M=256

N=15^15/3^15

Thông cảm vì mình ko giải ra chi tiết vì nó lâuuuu

N = \(\dfrac{3^{30}.5^{30}}{3^{30}.5^{15}}=\dfrac{5^{30}}{5^{15}}=5^{15}\)

\(M=\dfrac{2^{30}+2^{20}}{2^{12}+2^{22}}=\dfrac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}=2^8=256\\ N=\dfrac{\left(3\cdot5\right)^{30}}{\left(3^2\cdot5\right)^{15}}=\dfrac{3^{30}\cdot5^{30}}{3^{30}\cdot5^{15}}=5^{15}=...\)

Ta tính nhẩm như sau:

3x4=12     2x5=10    5x3=15     4x2=8

12:3=4      10:2=5    15:3=5      8:2=4

12:4=3      10:5=2     15:5=3     8:4=2

3x4 = 12               2x5 = 10        5x3 = 15                  4x2 =8

12 : 3 = 4              10 : 2 = 5       15 : 3 = 5                8 : 2 = 4

12 : 4 =3               10 :5 = 2        15 : 5 = 3                8 : 4 = 2

a: \(\dfrac{15}{8}-\dfrac{13}{8}=\dfrac{15-13}{8}=\dfrac{2}{8}=\dfrac{1}{4}\)

b: \(\dfrac{7}{15}-\dfrac{2}{15}=\dfrac{7-2}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)

c: \(\dfrac{11}{12}-\dfrac{2}{12}=\dfrac{11-2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)

d: \(\dfrac{19}{7}-\dfrac{5}{7}=\dfrac{19-5}{7}=\dfrac{14}{7}=2\)

Làm 1 câu

10/3...>...1/3

Mấy câu ssu tự quy đồng hay gì gì đấy!

10/3.>..1/3.     10/6.>..3/8.     17/4.>..5/6.    1/3..<.5/9     5/8.<..7/5    3/4..<.15/6      10/9.>..10/11   7/8..<.7/2       15/9.>..3/8  15/7.>..15/8      16/5.>..16/7    25/11..<.1

3.... + .....2 = 86 số thích hợp điền vảo chỗ chấm là

\(\frac{3}{9}+\frac{9}{15}+\frac{8}{12}=\frac{1}{3}+\frac{3}{5}+\frac{2}{3}=\left(\frac{1}{3}+\frac{2}{3}\right)+\frac{3}{5}=1+\frac{3}{5}=\frac{8}{5}\)

\(\frac{4}{10}+\frac{12}{32}+\frac{9}{15}=\frac{2}{5}+\frac{3}{8}+\frac{3}{5}=\left(\frac{2}{5}+\frac{3}{5}\right)+\frac{3}{8}=1+\frac{3}{8}=\frac{11}{8}\)

\(\left(\frac{4}{10}-\frac{8}{30}\right)\times\frac{6}{8}=\frac{2}{15}\times\frac{3}{4}=\frac{6}{60}=\frac{1}{10}\)

a, 3/9+9/15+8/12=1/3+3/5+2/3=14/15+2/3=8/5

<         ;           <            ;           >
=         ;           >            ;           <

5/6<11/12,4/7<3/4,5/8<3/8,10/14=5/7,10/9>10/11,8/15<8/9

5/6 <11/12       4/7<3/4               5/8>3/8

10/14=5/7       10/9>10/11         8/15<8/9              

a: =-5/9-4/9+8/15+7/15-2/11=-2/11

b: =10/17+7/17-5/13-8/13+11/25

=11/25

c: =(9/12-2/12)*3/2=7/12*3/2=21/24=7/8

d: =(31/10-25/10)*3-2

=3/5*3-2

=9/5-2

=-1/5

a) \(\dfrac{4}{5}< \dfrac{5}{a}< \dfrac{10}{7}\) \(\left(a\inℤ\right)\)

\(\Leftrightarrow\dfrac{7}{10}< \dfrac{a}{5}< \dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{7.5}{10}< a< \dfrac{5}{4}.5\)

\(\Leftrightarrow\dfrac{7}{2}< a< \dfrac{25}{4}\)

\(\Leftrightarrow a\in\left\{4;5;6\right\}\)

b) \(\dfrac{2}{5}< \dfrac{a-1}{10}< \dfrac{8}{15}\)

\(\Leftrightarrow\dfrac{2.10}{5}< a-1< \dfrac{8.10}{15}\)

\(\Leftrightarrow4< a-1< \dfrac{16}{3}\)

\(\Leftrightarrow5< a< \dfrac{19}{3}\)

\(\Leftrightarrow a\in\left\{6\right\}\)

c) \(\dfrac{12}{7}< \dfrac{4}{a}< \dfrac{8}{3}\)

\(\Leftrightarrow\dfrac{3}{8}< \dfrac{a}{4}< \dfrac{7}{12}\)

\(\Leftrightarrow\dfrac{3.4}{8}< a< \dfrac{7.4}{12}\)

\(\Leftrightarrow\dfrac{3}{2}< a< \dfrac{7}{3}\)

\(\Leftrightarrow a\in\left\{2\right\}\)

d) \(5< a^2-15< 16\)

\(\Leftrightarrow10< a^2< 31\)

\(\Leftrightarrow\sqrt[]{10}< a< \sqrt[]{31}\)

\(\Leftrightarrow a\in\left\{4;5\right\}\)

3) \(-12+2x-9+x=0\\ -21+3x=0\\ 3x=21\\ x=7\)

4)
\(11+\left(15-x\right)=1\)
\(15-x=1-11\)
\(15-x=-10\)
\(x=15-\left(-10\right)\)
\(x=25\)

5) 
\(4-\left(27-3\right)=x-\left(13-4\right)\)
\(4-24=x-9\)
\(x-9=-20\)
\(x=-20+9\)
\(x=-11\)
 

\(3.-12+\left(2x-9\right)+x=0.\)

\(\Leftrightarrow-12+2x-9+x=0.\Leftrightarrow3x=21.\Leftrightarrow x=7.\)

Vậy \(x=7.\)

\(4.11+\left(15-x\right)=1.\Leftrightarrow11+15-x=1.\Leftrightarrow26-x=1.\Leftrightarrow x=25.\)

Vậy \(x=25.\)

\(5.4-\left(27-3\right)=x-\left(13-4\right).\Leftrightarrow4-24=x-9.\Leftrightarrow-20=x-9.\Leftrightarrow x=-11.\)

Vậy \(x=-11.\)

\(6.8-\left(x-10\right)=23-\left(-4+12\right).\Leftrightarrow8-x+10=23-8.\Leftrightarrow18-x=15.\Leftrightarrow x=3.\)

Vậy \(x=3.\)

\(7.105-5\left(10-5x\right)=-20.\Leftrightarrow105-50+25x=-20.\Leftrightarrow25x=-75.\Leftrightarrow x=-3.\)

Vậy \(x=-3.\)

\(8.\left(x-1\right)\left(8-2x\right)\left(3x+123\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0.\\8-2x=0.\\3x+123=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=4.\\x=-41.\end{matrix}\right.\)

Vậy \(x\in\left\{1;4;-41\right\}.\)

\(9.\left(x^2-25\right)\left(x+10\right)=0.\)

\(\Leftrightarrow\left(x-5\right)\left(x+5\right)\left(x+10\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0.\\x+5=0.\\x+10=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5.\\x=-5.\\x=-10.\end{matrix}\right.\)

Vậy \(x\in\left\{5;-5;-10\right\}.\)

\(10.x\left(x^2+5\right)=0.\Leftrightarrow x=0.\)