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Nếu với \(1>a>0\)thì biểu thức dưới căn không xác định bạn nhé! =====> đề sai rồi.
1/\(A=\dfrac{x^2-2x+2014}{x^2}\)
\(\Leftrightarrow A=\dfrac{2014x^2-2.x.2014+2014^2}{2014x^2}\)
\(\Leftrightarrow A=\dfrac{2013x^2+x^2-2.x.2014+2014^2}{2014x^2}\)
\(\Leftrightarrow A=\dfrac{2013x^2+\left(x-2014\right)^2}{2014x^2}\)
\(\Leftrightarrow A=\dfrac{2013}{2014}+\dfrac{\left(x-2014\right)^2}{2014x^2}\)
Có: \(\left(x-2014\right)^2\ge0\forall x\)
\(2014x^2>0\forall xvìx\ne0\)
\(\Rightarrow\dfrac{\left(x-2014\right)^2}{2014x^2}\ge0\)
\(\Rightarrow\dfrac{2013}{2014}+\dfrac{\left(x-2014\right)^2}{2014x^2}\ge\dfrac{2013}{2014}\)
\(\Rightarrow A\ge\dfrac{2013}{2014}\)
dấu "=" xảy ra khi và chỉ khi x - 2014 =0 <=> x = 2014
Vậy \(min_A=\dfrac{2013}{2014}\Leftrightarrow x=2014\)
2) Ta có:
\(x=\sqrt{a+\sqrt{a^2-1}}+\sqrt{a-\sqrt{a^2-1}}\)
\(\Leftrightarrow x^2=a-\sqrt{a^2-1}+2\sqrt{a-\sqrt{a^2-1}}.\sqrt{a+\sqrt{a^2-1}}+a+\sqrt{a^2-1}\)
\(\Leftrightarrow x^2=2a+2.\sqrt{\left(a-\sqrt{a^2-1}\right)\left(a+\sqrt{a^2-1}\right)}\)
\(\Leftrightarrow x^2=2a+2\sqrt{a^2-\left(a^2-1\right)}\)
\(\Leftrightarrow x^2=2a+2=2\left(a+1\right)\)
\(\Leftrightarrow-x^3=-2\left(a+1\right)x\)
Đặt \(A=x^3-2x^2-2\left(a+1\right)x+4x+2021\)
\(\Leftrightarrow A=x^3-2\left(2a+2\right)-x^3+4a+2021\)
\(\Leftrightarrow A=-4a-4+4a+2021\)
\(\Leftrightarrow A=2017\)
\(B=\frac{-2a\sqrt{a}+2a^2}{\left(\sqrt{a}-\right)\left(a-1\right)}\)
\(C=-x\sqrt{x}+x+\sqrt{x}-1\)
\(D=x-\sqrt{x}+1\)
a)
\(\dfrac{\left(\sqrt{x^2+4}-2\right)\left(\sqrt{x^2+4}-2\right)\left(x+\sqrt{x}+1\right)\sqrt{x-2\sqrt{x}+1}}{x\left(x\sqrt{x}-1\right)}\\=\dfrac{\left(\left(\sqrt{x^2+4}\right)^2-4\right)\left(\left(x+\sqrt{x}+1\right)\sqrt{\left(x-1\right)^2}\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{\left(x^2+4-4\right)\left(\left(x+\sqrt{x}+1\right)\left(x-1\right)\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{x^2\left(x^3-1\right)}{x\left(x\sqrt{x}-1\right)}=x^2\sqrt{x}\)
b)
\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{a}{\sqrt{a}}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\right)\left(\dfrac{a-4}{\sqrt{a}}\right)\\ =\dfrac{-8\sqrt{a}}{a-4}\cdot\dfrac{a-4}{\sqrt{a}}=-8\)
c)
\(\left(\dfrac{\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)}\right)\left(1-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}}{\sqrt{a}}-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\\ =\dfrac{2a+2}{a-1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(a+1\right)}{a+1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
d)
\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}^3+1\right)}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\\ =\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)+x+1\\ =x-\sqrt{x}-x-\sqrt{x}+x+1\\ =x-2\sqrt{x}+1\\ =\left(x-1\right)^2\)
Ta có: \(x^2=\left(\sqrt{a+\sqrt{a^2-1}}+\sqrt{a-\sqrt{a^2-1}}\right)^2\)
\(=a+\sqrt{a^2-1}+2\sqrt{a+\sqrt{a^2-1}}\cdot\sqrt{a-\sqrt{a^2-1}}+a-\sqrt{a^2-1}\)
\(=2a+2\sqrt{a^2-a^2+1}=2a+2=2\left(a+1\right)\)
Suy ra: \(x^3=x^2\cdot x=2\left(a+1\right)x\)
\(4a=2\cdot2a=2\left(2a+2\right)-4=2x^2-4\)
Nên \(P=x^3-2x^2-2\left(a+1\right)x+4a+2021\)
\(=x^3-2x^2-x^3+2x^2-4+2021=2021-4=2017\)