Chương I - Căn bậc hai. Căn bậc ba

$DK:..$

\(...< =>x^2=5-\sqrt{x+5}< =>x^2+x+\dfrac{1}{4}=x+5-\sqrt{x+5}+\dfrac{1}{4}< =>\left(x+\dfrac{1}{2}\right)^2=\left(\sqrt{x+5}-\dfrac{1}{2}\right)^2< =>\left(x+\sqrt{x+5}\right)\left(x-\sqrt{x+5}+1\right)=0=>...\)

Lời giải:

ĐKXĐ:......

a) Ta có:

\(\frac{3+\sqrt{x}}{3-\sqrt{x}}-\frac{3-\sqrt{x}}{3+\sqrt{x}}-\frac{4x}{x-9}=\frac{(3+\sqrt{x})^2-(3-\sqrt{x})^2}{(3-\sqrt{x})(3+\sqrt{x})}-\frac{4x}{x-9}\)

\(=\frac{9+x+6\sqrt{x}-(9+x-6\sqrt{x})}{9-x}-\frac{4x}{x-9}=\frac{-12\sqrt{x}}{x-9}-\frac{4x}{x-9}=\frac{-4\sqrt{x}(3+\sqrt{x})}{(\sqrt{x}-3)(\sqrt{x}+3)}=\frac{4\sqrt{x}}{3-\sqrt{x}}\)

Và:

\(\frac{5}{3-\sqrt{x}}-\frac{4\sqrt{x}+2}{3\sqrt{x}-x}=\frac{5\sqrt{x}}{3\sqrt{x}-x}-\frac{4\sqrt{x}+2}{3\sqrt{x}-x}=\frac{\sqrt{x}-2}{\sqrt{x}(3-\sqrt{x})}\)

Do đó:
\(C=\frac{4\sqrt{x}}{3-\sqrt{x}}: \frac{\sqrt{x}-2}{\sqrt{x}(3-\sqrt{x})}=\frac{4\sqrt{x}}{3-\sqrt{x}}.\frac{\sqrt{x}(3-\sqrt{x})}{\sqrt{x}-2}=\frac{4x}{\sqrt{x}-2}\)

b)

Nếu $C\leq 0$ thì \(|C|=-C\) (không thỏa mãn)

Nếu $C>0$ thì \(|C|=C>0>-C\) (thỏa mãn)

Vậy để \(|C|> -C\) thì \(C>0\Leftrightarrow \frac{4x}{\sqrt{x}-2}>0\Leftrightarrow \sqrt{x}-2>0\) (do \(x>0)\)

\(\Leftrightarrow x> 4\)

Kết hợp đkxđ suy ra điều kiện của $x$ là \(x>4; x\neq 9\)

c)

\(C^2=40C\Leftrightarrow C(C-40)=0\Leftrightarrow \left[\begin{matrix} C=0\\ C=40\end{matrix}\right.\)

Nếu $C=0$ thì \(\frac{4x}{\sqrt{x}-2}=0\Rightarrow x=0\) (không t/m ĐKXĐ)

Nếu \(C=40\Leftrightarrow \frac{4x}{\sqrt{x}-2}=40\Leftrightarrow x=10(\sqrt{x}-2)\)

\(\Rightarrow \sqrt{x}=5\pm \sqrt{5}\Rightarrow x=(5\pm \sqrt{5})^2\)

Cô Akai Haruma giúp e với

điều kiện xác định : \(x>0;x\ne9\)

a) ta có : \(C=\left(\dfrac{3+\sqrt{x}}{3-\sqrt{x}}-\dfrac{3-\sqrt{x}}{3+\sqrt{x}}-\dfrac{4x}{x-9}\right):\left(\dfrac{5}{3-\sqrt{x}}-\dfrac{4\sqrt{x}+2}{3\sqrt{x}-x}\right)\)

b) để \(\left|C\right|>-C\) \(\Leftrightarrow C< 0\) \(\Leftrightarrow\dfrac{4x}{\sqrt{x}-2}< 0\) \(\Leftrightarrow\sqrt{x}-2< 0\Leftrightarrow\sqrt{x}< 2\Leftrightarrow0< x< 4\)

c) để \(C^2=40C\Leftrightarrow C^2-40C=0\Leftrightarrow C\left(C-40\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}C=0\\C=40\end{matrix}\right.\)

+) \(C=0\Leftrightarrow\dfrac{4x}{\sqrt{x}-2}=0\) \(\Leftrightarrow x=0\left(loại\right)\)

+) \(C=40\Leftrightarrow\dfrac{4x}{\sqrt{x}-2}=40\Leftrightarrow x=10\sqrt{x}-20\)

\(\Leftrightarrow x-10\sqrt{x}+20=0\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=5+3\sqrt{5}\left(N\right)\\\sqrt{x}=5-3\sqrt{5}\left(L\right)\end{matrix}\right.\)

ta có : \(\sqrt{x}=5+3\sqrt{5}\Leftrightarrow x=70+30\sqrt{5}\)

vậy ..............................................................................................................................

Mysterious Person giúp e

\(\dfrac{1}{\sqrt{x^2-x+1}}=\dfrac{1}{\sqrt{x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}}}=\dfrac{1}{\sqrt{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}}\)Ta có \(\left(x+\dfrac{1}{2}\right)^2\ge0\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\Rightarrow\sqrt{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\ge\sqrt{\dfrac{3}{4}}\)Vậy \(\dfrac{1}{\sqrt{x^2-x+1}}\)luôn có nghĩa với x\(\in\)R

\(P=\left(\dfrac{a+2\sqrt{a}}{\sqrt{a}+2}-1\right):\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}+1\right)\)

\(=\left[\dfrac{\sqrt{a}\left(\sqrt{a}+2\right)}{\sqrt{a}+2}-1\right]:\)\(\left[\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}+1\right]\)

\(=\dfrac{\sqrt{a}-1}{\sqrt{a}+1}\)

Mysterious Person giúp e

a) điều kiện xác định : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\\x^2-1\ge0\end{matrix}\right.\Leftrightarrow x>1\)

b) ta có : \(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)^2.\dfrac{x^2-1}{2}-\sqrt{x^2-1}\)

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

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

Bài 1:

a) Để \(\sqrt[]{9x^2-6x+1}\) xác định \(\Leftrightarrow9x^2-6x+1\ge0\Leftrightarrow\left(3x-1\right)^2\ge0\) ( luôn đúng )

Vậy \(\sqrt[]{9x^2-6x+1}\) luôn xác định với mọi giá trị của x

b) Để \(\sqrt[]{\dfrac{2x-1}{2-x}}\) xác định \(\Leftrightarrow\dfrac{2x-1}{2-x}\ge0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-1\ge0\\2-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-1\le0\\2-x< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\dfrac{1}{2}\le x< 2\)

Bài 2:

a) \(\sqrt{\left(\dfrac{1}{\sqrt[]{2}}-\dfrac{1}{\sqrt{3}}\right)^2}=\left|\dfrac{1}{\sqrt[]{2}}-\dfrac{1}{\sqrt[]{3}}\right|=\dfrac{1}{\sqrt[]{2}}-\dfrac{1}{\sqrt[]{3}}\)

\(=\dfrac{\sqrt[]{2}}{2}-\dfrac{\sqrt[]{3}}{3}=\dfrac{3\sqrt[]{2}-2\sqrt[]{3}}{6}\)

b) Câu b có sai đề không vậy?

Bài 3: ( Sai đề ) Điều kiện: \(x\le3\)

\(\sqrt{\left(x-3\right)^2}=3-x\Leftrightarrow\left|x-3\right|=3-x\)

\(\Leftrightarrow3-x\ge0\) \(\Leftrightarrow x\le3\)

a) ĐK: \(x\ge1\)

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

\(\Rightarrow x+1=x^2-2x+1\)

\(\Rightarrow3x=x^2\) (1)

Do \(x\ne0\) nên chia mỗi vế của (1) cho x, ta được x = 3 (^t/m)