a) \(B=\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{9-\sqrt{9}+1}{\sqrt{9}-1}=\dfrac{9-3+1}{3-1}=\dfrac{7}{2}\)

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

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

c) \(A=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}>0\Leftrightarrow\sqrt{x}-1>0\left(do.\sqrt{x}+3>0\right)\)

\(\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)

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

Do \(\sqrt{x}>1\Leftrightarrow\sqrt{x}-1>0\)

Áp dụng BĐT Cauchy cho 2 số k âm:

\(B=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+1\ge2\sqrt{\left(\sqrt{x}-1\right).\dfrac{1}{\sqrt{x}-1}}+1=2+1=3\)

Dấu "=" xảy ra \(\Leftrightarrow\left(\sqrt{x}-1\right)^2=1\Leftrightarrow x=4\)

Câu 1:

a) \(A=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}.\left(\dfrac{x+1}{3x}-x-1\right)\right]:\dfrac{x-1}{x}\)

        \(=\left[\dfrac{2}{3x}-\dfrac{2}{3x}+\dfrac{2x}{x+1}+\dfrac{2}{x+1}\right]\dfrac{x}{x-1}\)

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

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

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

        \(=2.\dfrac{x}{x-1}\)

        \(=\dfrac{2x}{x-1}\)

Câu 1: 

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

a) Ta có: \(A=\left(\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\left(\dfrac{x+1}{3x}-x-1\right)\right):\dfrac{x-1}{x}\)

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

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

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

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

\(=\dfrac{2x+2+6x^2+4x-2}{3x\left(x+1\right)}:\dfrac{x-1}{x}\)

\(=\dfrac{6x^2+6x}{3x\left(x+1\right)}:\dfrac{x-1}{x}\)

\(=\dfrac{6x\left(x+1\right)}{3x\left(x+1\right)}:\dfrac{x-1}{x}\)

\(=2\cdot\dfrac{x}{x-1}=\dfrac{2x}{x-1}\)

b) Để A nguyên thì \(2x⋮x-1\)

\(\Leftrightarrow2x-2+2⋮x-1\)

mà \(2x-2⋮x-1\)

nên \(2⋮x-1\)

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

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

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

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

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

1) Khi x = 49 thì:

\(A=\frac{4\sqrt{49}}{\sqrt{49}-1}=\frac{4\cdot7}{7-1}=\frac{28}{6}=\frac{14}{3}\)

2) Ta có:

\(B=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-1}\)

\(B=\frac{\sqrt{x}-1+x+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(B=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

c) \(P=A\div B=\frac{4\sqrt{x}}{\sqrt{x}-1}\div\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{4\sqrt{x}}{\sqrt{x}+1}\)

Ta có: \(P\left(\sqrt{x}+1\right)=x+4+\sqrt{x-4}\)

\(\Leftrightarrow\frac{4\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=x+4+\sqrt{x-4}\)

\(\Leftrightarrow4\sqrt{x}=x+4+\sqrt{x-4}\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\sqrt{x-4}=0\)

Mà \(VT\ge0\left(\forall x\ge0,x\ne1\right)\)

\(\Rightarrow\hept{\begin{cases}\left(\sqrt{x}-2\right)^2=0\\\sqrt{x-4}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}=2\\x-4=0\end{cases}}\Rightarrow x=4\)

Vậy x = 4

a) Rút gọn E Þ đpcm.

b) Điều kiện xác định E là: x ≠    ± 1  

Rút gọn F ta thu được F = 4 Þ đpcm

a: \(A=\sqrt{x-2\sqrt{x}+1}=\left|\sqrt{x}-1\right|\)

Khi x=25 thì A=|5-1|=4

b: \(B=\dfrac{-x+\sqrt{x}+2+\sqrt{x}-2}{x-4}=\dfrac{-x+2\sqrt{x}}{x-4}=\dfrac{-\sqrt{x}}{\sqrt{x}+2}\)

ta thấy rằng 5 phải chia hết cho a tức là 

a(U)5=1,-1;5,-5

vậy a 1,-1,5,-5 thì x có giá trị nguyên 

Bài 5:

\(x^3=18+3\sqrt[3]{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\\ \Leftrightarrow x^3=18+3x\sqrt[3]{1}\\ \Leftrightarrow x^3-3x=18\\ y^3=6+3\sqrt[3]{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\\ \Leftrightarrow y^3=6+3y\sqrt[3]{1}\\ \Leftrightarrow y^3-3y=6\\ P=x^3+y^3-3\left(x+y\right)+1993\\ P=\left(x^3-3x\right)+\left(y^3-3y\right)+1993\\ P=18+6+1993=2017\)

x3=18+33√(9+4√5)(9−4√5)(3√9+4√5+3√9−4√5)⇔x3=18+3x3√1⇔x3−3x=18y3=6+33√(3−2√2)(3+2√2)(3√3+2√2+3√3−2√2)⇔y3=6+3y3√1⇔y3−3y=6P=x3+y3−3(x+y)+1993P=(x3−3x)+(y3−3y)+1993P=18+6+1993=2017

1:

\(A=\dfrac{9}{x-\sqrt{x}-2}+\dfrac{2\sqrt{x}+5}{\sqrt{x}+1}-\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)

\(=\dfrac{9}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}+\dfrac{2\sqrt{x}+5}{\sqrt{x}+1}-\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)

\(=\dfrac{9+\left(2\sqrt{x}+5\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{9+2x-4\sqrt{x}+5\sqrt{x}-10-x+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

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

Để A là số nguyên thì \(\sqrt{x}⋮\sqrt{x}-2\)

=>\(\sqrt{x}-2+2⋮\sqrt{x}-2\)

=>\(\sqrt{x}-2\in\left\{1;-1;2;-2\right\}\)

=>\(\sqrt{x}\in\left\{3;1;4;0\right\}\)

=>\(x\in\left\{9;1;16;0\right\}\)

2:

\(\text{Δ}=\left(-2m-3\right)^2-4m\)

\(=4m^2+12m+9-4m\)

\(=4m^2+5m+9\)

\(=\left(2m\right)^2+2\cdot2m\cdot\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{56}{16}\)

\(=\left(2m+\dfrac{5}{4}\right)^2+\dfrac{56}{16}>=\dfrac{56}{16}>0\)

=>Phương trình luôn có hai nghiệm phân biệt

\(x_1^2+x_2^2=9\)

=>\(\left(x_1+x_2\right)^2-2x_1x_2=9\)

=>\(\left(2m+3\right)^2-2m=9\)

=>\(4m^2+12m+9-2m-9=0\)

=>4m^2+10m=0

=>2m(2m+5)=0

=>m=0 hoặc m=-5/2

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