a) ĐK: `a >0`

`P=(a^2+\sqrta)/(a-\sqrta+1)-(2a+\sqrta)/(\sqrta)+1`

`=(\sqrta(\sqrt(a^3)+1^3))/(a-\sqrta+1)-(\sqrta(2\sqrta+1))/(\sqrta)+1`

`=(\sqrta(\sqrta+1)(a-\sqrta+1))/(a-\sqrta+1)-(2\sqrta+1)+1`

`=a+\sqrta-2\sqrta-1+1`

`=a-\sqrta`

b) `P=a-\sqrta`

`=(\sqrta)^2-2.\sqrta .1/2 + (1/2)^2 -1/4`

`=(\sqrta-1/2)^2 -1/4 ≥ -1/4`

`=> P_(min) =-1/4 <=> a=1/4`

a)Vì |4x - 2| = 6 <=> 4x - 2 ϵ {6,-6} <=> x ϵ {2,-1}

Thay x = 2, ta có B không tồn tại

Thay x = -1, ta có B = \(\dfrac{1}{3}\)

b)ĐKXĐ:x ≠ 2,-2

Ta có \(A=\dfrac{5}{x+2}+\dfrac{3}{2-x}-\dfrac{15-x}{4-x^2}=\dfrac{10-5x+3x+6}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{16-2x}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{\left(x+2\right)\left(x-2\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{x^2-4}+\dfrac{15-x}{x^2-4}=\dfrac{x-1}{x^2-4}\)c)Từ câu b, ta có \(A=\dfrac{x-1}{x^2-4}\)\(\Rightarrow\dfrac{2A}{B}=\dfrac{\dfrac{\dfrac{2x-2}{x^2-4}}{2x+1}}{x^2-4}=\dfrac{2x-2}{2x+1}< 1\) với mọi x

Do đó không tồn tại x thỏa mãn đề bài

a) ĐKXĐ: a²-1 ≠0 ⇔ (a-1)(a+1)≠0 ⇔\(\left[{}\begin{matrix}a-1\ne0\\a+1\ne0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a\ne1\\a\ne-1\end{matrix}\right.\)

b) A=\(\dfrac{2a^2}{a^2-1}-\dfrac{a}{a+1}+\dfrac{a}{a-1}\) , a≠1, -1

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

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

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

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

vậy A =\(\dfrac{2a}{a-1}\) với a≠1,-1.

c) Có:A= \(\dfrac{2a}{a-1}\) = \(\dfrac{2a-2+2}{a-1}=\dfrac{2\left(a-1\right)+2}{a-1}=2+\dfrac{2}{a-1}\)

Để a∈Z thì a-1 ∈ Z ⇒ (a-1) ∈ Ư(2) =\(\left\{1;-1;2;-2\right\}\)

Ta có bảng sau:

Vậy để biểu thức A có giá trị nguyên thì a∈\(\left\{2;0;3\right\}\)

a) ĐKXĐ: \(a\notin\left\{1;-1\right\}\)

b) Ta có: \(A=\dfrac{2a^2}{a^2-1}-\dfrac{a}{a+1}+\dfrac{a}{a-1}\)

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

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

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

\(=\dfrac{2a\left(a+1\right)}{\left(a+1\right)\left(a-1\right)}\)

\(=\dfrac{2a}{a-1}\)

c) Để A nguyên thì \(2a⋮a-1\)

\(\Leftrightarrow2a-2+2⋮a-1\)

mà \(2a-2⋮a-1\)

nên \(2⋮a-1\)

\(\Leftrightarrow a-1\inƯ\left(2\right)\)

\(\Leftrightarrow a-1\in\left\{1;-1;2;-2\right\}\)

\(\Leftrightarrow a\in\left\{2;0;3;-1\right\}\)

Kết hợp ĐKXĐ, ta được: \(a\in\left\{0;2;3\right\}\)

Vậy: Để A nguyên thì \(a\in\left\{0;2;3\right\}\)

a: Ta có: \(P=\left(\dfrac{1}{a+\sqrt{a}}+\dfrac{1}{\sqrt{a}+1}\right):\dfrac{\sqrt{a}-1}{a+2\sqrt{a}+1}\)

\(=\dfrac{a+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}-1}\)

\(=\dfrac{\left(a+1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\)

a ĐKXĐ \(a\ge0,a\ne\dfrac{1}{4},a\ne1\)

\(\Rightarrow P=1+\left(\dfrac{\left(2\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\dfrac{\sqrt{a}\left(2\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}\right)\cdot\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)

= \(1+\left(\dfrac{\left(-1\right)\left(2\sqrt{a}-1\right)}{\sqrt{a}-1}+\dfrac{\sqrt{a}\left(2\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\right)\cdot\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{2\sqrt{a}-1}\)

= \(1+\left(-1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{a+\sqrt{a}+1}\right)\sqrt{a}\)

= \(1-\sqrt{a}+\dfrac{a\sqrt{a}+a}{a+\sqrt{a}+1}\) = \(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)+a\sqrt{a}+a}{a+\sqrt{a}+1}=\dfrac{1-a\sqrt{a}+a\sqrt{a}+a}{a+\sqrt{a}+1}=\dfrac{a+1}{a+\sqrt{a}+1}\)

b Xét hiệu \(P-\dfrac{2}{3}=\dfrac{a+1}{a+\sqrt{a}+1}-\dfrac{2}{3}=\dfrac{3a+3-2a-2\sqrt{a}-2}{a+\sqrt{a}+1}=\dfrac{a-2\sqrt{a}+1}{a+\sqrt{a}+1}=\dfrac{\left(\sqrt{a}-1\right)^2}{a+\sqrt{a}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}>0\) \(\Rightarrow P>\dfrac{2}{3}\) 

c Ta có \(P=\dfrac{\sqrt{6}}{\sqrt{6}+1}\Rightarrow\dfrac{a+1}{a+\sqrt{a}+1}=\dfrac{\sqrt{6}}{\sqrt{6}+1}\) \(\Rightarrow\left(a+1\right)\left(\sqrt{6}+1\right)=\sqrt{6}\left(a+\sqrt{a}+1\right)\Leftrightarrow a\sqrt{6}+a+\sqrt{6}+1=a\sqrt{6}+\sqrt{6a}+\sqrt{6}\)

\(\Leftrightarrow a-\sqrt{6a}+1=0\Leftrightarrow a-\sqrt{6a}+\dfrac{6}{4}-\dfrac{2}{4}=0\Leftrightarrow\left(\sqrt{a}-\dfrac{\sqrt{6}}{2}\right)^2=\dfrac{1}{2}\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{a}=\dfrac{\sqrt{6}+1}{2}\\\sqrt{a}=\dfrac{1-\sqrt{6}}{2}\left(L\right)\end{matrix}\right.\) (Do \(\sqrt{a}\ge0\))  \(\Rightarrow a=\dfrac{\left(\sqrt{6}+1\right)^2}{4}=\dfrac{7+2\sqrt{6}}{4}\left(TM\right)\) 

Vậy...

a: ĐKXĐ: \(x\notin\left\{0;1;-1\right\}\)

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

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

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

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

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

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

Dấu '=' xảy ra khi (x-1)²=1

=>x-1=1 hoặc x-1=-1

=>x=0(loại) hoặc x=2(nhận)

Vậy: \(A_{min}=4\) khi x=2

\(a,P=\dfrac{2+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}:\dfrac{2+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{\left(1-a\right)\left(1+a\right)}}\left(-1< a< 1\right)\\ P=\dfrac{2+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}}{2+\sqrt{\left(1-a\right)\left(1+a\right)}}\\ P=\sqrt{1-a}\\ b,a=\dfrac{24}{49}\Leftrightarrow1-a=\dfrac{25}{49}\\ \Leftrightarrow P=\sqrt{1-a}=\sqrt{\dfrac{25}{49}}=\dfrac{5}{7}\\ c,P=2\Leftrightarrow1-a=4\Leftrightarrow a=-3\left(ktm\right)\Leftrightarrow a\in\varnothing\)

Sửa đề: \(P=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{a-1}+\dfrac{\sqrt{a}}{\sqrt{a}+1}\right):\dfrac{\sqrt{a}}{\sqrt{a}+1}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

a) Ta có: \(P=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{a-1}+\dfrac{\sqrt{a}}{\sqrt{a}+1}\right):\dfrac{\sqrt{a}}{\sqrt{a}+1}\)

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

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

\(=\dfrac{a-2\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}+1}{\sqrt{a}}\)

\(=\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}\left(\sqrt{a}-1\right)}\)

\(=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

b) Để \(P< \dfrac{1}{2}\) thì \(P-\dfrac{1}{2}< 0\)

\(\Leftrightarrow\dfrac{\sqrt{a}-1}{\sqrt{a}}-\dfrac{1}{2}< 0\)

\(\Leftrightarrow\dfrac{2\left(\sqrt{a}-1\right)}{2\sqrt{a}}-\dfrac{\sqrt{a}}{2\sqrt{a}}< 0\)

\(\Leftrightarrow\dfrac{2\sqrt{a}-2-\sqrt{a}}{2\sqrt{a}}< 0\)

\(\Leftrightarrow\dfrac{\sqrt{a}-2}{2\sqrt{a}}< 0\)

mà \(2\sqrt{a}>0\forall a\) thỏa mãn ĐKXĐ

nên \(\sqrt{a}-2< 0\)

\(\Leftrightarrow\sqrt{a}< 2\)

hay a<4

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< a< 4\\a\ne1\end{matrix}\right.\)

Vậy: Để \(P< \dfrac{1}{2}\) thì \(\left\{{}\begin{matrix}0< a< 4\\a\ne1\end{matrix}\right.\)

Bạn xem thử tại đây:

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