Ta có: 

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

Để \(2x+1\) chia hết cho x-1 thì:

\(x-1\in U\left(3\right)=\left\{1;-1;3;-3\right\}\)

Ta có bảng:

Vậy: \(x\in\left\{0;2;-2;4\right\}\)

Bài 1 :

a ) Ta có :

\(\left(x+y\right)^2=x^2+y^2+2xy=20+16=36\)

b ) Ta có :

\(x^2+y^2=\left(x+y\right)^2-2xy=64-30=34\)

dễ 

dễ trong ngoặc đơn

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2) Ta có:

\(\frac{1}{xy}+\frac{2}{x^2+y^2}=2\left(\frac{1}{2xy}+\frac{1}{x^2+y^2}\right)\)

Áp dụng BĐT Schwarz:

\(\frac{1}{2xy}+\frac{1}{x^2+y^2}\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}=\frac{4}{\left(x+y\right)^2}\)

Mà x+y=1 nên suy ra:

\(\frac{1}{2xy}+\frac{1}{x^2+y^2}\ge4\)

\(\Rightarrow2\left(\frac{1}{2xy}+\frac{1}{x^2+y^2}\right)\ge8\)

=>đpcm.

Dấu ''='' xảy ra khi x=y=1/2

giúp minh với nha

\(A=\left(\dfrac{x+8}{x\sqrt{x}+8}-\dfrac{2}{x-2\sqrt{x}+4}\right):\dfrac{1}{\sqrt{x}-1}\left(ĐKXĐ:x\ge0;x\ne1\right)\)

\(=\left[\dfrac{x+8}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}-\dfrac{2}{x-2\sqrt{x}+4}\right].\left(\sqrt{x}-1\right)\)

\(=\dfrac{x+8-2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}.\left(\sqrt{x}-1\right)\)

\(=\dfrac{x+8-2\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}.\left(\sqrt{x}-1\right)\)

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

Vậy \(A=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\) , với  \(x\ne1;x\ge0\)

\(A=\left(\dfrac{x+8}{\left(\sqrt{x}\right)^3+8}-\dfrac{2}{x-2\sqrt{x}+4}\right):\dfrac{1}{\sqrt{x}-1}\\ =\dfrac{x+8-2.\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\times\dfrac{\sqrt{x}-1}{1}\\ =\dfrac{x+8-2\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}.\dfrac{\sqrt{x}-1}{1}\\ =\dfrac{x-2\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}.\left(\sqrt{x}-1\right)\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)

 \(x^2+\frac{1}{x^2}=7\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)^2=49\)

\(\Leftrightarrow x^4+\frac{1}{x^4}+2=49\Leftrightarrow x^4+\frac{1}{x^4}=47\)

\(\Leftrightarrow\left(x^4+\frac{1}{x^4}\right)^2=2209\Leftrightarrow x^8+\frac{1}{x^8}+2=2209\Leftrightarrow x^8+\frac{1}{x^8}=2207\)

đây là phân số hay sao bn