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Câu a :
ĐKXĐ \(\left\{{}\begin{matrix}x\ne1\\x\ge0\end{matrix}\right.\)
Câu b :
\(A=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{\sqrt{x}-x}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x^3}-1\right)}{x+\sqrt{x}+1}+\dfrac{\sqrt{x}-x}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}+\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\)
\(=\sqrt{x}\left(\sqrt{x}-2\right)\)
Câu c :
\(A=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=4\)
Ta có A=\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\) với x≥ 9, x ∈ R
Để A > 0 \(\Leftrightarrow\) \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\) > 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-2>0\\\sqrt{x}+1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-2< 0\\\sqrt{x}+1>0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}>2\\\sqrt{x}< -1\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}< 2\\\sqrt{x}>-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>4\\x< 1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 4\\x>1\end{matrix}\right.\end{matrix}\right.\)
Kết hợp với ĐKXĐ\(\Rightarrow\) x ∈ ∅
ĐKXĐ: x≥9, x∈R
Ta có:
A= \(\left[\dfrac{1+\sqrt{x}-\sqrt{x}}{1+\sqrt{x}}\right]\):\(\left[\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{x-2\sqrt{x}-3\sqrt{x}+6}\right]\)
= \(\left[\dfrac{1}{1+\sqrt{x}}\right]\):\(\left[\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
=\(\left[\dfrac{1}{1+\sqrt{x}}\right]\):\(\left[\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
=\(\left[\dfrac{1}{1+\sqrt{x}}\right]\):\(\left[\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
=\(\dfrac{1}{1+\sqrt{x}}\):\(\dfrac{1}{\sqrt{x}-2}\)
=\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
1) a) Căn thức có nghĩa \(\Leftrightarrow4-2x\ge0\Leftrightarrow2x\le4\Leftrightarrow x\le2\)
b) Thay x = 2 vào biểu thức A, ta được: \(A=\sqrt{4-2.2}=\sqrt{0}=0\)
Thay x = 0 vào biểu thức A, ta được: \(A=\sqrt{4-2.0}=\sqrt{4}=2\)
Thay x = 1 vào biểu thức A, ta được: \(A=\sqrt{4-2.1}=\sqrt{2}\)
Thay x = -6 vào biểu thức A, ta được: \(A=\sqrt{4-2.\left(-6\right)}=\sqrt{16}=4\)
Thay x = -10 vào biểu thức A, ta được: \(A=\sqrt{4-2.\left(-10\right)}=\sqrt{24}=2\sqrt{6}\)
c) \(A=0\Leftrightarrow\sqrt{4-2x}=0\Leftrightarrow4-2x=0\Leftrightarrow x=2\)
\(A=5\Leftrightarrow\sqrt{4-2x}=5\Leftrightarrow4-2x=25\Leftrightarrow x=\frac{-21}{2}\)
\(A=10\Leftrightarrow\sqrt{4-2x}=10\Leftrightarrow4-2x=100\Leftrightarrow x=-48\)
\(b,P< 0\Leftrightarrow\dfrac{x-1}{\sqrt{x}}< 0\)
Mà: \(\sqrt{x}\ge0\)
\(\Rightarrow x-1< 0\\ \Leftrightarrow x< 1\)
Ta có: \(x=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)
\(\Rightarrow P=\dfrac{x-1}{\sqrt{x}}=\dfrac{\left(\sqrt{3}-1\right)^2-1}{\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\dfrac{\left(\sqrt{3}-1+1\right)\left(\sqrt{3}-1-1\right)}{\sqrt{3}-1}\\ =\dfrac{\sqrt{3}\left(\sqrt{3}-2\right)}{\sqrt{3}-1}\)
\(a.A=\left(\dfrac{1}{1-\sqrt{3}}-\dfrac{1}{1+\sqrt{3}}\right):\dfrac{1}{\sqrt{3}}\)
\(A=\left(\dfrac{1+\sqrt{3}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}-\dfrac{1-\sqrt{3}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}\right):\dfrac{1}{\sqrt{3}}\)
\(A=\left(\dfrac{1+\sqrt{3}-1-\sqrt{3}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}\right):\dfrac{1}{\sqrt{3}}\)
\(A=\left(\dfrac{0}{1-3}\right):\dfrac{1}{\sqrt{3}}\) \(=0:\dfrac{1}{\sqrt{3}}=0\)
b. B được xác định ⇔ x > 0 ; \(x\ne1\)
\(B=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{x-\sqrt{x}}\)
\(B=\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\).
\(B=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(B=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
c. Giả Sử A = \(\dfrac{1}{6}B\)
⇔ 0 = \(\dfrac{1}{6}\times\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
⇔ 0=\(\dfrac{\sqrt{x}-1}{6\sqrt{x}}\)
⇔0 = \(\sqrt{x}-1\)
⇔x = 1(không thỏa mãn)
⇒ A ≠ \(\dfrac{1}{6}B\)
Vậy A ≠ \(\dfrac{1}{6}B\) (Do x không có giá trị nào thỏa mãn)
a) ĐKXĐ: \(2-x^2\ge0\Leftrightarrow\left|x\right|< \sqrt{2}\Leftrightarrow-\sqrt{2}\le x\le\sqrt{2}\)
b) ĐKXĐ: \(5x^2-3>0\Leftrightarrow\left|x\right|>\sqrt{\dfrac{3}{5}}\Leftrightarrow x>\sqrt{\dfrac{3}{5}}\) hoặc \(x< -\sqrt{\dfrac{3}{5}}\)
c) ĐKXĐ: \(-\left(2x-1\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)
d) ĐKXĐ: \(\left(x-1\right)\left(x+2\right)>0\Leftrightarrow x>1\) hoặc \(x< -2\)
2. \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}=\)
\(\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}=\sqrt{4+\sqrt{25}}=\sqrt{4+5}=3\)
3. Ta có: VT=\(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}:\sqrt{a}\right).\left(\dfrac{1-\sqrt{a}}{1-a}\right)=\left[\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}.\dfrac{1}{\sqrt{a}}\right].\left[\dfrac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right]=\dfrac{1+\sqrt{a}+a}{\sqrt{a}}.\dfrac{1}{1+\sqrt{a}}=\dfrac{1+\sqrt{a}+a}{\sqrt{a}+a}=\dfrac{1}{\sqrt{a}+a}+1\)
??? Sao rút gọn rồi ra kì vậy nhờ =="
1,
a.
\(\left[{}\begin{matrix}x-5\sqrt{x}+6\ne0\\\sqrt{x}-2\ne0\\3-\sqrt{x}\ne0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)\ne0\\\sqrt{x}\ne2\\\sqrt{x}\ne3\end{matrix}\right.\)
\(\left[{}\begin{matrix}\sqrt{x}\ne3\\\sqrt{x}\ne2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ne9\\x\ne4\end{matrix}\right.\)
Vậy ĐKXĐ : \(\left[{}\begin{matrix}x\ne9\\x\ne4\end{matrix}\right.\)
ĐKXĐ: x² ≠ 0
⇔ x ≠ 0
Chọn B