hâm rồi à gia linh

\(S< \dfrac{1}{15}+\dfrac{1}{15}+\dfrac{1}{15}+\dfrac{1}{15}+\dfrac{1}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)

`@` `\text {Ans}`

`\downarrow`

`x - 32 \div 16 = 18`

`=> x - 2 = 18`

`=> x = 18 + 2`

`=> x = 20`

Vậy, `x = 20.`

`15 + 2x = 17`

`=> 2x = 17 - 15`

`=> 2x = 2`

`=> x = 2 \div 2`

`=> x = 1`

Vậy, `x = 1`

`324 - 13x = 57*5`

`=> 324 - 13x = 285`

`=> 13x = 324 - 285`

`=> 13x = 39`

`=> x = 39 \div 13`

`=> x = 3`

Vậy, `x = 3.`

`@` `\text {Kaizuu lv uuu}`

\(VT=12^8.9^{12}=\left(2^2.3\right)^8.\left(3^2\right)^{12}=2^{16}.3^8.3^{24}=2^{16}.3^{32}=2^{16}.\left(3^2\right)^{16}=2^{16}.9^{16}=\left(2.9\right)^{16}=18^6=VP\)

Xét: \(12^8\cdot9^{12}=\left(2^2\cdot3\right)^8\cdot\left(3^2\right)^{12}\)

\(=2^{16}\cdot3^8\cdot3^{24}=2^{16}\cdot3^{32}\)

\(=2^{16}\cdot\left(3^2\right)^{16}=2^{16}\cdot9^{16}=18^{16}\)

Ta có:

\(12^8.9^{12}=2^{16}.9^4.9^{12}=2^{16}.9^{16}=\left(2.9\right)^{16}=18^{16}\left(đpcm\right)\)

A = 3/4 + 8/9 + 15/16 + ... + 399/400

A = 1 - 1/4 + 1 - 1/9 + 1 - 1/16 + ... + 1 - 1/400

A = (1 + 1 + 1 + ... +1) - (1/4 + 1/9 + 1/16 + ... + 1/400)

A = 19 - (1/2.2 + 1/3.3 + 1/4.4 + ... + 1/20.20)

đặt b = 1/2.2 + 1/3.3 + 1/4.4 + ... + 1/20.20

có 1/2.2 < 1/1.2 ; 1/3.3 < 1/2.3 ; ... 1/20.20 < 1/19.20

=> b < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/19.20

=> b < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/19 - 1/20

=> b < 1 - 1/20

=> b < 1

mà A = 19 - b

=> A > 18

\(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{399}{400}\)

\(=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+.....+\frac{20^2-1}{20^2}\)

\(=19-\left(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{20^2}\right)\)

\(>19-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{19\cdot20}\right)\)

\(=19-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{19}-\frac{1}{20}\right)\)

\(=19-\left(1-\frac{1}{20}\right)\)

\(>19-1=18\)

\(A=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+...+\frac{20^2-1}{20^2}\)

\(A=\frac{2^2}{2^2}-\frac{1}{2^2}+\frac{3^2}{3^2}-\frac{1}{3^2}+...+\frac{20^2}{20^2}-\frac{1}{20^2}\)

\(A=19-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{20^2}\right)\)

Đặt \(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{20^2}\)

\(B< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\)

\(B< 1-\frac{1}{20}< 1\)

\(B< 1\)

\(\Rightarrow A>19-1\)

\(\Rightarrow A>18\)

Gọi \(S=\frac{15}{15\cdot16}+\frac{15}{16\cdot17}+..+\frac{15}{19\cdot20}\)

\(\Leftrightarrow S=1-\frac{15}{16}+\frac{15}{16}-\frac{15}{17}+...+\frac{15}{19}-\frac{15}{20}\)

\(\Leftrightarrow S=1-\frac{15}{20}=\frac{1}{4}<\frac{1}{3}\)

Vậy S< \(\frac{1}{3}\)

--------------------Good luck------------------------

a: =5*1,4-4*1,5+3*1,3

=7-6+3,9=4,9

b: =1/3*7+3/4*18-2/3*20

=7/3+54/4-40/3

=-11+54/4

=2,5

c: =5/6*17-1/2*16+2/5*15

=85/6-8+6

=85/6-2

=73/6

d: =15*4/5+12*3/4-18*4/9

=12+9-8

=12+1=13

A

a

A