Bài 8: Rút gọn biểu thức chứa căn bậc hai

_tớ làm chi tiết rồi nhé .

-biểu thức: N=\(\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\left(\dfrac{\sqrt{x}-2}{2}\right)\Leftrightarrow\left(\dfrac{\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{1\cdot\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{\sqrt{x}-2}{2}\Leftrightarrow\dfrac{\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-2}{2}\Rightarrow\dfrac{2\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-2}{2}\Rightarrow\dfrac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-2}{2}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)

\(B^2=\left(2-\sqrt{3}\right)^2.\left(26+15\sqrt{3}\right)+\left(2+\sqrt{3}\right)^2.\left(26-15\sqrt{3}\right)-2\left(4-3\right)\sqrt{26^2-3.15^2}\)

\(B^2=\left(7-4\sqrt{3}\right).\left(26+15\sqrt{3}\right)+\left(7+4\sqrt{3}\right)\left(26-15\sqrt{3}\right)-2\)

\(B^2+2=\left(a-b\right)\left(c+d\right)+\left(a+b\right)\left(c-d\right)=ac+ad-bc-bd+ac-ad+bc-bd=2\left(ac-bd\right)\)\(B^2+2=2.\left(7.26-4.3.15\right)=2\left(182-180\right)\Rightarrow B^2=2\)

\(B>0\Rightarrow B=\sqrt{2}\)

Ta có:

\(x^2+y^2\ge2xy\)

\(\Rightarrow1=\left(x^2+y^2\right)^2\ge4x^2y^2\)

\(\Rightarrow0\le x^2y^2\le\dfrac{1}{4}\)

Ta có: \(M=x^6+y^6=\left(x^2+y^2\right)^3-3x^2y^2\left(x^2+y^2\right)=1-3x^2y^2\)

\(\Rightarrow1\ge M\ge\dfrac{1}{4}\)

Max là 1

\(\left[\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\right].\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)

\(=\left[\dfrac{\sqrt{b}.\sqrt{b}-\sqrt{a}.\sqrt{a}}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right].\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)

=\(\left[\dfrac{b-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right].\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)

\(=\dfrac{\left(b-a\right).\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\)

=b-a

\(A=\sqrt{2+2\sqrt{2}+1}-\sqrt{2-2\sqrt{2}+1}\)
\(A=\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(A=\sqrt{2}+1-\left|\sqrt{2}-1\right|\)
\(A=\sqrt{2}+1-\left(\sqrt{2}-1\right)\) ( vì căn 2 > 1)
\(A=2\)

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

để B nhận giá trị nguyên thì 2x-1\(⋮\)4x-1=>2(2x-1)\(⋮\)4x-1

=>4x-2\(⋮\)4x-1=>-1\(⋮\)4x-1

=>4x-1\(\in\)Ư(-1)=\(\left\{-1;1\right\}\)

*4x-1=-1=>x=0

với x=0 thay vào biểu thức thấy B dương=>nhận giá trị x=0

*4x-1=1=>x=\(\dfrac{1}{2}\)(loại)

vậy với x=0 thì B nguyên dương