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`26.`
`-` ĐKXĐ : `{ (x >= 0) , (x != 25) , (x != 9) :}`
`a)P = [ (x - 5\sqrt{x}) / (x - 25) - 1 ] : [ (25 - x) / (x + 2\sqrt{x} - 15) - (\sqrt{x} + 3) / (\sqrt{x} + 5) - (\sqrt{x} - 5) / (\sqrt{x} - 3) ]`
`= [ (sqrt(x)(sqrt(x) - 5)) / ((sqrt(x) - 5)(sqrt(x) + 5)) - 1 ] : [ (25 - x - (sqrt(x) + 3)(sqrt(x) - 3) + (sqrt(x) - 5)(sqrt(x) + 5)) / ((sqrt(x) + 5)(sqrt(x) - 3)) ]`
`= [ sqrt(x) / (sqrt(x) + 5) - 1 ] : [ (25 - x - (x - 9) + (x - 25)) / ((sqrt(x) + 5)(sqrt(x) - 3)) ]`
`= [ (sqrt(x) - (sqrt(x) + 5)) / (sqrt(x) + 5) ] : [ (25 - x - x + 9 + x - 25) / ((sqrt(x) + 5)(sqrt(x) - 3)) ]`
`= [ -5 / (sqrt(x) + 5) ] : [ (9 - x) / ((sqrt(x) + 5)(sqrt(x) - 3)) ]`
`= [ -5 / (sqrt(x) + 5) ] : [ ((3 - sqrt(x))(3 + sqrt(x))) / ((sqrt(x) + 5)(sqrt(x) - 3)) ]`
`= [ -5 / (sqrt(x) + 5) ] : [ (-(sqrt(x) - 3)(sqrt(x) + 3)) / ((sqrt(x) + 5)(sqrt(x) - 3)) ]`
`= [ -5 / (sqrt(x) + 5) ] : [ -(sqrt(x) + 3) / (sqrt(x) + 5) ]`
`= [ -5 / (sqrt(x) + 5) ] . [ (sqrt(x) + 5) / (-(sqrt(x) + 3)) ]`
`= 5 / (sqrt(x) + 3)`
`b)P < 1`
`-> 5 / (sqrt(x) + 3) < 1`
`-> 5 / (sqrt(x) + 3) - 1 < 0`
`-> (5 - (sqrt(x) + 3)) / (sqrt(x) + 3) < 0`
`-> (2 - sqrt(x)) / (sqrt(x) + 3) < 0`
Có `x >= 0->\sqrt(x) >= 0`
Do đó `2 - sqrt(x) < 0-> sqrt(x) > 2-> x > 4`
Kết hợp ĐKXĐ : `x > 4, x != 9` và `x != 25`
`27.`
`-` ĐKXĐ : `{ (a > 0) , (b > 0) , (a != b) :}`
`a)P = [ (3\sqrt{a}) / (a + \sqrt{ab} + b) - (3a) / ((\sqrt{a} - \sqrt{b})(a + \sqrt{ab} + b)) + 1 / (\sqrt{a} - \sqrt{b}) ] : [ [(a - 1)(\sqrt{a} - \sqrt{b})] / (2(a + \sqrt{ab} + b)) ]`
`= [ (3\sqrt{a}(\sqrt{a} - \sqrt{b}) - 3a + a + \sqrt{ab} + b) / ((\sqrt{a} - \sqrt{b})(a + \sqrt{ab} + b)) ] . [ (2(a + \sqrt{ab} + b)) / ((a - 1)(\sqrt{a} - \sqrt{b})) ]`
`= (3a - 3\sqrt{ab} - 2a + \sqrt{ab} + b) / (\sqrt{a} - \sqrt{b}) . 2 / ((a - 1)(\sqrt{a} - \sqrt{b}))`
`= (a - 2\sqrt{ab} + b) / (\sqrt{a} - \sqrt{b}) . 2 / ((a - 1)(\sqrt{a} - \sqrt{b}))`
`= (\sqrt{a} - \sqrt{b})^2 / (\sqrt{a} - \sqrt{b}) . 2 / ((a - 1)(\sqrt{a} - \sqrt{b}))`
`= (\sqrt{a} - \sqrt{b}) . 2 / ((a - 1)(\sqrt{a} - \sqrt{b}))`
`= 2 / (a - 1)`
`b)` Có `P in ZZ <=> 2 / (a - 1) in ZZ`
`-> a - 1\inƯ(2) = { -2 , -1 , 1 , 2 }`
Khi đó `a - 1 = -2 -> a = -1` (loại)
`a - 1 = -1 -> a = 0` (loại)
`a - 1 = 1 -> a = 2` (thỏa mãn)
`a - 1 = 2 -> a = 3` (thỏa mãn)
Hãy giúp mọi người biết câu trả lời này thế nào?

Bài `26:`
`a)`
`P=((x-5sqrtx)/(x-25)-1):((25-x)/(x+2sqrtx-15)-(sqrtx+3)/(sqrtx+5)+(sqrtx-5)/(sqrtx-3)) \ (xge0,xne9,xne25)`
`P=((sqrtx(sqrtx-5))/((sqrtx-5)(sqrtx+5))-1):(25-x(sqrtx+3)(sqrtx-3)+(sqrtx-5)(sqrtx+5))/((sqrtx+5)(sqrtx-3))`
`P=(sqrtx/(sqrtx+5)-1):(25-x(x-9)+(x-25))/((sqrtx+5)(sqrtx-3))`
`P=(sqrtx-(sqrtx+5))/(sqrtx+5):(25-x-x+9+x-25)/((sqrtx+5)(sqrtx-3))`
`P=(-5)/(sqrtx+5):(9-x)/((sqrtx+5)(sqrtx-3))`
`P=(-5)/(sqrtx+5):((3-sqrtx)(3+sqrtx))/((sqrtx+5)(sqrtx-3))`
`P=(-5)/(sqrtx+5):(-(sqrtx-3)(sqrtx+3))/((sqrtx+5)(sqrtx-3))`
`P=(-5)/(sqrtx+5):(-(sqrtx+3))/(sqrtx+5)`
`P=(-5)/(sqrtx+5) . (sqrtx+5)/(-(sqrtx+3))`
`P=5/(sqrtx+3)`
`b)`
Với `xge0,xne9,xne25`, ta có:
`P<1<=>5/(sqrtx+3)<1`
Do `sqrtx+3>0AAxge0` nên
`5<sqrtx+3<=>sqrtx>2<=>x>4`
Kết hợp điều kiện xác định, ta được:
`x>4,xne9,xne25`
Bài `27:`
`a)`
`P=((3sqrta)/(a+sqrt(ab)+b)-(3a)/(asqrta-bsqrtb)+1/(sqrta-sqrtb)):((a-1)(sqrta-sqrtb))/(2a+2sqrt(ab)+2b) \ (a>0,bne0,aneb,ane1)`
`P=(3sqrta(sqrta-sqrtb)-3a+(a+sqrt(ab)+b))/((sqrta-sqrtb)(a+sqrt(ab)+b)) :((a-1)(sqrta-sqrtb))/(2a+2sqrt(ab)+2b)`
`P=(3a-3sqrt(ab)-3a+a+sqrt(ab)+b)/((sqrta-sqrtb)(a+sqrt(ab)+b)):((a-1)(sqrta-sqrtb))/(2a+2sqrt(ab)+2b)`
`P=(a-2sqrt(ab)+b)/((sqrta-sqrtb)(a+sqrt(ab)+b)) : ((a-1)(sqrta-sqrtb))/(2a+2sqrt(ab)+2b)`
`P=(sqrta-sqrtb)^2/((sqrta-sqrtb)(a+sqrt(ab)+b)) : ((a-1)(sqrta-sqrtb))/(2a+2sqrt(ab)+2b)`
`P=(sqrta-sqrtb)/(a+sqrt(ab)+b):((a-1)(sqrta-sqrtb))/(2a+2sqrt(ab)+2b)`
`P=(sqrta-sqrtb)/(a+sqrt(ab)+b) . (2(a+sqrt(ab)+b))/((a-1)(sqrta-sqrtb))`
`P=2/(a-1)`
`b)`
Để `P=2/(a-1) in ZZ` thì `(a-1) in Ư(2)`
`U(2)={+-1;+-2}`
`a-1=1=>a=2` (tmdk)
`a-1=-1=>a=0` (l)
`a-1=2=>a=3` (tmdk)
`a-1=-2=>a=-1` (l)
Vậy `a in {2,3}` thì `P` nguyên.
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Hãy giúp mọi người biết câu trả lời này thế nào?
nhm e nghi cj con luy ong:<
cj kh lụy em ậ, cj iu Long
ahhahahahah 🤣🤣
:>
:))
chac e sang nhom khac._.khum bic coa dua dut hog nua - sang nhom nào z
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