a) đk: x khác 0;1

 \(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)

= \(\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left[\dfrac{\left(x+1\right)\left(x-1\right)+x+2-x^2}{x\left(x-1\right)}\right]\)

= \(\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)

= \(\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}.\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)

b) Để \(\left|2x-5\right|=3\)

<=>  \(\left[{}\begin{matrix}2x-5=3< =>2x=8< =>x=4\left(c\right)\\2x-5=-3< =>2x=2< =>x=1\left(l\right)\end{matrix}\right.\)

Thay x = 4 vào A, ta có: 

\(A=\dfrac{4^2}{4-1}=\dfrac{16}{3}\)

c) Để A = 4

<=> \(\dfrac{x^2}{x-1}=4\)

<=> \(\dfrac{x^2}{x-1}-4=0< =>\dfrac{x^2-4x+4}{x-1}=0\)

<=> \(\left(x-2\right)^2=0\)

<=> x = 2 (T/m)

d) Để A < 2

<=> \(\dfrac{x^2}{x-1}< 2< =>\dfrac{x^2}{x-1}-2< 0< =>\dfrac{x^2-2x+2}{x-1}< 0\)

<=> \(\dfrac{\left(x-1\right)^2+1}{x-1}< 0\)

Mà \(\left(x-1\right)^2+1>0\)

<=> x - 1 < 0 <=> x < 1

KHĐK: x < 1 ( x khác 0)

e) Để A thuộc Z

<=> \(\dfrac{x^2}{x-1}\in Z\)

<=> \(x^2⋮x-1\)

<=> \(x^2-x\left(x-1\right)-\left(x-1\right)⋮x-1\) 

<=> \(1⋮x-1\)

Ta có bảng: 

KL: Để A thuộc Z <=> \(x\in\left\{2;0\right\}\) 

f) Để A thuộc N <=> \(x\in\left\{2;0\right\}\) 

\(\left(x+1\right)^{2020}\ge0\forall x\\ \Rightarrow-\left(x+1\right)^{2020}\le0\forall x\\ \Rightarrow2019-\left(x+1\right)^{2020}\le2019\forall x\\ \Rightarrow P\le2019\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x+1\right)^{2020}=0\\ \Leftrightarrow x+1=0\\ \Leftrightarrow x=-1\)

Vậy GTLN của P = 2019

vì -(x+1)^2<0 => -(x+1)^2. (x+1)^2018<0 với mọi x => 2019-(x+1)^2020<2019

dấu "=" xảy ra khi x+1=0 <=> x= -1

vậy Pmax= 2019 khi x= -1

Lời giải:

Để $B$ nguyên thì $x^2+19x+93$ là scp.

Đặt $x^2+19x+93=t^2$ với $t\in\mathbb{N}$

$\Leftrightarrow 4x^2+76x+372=4t^2$

$\Leftrightarrow (2x+19)^2+11=(2t)^2$

$\Leftrightarrow 11=(2t-2x-19)(2t+2x+19)$

Đến đây là dạng pt tích cơ bản với $2t-2x-19, 2t+2x+19$ là các số nguyên.

a/\(2\cdot x^2-3=29\)

\(2\cdot x^2=29+3=32\)

\(x^2=\frac{32}{2}=16\)

=> \(x=\sqrt{16}=4\)

b/\(\left(-6\right)x-\left(-7\right)=25\)

\(\left(-6\right)x+7=25\)

\(\left(-6\right)x=25-7=18\)

\(x=\frac{18}{-6}=-3\)

c/ \(46-\left(x-11\right)=-48\)

\(46-x+11=-48\)

\(x=46-\left[\left(-48\right)-11\right]=105\)

a) 2x^2 -3= 29

   2x^2     = 32

     x^2     = 16

     x        = +4 ; -4

b) -6x - (-7) =25

    -6x +7    =25

    -6x         = 18

       x         = -3

c) 46 - (x-11) = -48

    46 -x +11 = -48

    46 -x        = -59

         x         = 105

a/\(2x^2-3=29\)

\(2x^2=29+3=32\)

\(x^2=\frac{32}{2}=16\)

\(x=\sqrt{16}=4\)

b/\(\left(-6\right)x-\left(-7\right)=25\)

\(\left(-6\right)x+7=25\)

\(\left(-6\right)x=25-7=18\)

\(x=\frac{18}{-6}=-3\)

c/\(46-\left(x-11\right)=-48\)

\(46-x+11=-48\)

\(x=46-\left[\left(-48\right)-11\right]=105\)

`a)` Với `x >= 0,x ne 4` có:

`Q=[2(2-\sqrt{x})+2+\sqrt{x}-2\sqrt{x}]/[(2+\sqrt{x})(2-\sqrt{x})]`

`Q=[4-2\sqrt{x}+2+\sqrt{x}-2\sqrt{x}]/[(2+\sqrt{x})(2-\sqrt{x})]`

`Q=[6-3\sqrt{x}]/[(2+\sqrt{x})(2-\sqrt{x})]`

`Q=3/[2+\sqrt{x}]`

`b)` Với `x >= 0,x ne 4` có:

`Q=6/5<=>3/[2+\sqrt{x}]=6/5`

      `=>12+6\sqrt{x}=15`

   `<=>x=1/4` (t/m)

`c)` Với `x >= 0,x ne 4` có:

`Q in Z<=>3/[2+\sqrt{x}] in ZZ`

   `=>2+\sqrt{x} in Ư_{3}`

  Mà `Ư_{3}={+-1;+-3}`

`@2+\sqrt{x}=1=>\sqrt{x}=-1` (Vô lý)

`@2+\sqrt{x}=-1=>\sqrt{x}=-3` (Vô lý)

`@2+\sqrt{x}=-2=>\sqrt{x}=-4` (Vô lý)

`@2+\sqrt{x}=2=>\sqrt{x}=0<=>x=0` (t/m)

Vậy `x=0`

a) 2x²-3=29

2x²=32

x²=16

=>x=±4

b)-6x-(-7)=25

=>-6x=18

=>x=-3

c)46-(x-11)=-48

=> x-11=46+48

=>x-11=94

=>x=94+11=105

a) 2 . x² - 3 = 29

= 2 . x² = 29 + 3 = 32

= x² = 32 : 2 = 16

= x² = 4² ; -4²

= x = 4 ; -4

a/ 2x^2 - 3 = 29

2x^2 = 32

x^2 = 16

x = 4;-4

b/ -6x - (-7) = 25

‐6x +7 =25

‐6x = 18

x = ‐3

c/ 46 ‐ ﴾x‐11﴿ = ‐48

46 ‐x +11 = ‐48

46 ‐x = ‐59

x = 105

a) \(P=\dfrac{3x+3\sqrt{x}-9}{x+\sqrt{x}-2}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)

\(=\dfrac{3x+3\sqrt{x}-9}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

\(=\dfrac{3x+3\sqrt{x}-9+\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)

b) \(P=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}=\dfrac{3\sqrt{x}+6+2}{\sqrt{x}+2}=3+\dfrac{2}{\sqrt{x}+2}\)

Để \(P\in Z\Rightarrow2⋮\sqrt{x}+2\Rightarrow\sqrt{x}+2=2\left(\sqrt{x}+2\ge2\right)\)

\(\Rightarrow x=0\)

c) Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\dfrac{2}{\sqrt{x}+2}\le1\Rightarrow3+\dfrac{2}{\sqrt{x}+2}\le4\)

\(\Rightarrow P_{max}=4\) khi \(x=0\)

a) ĐKXĐ: \(x\ge0,x\ne1\)

\(P=\left(\dfrac{3}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{x-1}\right):\left(\dfrac{x+2}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)

\(=\left(\dfrac{3}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{x+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)

\(=\dfrac{3\left(\sqrt{x}+1\right)+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{1}{\sqrt{x}-1}=\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)

\(=\dfrac{4\sqrt{x}}{\sqrt{x}+1}\)

b) \(P=\sqrt{x}-1\Rightarrow\dfrac{4\sqrt{x}}{\sqrt{x}+1}=\sqrt{x}-1\Rightarrow4\sqrt{x}=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)

\(\Rightarrow4\sqrt{x}=x-1\Rightarrow x-4\sqrt{x}-1=0\)

\(\Delta=\left(-4\right)^2-4.\left(-1\right)=20\Rightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{4-2\sqrt{5}}{2}=2-\sqrt{5}\\\sqrt{x}=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{4+2\sqrt{5}}{2}=2+\sqrt{5}\end{matrix}\right.\)

mà \(\sqrt{x}\ge0\Rightarrow\sqrt{x}=2+\sqrt{5}\Rightarrow x=9+4\sqrt{5}\)

c) \(P=\dfrac{4\sqrt{x}}{\sqrt{x}+1}=\dfrac{4\left(\sqrt{x}+1\right)-4}{\sqrt{x}+1}=4-\dfrac{4}{\sqrt{x}+1}\)

Để \(P\in Z\Rightarrow4⋮\sqrt{x}+1\Rightarrow\sqrt{x}+1\in\left\{1;2;4\right\}\left(\sqrt{x}+1\ge1\right)\)

\(\Rightarrow x\in\left\{0;1;9\right\}\) mà \(x\ne1\Rightarrow x\in\left\{0;9\right\}\)

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