Ta có `: 0 < a_1 < a_2 < a3<....<a15`

`->` \(\begin{cases} a_1+ a_2 + a_3 + a_4 + a_5 < 5a_5\\a_6+ a_7 + a_8 + a_9 + a_10 < 5a_{10}\\ a_{11}+ a_{12} + a_{13}+ a_{14} + a_{15} < 5a_{15}\end{cases} \) 

`-> a_1 + a_2 + ..... + a_{15} < 5( a_5 + a_{10} + a_{15} )`

`-> ( a_1 + a_2 + ..... + a_{15} )/( a_5 + a_{10}+a_15 ) <5` (đpcm)

CM gì vậy bạn 

`a)`

Ta có dạng tổng quát `: |f(x)| = |g(x)|`

`-> f(x) = g(x)` hoặc `f(x) = -g(x)` 

`b)`

`x + xy + y = 9`

`<=> x( y + 1 ) + ( y + 1 ) = 10`

`<=> ( x + 1 )( y + 1 ) = 10`

Do `x;y in ZZ`

`-> x + 1 ; y + 1 in Ư_{(10)} = { +-1 ; +-2 ; +-5 ; +-10 }`

Th1 `:`

`{(x+1=1),(y+1=10):}`

`<=>` `{(x=0),(y=9):}`

Th2 `:`

`{(x+1=2),(y+1=5):}`

`<=>` `{(x=1),(y=4):}`

Th3 `:`

`{(x+1=5),(y+1=2):}`

`<=>` `{(x=4),(y=1):}`

Th4 `:`

`{(x+1=10),(y+1=11):}`

`<=>` `{(x=9),(y=0):}`

Th5 `;`

`{(x+1=-1),(y+1=-10):}`

`<=>` `{(x=-2),(y=-11):}`

Th6 `:`

`{(x+1=-2),(y+1=-5):}`

`<=>` `{(x=-3),(y=-6):}`

Th7 `;`

`{(x+1=-5),(y+1=-2):}`

`<=>` `{(x=-6),(y=-3):}`

Th8 `:`

`{(x+1=-10),(y+1=-1):}`

`<=>` `{(x=-11),(y=-2):}`

Vậy `....` 

`a)`

`|5x-4| = |x+2|`

`<=> 5x - 4 = x+2` hoặc `5x - 4 = -( x+2)`

`<=> 4x = 6` hoặc `5x - 4 = -x - 2`

`<=> x = 3/2` hoặc `x = 1/3`

Vậy `S={3/2 ; 1/3}`

Sửa `:`

`23-(12-4^2)+|15|`

`=23-(12-16)+15`

`=23+4+15`

`=42` 

`23-(12-4^2)+|15|`

`=23-(12-16)+15`

`=24+4+15`

`= 43`

`a) 0,12 xx x = 6`

`x = 6 : 0,12`

`x = 50`

`b) x:2,5=4`

`x=4 xx2,5`

`x=10`

`c) 5,6:x=4`

`x=5,6:4`

`x=1,4`

`d) x xx 0,1 = 2/5`

`x = 2/5 : 0,1`

`x=4`

`x + 25%x = 1 3/8`

`=> 5/4x = 11/8`

`=> 10/8x = 11/8`

`=> x = 11/8:10/8`

`=> x = 11/10`

Vậy `...`

`a)`

`| x-36|=1.4`

`<=> x-36 = 1.4` hoặc `x-36=-1,4`

`<=> x = 37,4` hoặc `x = 34,6`

Vậy `...`

`b)`

`1/( x+3)=5/4`

`<=> 5/(5(x+3)) = 5/4`

`=> 5( x + 3 ) = 4`

`<=> 5x+15=4`

`<=> 5x=-11`

`<=> x = -11/5`

Vậy `...`

`c)`

`1/(1.3)+1/(3.5)+....+1/(x(x+2)) = 8/17`

`<=>` `2/(1.3)+2/(3.5)+....+2/(x(x+2)) = 16/17`

`<=> 1 - 1/3 + 1/3 - 1/5 +....+1/x - 1/(x+2) = 16/17`

`<=> 1 - 1/(x+2) = 16/17`

`<=> 1/(x+2)=1/17`

`=>x+2=17`

`<=> x=15`

Vậy `....`

`d)`

\(\sqrt{2+\sqrt{2+\sqrt{x}}}=2\)

`<=> 2 +` \(\sqrt{2+\sqrt{x}}=4\)

`<=>` \(\sqrt{2+\sqrt{x}}=2\) 

`<=>` \(2+\sqrt{x}=4\)

`<=>` \(\sqrt{x}=2\)

`<=> x=4`

Vậy `...`