Bài 8: Rút gọn biểu thức chứa căn bậc hai

\(a.A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}=\dfrac{2}{x+\sqrt{x}+1}\) ( x ≥ 0 ; x # 1 )

\(b.\dfrac{2}{x+\sqrt{x}+1}=\dfrac{2}{x+2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{2}{\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}>0\) \(c.\) \(\dfrac{2}{x+\sqrt{x}+1}\) ≤ \(\dfrac{2}{1}=2\left(x\text{≥ }0\right)\)

⇒ \(A_{Max}=2."="\) ⇔ \(x=0\left(TM\right)\)

a, ĐKXĐ: \(2-4x\ge0\)

\(\Rightarrow x\le\dfrac{1}{2}\)

b, ĐKXĐ: \(\left\{{}\begin{matrix}\dfrac{-3}{x-1}>0\\x^2+4\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1< 0\\x\in R\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 1\\x\in R\end{matrix}\right.\)

(Do ta có: \(x^2+4\ge0\) \(\left(\forall x\in R\right)\))

c, ĐKXĐ: \(4x^2-12x+9>0\) (do biểu thức căn dưới mẫu)

\(\Rightarrow\left(2x-3\right)^2>0\)

\(\Rightarrow x\ne\dfrac{3}{2}\)

Câu (A) đề có sao không nhỉ?

\(B=\dfrac{1}{a^2-\sqrt{x}}:\dfrac{\sqrt{a}+1}{a\sqrt{a}+a+\sqrt{a}}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{x}.\left(a\sqrt{a}-1\right)}.\dfrac{a\sqrt{a}+1+\sqrt{a}}{\sqrt{a}+1}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{a}.\left(\sqrt{a}-1\right).\left(a+\sqrt{a}+1\right)}.\dfrac{\sqrt{a}.\left(a+\sqrt{a}+1\right)}{\sqrt{a}+1}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{a}-1}.\dfrac{1}{\sqrt{a}+1}\)

\(\Leftrightarrow\dfrac{1}{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}\)

\(\Leftrightarrow\dfrac{1}{a-1}\)

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

\(\Leftrightarrow\dfrac{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right).\left(x-\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}+1\right)}+\dfrac{x+1}{\sqrt{x}}\)

\(\Leftrightarrow\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}\)

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

\(\Leftrightarrow\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}\)

\(\Leftrightarrow\dfrac{2\sqrt{x}+x+1}{\sqrt{x}}\)

sử dụng bđt bunhia

Áp dụng BDT Bu-nhi-a-cốp-xki:

\(\left(\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\right)^2\le\left(c+b-c\right)\left(a-c+c\right)=ab\\ \Rightarrow\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\le\sqrt{ab}\)

Đẳng thức xảy ra khi: \(\dfrac{c}{b-c}=\dfrac{a-c}{c}\)

\(\Rightarrow c^2=\left(b-c\right)\left(a-c\right)\\ \Rightarrow c^2=ab-ac-bc+c^2\\ \Rightarrow ab-ac-bc=0\)

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

điều kiện xác định : \(x\ge0;x\ne4\)

ta có : \(B=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\)

\(a.P=\dfrac{2\sqrt{x}-5}{x-5\sqrt{x}+4}+\dfrac{2}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-4}=\dfrac{2\sqrt{x}-5+2\sqrt{x}-8-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-4\right)}=\dfrac{3}{\sqrt{x}-1}\) ( x ≥ 0 ; x # 1 ; x # 16 )

\(b.\) \(P\text{∈}Z\) ⇔ \(\dfrac{3}{\sqrt{x}-1}\text{∈}Z\) ⇔ \(\sqrt{x}-1\text{∈}\left\{1;-1;3;-3\right\}\)

+) \(\sqrt{x}-1=1\text{⇔}x=4\left(TM\right)\)

+) \(\sqrt{x}-1=-1\text{⇔}x=0\left(TM\right)\)

+) \(\sqrt{x}-1=3\text{⇔}x=16\left(KTM\right)\)

+) \(\sqrt{x}-1=-3\text{⇔}vo-nghiem\)

KL............

\(a.\sqrt{4x^2-4x+1}=3\)

⇔ \(\sqrt{\left(2x-1\right)^2}=3\)

⇔ \(\text{ |}2x-1\text{ |}=3\)

⇔ \(2x-1=3or2x-1=-3\)

⇔ \(x=2orx=-1\)

\(b.3\left(\sqrt{x}+2\right)+5=4\sqrt{4x}+1\) ( x ≥ 0 )

⇔ \(3\sqrt{x}+11=8\sqrt{x}+1\)

⇔ \(x=4\left(TM\right)\)

KL.........

a) điều kiện : \(x\ge0;x\ne1\)

ta có : \(Q=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)

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

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

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

b) thế \(x=9\) vào \(Q\) ta có : \(Q=\dfrac{\sqrt{9}}{9+\sqrt{9}+1}=\dfrac{3}{13}\)

c) ta có : \(Q=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\Leftrightarrow\sqrt{x}=Q\left(x+\sqrt{x}+1\right)\)

\(\Leftrightarrow Qx+\left(Q-1\right)\sqrt{x}+Q=0\)

vì phương trình này luôn có nghiệm \(\Rightarrow\Delta\ge0\)

\(\Rightarrow\left(Q-1\right)^2-4Q^2\ge0\Leftrightarrow Q^2-2Q+1-4Q^2\ge0\)

\(\Leftrightarrow\left(Q+1\right)\left(1-3Q\right)\ge0\) \(\Leftrightarrow-1\le Q\le\dfrac{1}{3}\)

\(\Rightarrow Q_{max}=\dfrac{1}{3}\) dấu "=" xảy ra khi \(\sqrt{x}=\dfrac{1-Q}{2Q}=\dfrac{1-\dfrac{1}{3}}{\dfrac{2}{3}}=1\Leftrightarrow x=1\)

nhận xét : ta thấy \(Q=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\ge0\)

\(\Rightarrow Q_{min}=0\) dấu "=" xảy ra khi \(\sqrt{x}=0\Leftrightarrow x=0\)

vậy \(Q_{min}=0\) khi \(x=0\) ; \(Q_{max}=\dfrac{1}{3}\) khi \(x=1\)