050

2 + 4 + 6 + ... + 98 + 100

số số hạng là : ( 100 - 2 ) : 2 + 1 = 50 số

tổng là : ( 100 + 2 ) x 50 : 2 = 2550

a) 1 + 2 + 3 + ......+ 98 + 99 + 100

có số số hạng là

( 100 - 1 ) : 1 + 1 = 100 số

tổng dãy trên là

( 100 + 1 ) x 100 :2 = 5050

đ/s: 5050

b) 2 + 4 + 6 + .......+ 98 + 100

có số số hạng là

( 100 - 2 ) : 2  + 1 = 50 số

tổng dãy trên là

( 100 + 2 ) x 50 : 2 = 2550

đ/s: 2550

c) 1 + 6 + 11 + 16 + .......+ 45 + 51 

có số số hạng là

( 51 - 1 ) : 5 + 1 = 11 số

tổng dãy trên là

( 51 + 1 ) x 11 : 2 = 286

đ/s: 286

d) 2 + 5 + 7 + 12 + .....+ 81 + 131

= 2 + 5 + 7 + 12 + 19 + 31+ 50 + 81 + 131

= 14           +    62            + 262

= 76 + 262

= 338

chúc bạn học tốt ^ _ ^ 

\(\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}\)

\(=2\times\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{96}+\dfrac{1}{192}\right)\)

\(=2\times\left(\dfrac{64+32+16+8+4+2+1}{192}\right)\)

\(=\dfrac{127}{96}\)

=2/3+1/3+1/6+1/12+1/24+1/48+1/96

=\(\dfrac{127}{96}\)

\(1,\left(3x+2\right)\left(5-x^2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\5-x^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\-x^2=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=\pm\sqrt{5}\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{2}{3};-\sqrt{5};\sqrt{5}\right\}\)

\(2,-2x-\dfrac{2}{3}\left(\dfrac{3}{4}-\dfrac{1}{8}x\right)=\left(-\dfrac{1}{2}\right)^3\)

\(\Leftrightarrow-2x-\dfrac{1}{2}+\dfrac{1}{12}x=-\dfrac{1}{8}\)

\(\Leftrightarrow-2x+\dfrac{1}{12}x=-\dfrac{1}{8}+\dfrac{1}{2}\)

\(\Leftrightarrow-\dfrac{23}{12}=\dfrac{3}{8}\)

\(\Leftrightarrow x=-\dfrac{9}{46}\)

Vậy \(S=\left\{-\dfrac{9}{46}\right\}\)

\(3,\dfrac{1}{12}:\dfrac{4}{21}=3\dfrac{1}{2}:\left(3x-2\right)\)

\(\Leftrightarrow\dfrac{1}{12}.\dfrac{21}{4}=\dfrac{7}{2}.\dfrac{1}{3x-2}\)

\(\Leftrightarrow\dfrac{7}{16}=\dfrac{7}{6x-4}\)

\(\Leftrightarrow6x-4=7:\dfrac{7}{16}\)

\(\Leftrightarrow6x-4=16\)

\(\Leftrightarrow x=\dfrac{10}{3}\)

Vậy \(S=\left\{\dfrac{10}{3}\right\}\)

\(4,\dfrac{x-1}{x+2}=\dfrac{4}{5}\left(dk:x\ne-2\right)\)

\(\Rightarrow5\left(x-1\right)=4\left(x+2\right)\)

\(\Rightarrow5x-5=4x+8\)

\(\Rightarrow x=13\left(tmdk\right)\)

Vậy \(S=\left\{13\right\}\)

\(1+2+3+...+n=\frac{n\left(n+1\right)}{2}\)

\(1+3+5+7+...+\left(2n-1\right)=n^2\)

\(2+4+6+8+...+2n=n\left(n+1\right)\)

( 1/2 + 1/6 + 1/12 + 1/30 ) : ( 1/3 + 1/6 + 1/12 + 1/ 24 + 1/48 )=(1/1.2 + 1/2.3 + 1/3.4 +1/5.6 ):(1/3.1 +1/3.2 + 1/3.4 + 1/3.8 +1/3.16)=....

\(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{30}\right):\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}\right)\)

\(=\left(\frac{12}{12}+\frac{2}{12}+\frac{1}{12}+\frac{1}{30}\right):\left(\frac{16}{16}+\frac{8}{48}+\frac{4}{48}+\frac{2}{48}+\frac{1}{48}\right)\)

\(=\left(\frac{15}{12}+\frac{1}{30}\right):\frac{31}{48}\)

\(=\left(\frac{225}{180}+\frac{6}{180}\right):\frac{31}{48}\)

\(=\frac{231}{180}:\frac{31}{48}\)

\(=\frac{231}{180}.\frac{48}{31}=\frac{231.24}{90.31}\)

\(=\frac{5544}{2790}\)nếu rút gọn được thì bạn rút gon luôn nhé

Đề ???

\(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\right):\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}\right)\)

\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}\right):\left(\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.6}+\frac{1}{6.8}\right)\)

\(=\left(1-\frac{1}{2}+...+\frac{1}{4}-\frac{1}{5}\right):\left[\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{4.6}+\frac{2}{6.8}\right)+(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4})\right]\)

\(=\left(1-\frac{1}{5}\right):\left[\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}\right)+(\frac{1}{2}-\frac{1}{4})\right]\)

\(=\frac{4}{5}:\left[\frac{1}{2}\left(\frac{2}{3}+(\frac{1}{4}-\frac{1}{8})\right)+\frac{1}{4}\right]\)

\(=\frac{4}{5}:\left(\frac{1}{2}.\frac{19}{24}+\frac{1}{4}\right)\)

\(=\frac{4}{5}:\frac{31}{48}\)

\(=\frac{192}{155}\)