Bài 1: 

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

b: Để A=3 thì 3x-9=x+1

=>2x=10

hay x=5

Bài 2: 

a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)

\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)

b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{3;1;5;-1\right\}\)

đề bài lỗi bn ơi

Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề và hỗ trợ bạn tốt hơn nhé.

1:

\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)

2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)

\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)

\(=\dfrac{20-6}{2}=7\)

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

1: ĐKXĐ: \(a\ge0\)

a: \(A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)

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

\(=\sqrt{x}-1\)

a) \(A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)

                        Đk: \(x>0\) và \(x\ne1\)

\(\Rightarrow A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)

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

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

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

b) Thay \(x=3+2\sqrt{2}\) vào A ta được:

  \(A=\sqrt{3+2\sqrt{2}}-1=\sqrt{\left(\sqrt{2}+1\right)^2}-1\)

      \(=\sqrt{2}+1-1=\sqrt{2}\)

(Vì \(\sqrt{2}+1>0\Rightarrow\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\))

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