B=\(\left(\dfrac{3}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\dfrac{3}{\sqrt{1-a^2}}+1\right)\)
a) Rút gọn
b) Tìm B khi a=\(\dfrac{\sqrt{3}}{2+\sqrt{3}}\)
c) Tìm a để \(\sqrt{B}>B\)
K
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Y
25 tháng 5 2023
\(a,\) Rút gọn
\(A=\dfrac{3}{\sqrt{7}-2}+\sqrt{\left(\sqrt{7}-3\right)^2}\)
\(=\dfrac{3}{\sqrt{7}-2}+\left|\sqrt{7}-3\right|\)
\(=\dfrac{3}{\sqrt{7}-2}+3-\sqrt{7}\)
\(=\dfrac{3+\left(3-\sqrt{7}\right)\left(\sqrt{7}-2\right)}{\sqrt{7}-2}\)
\(=\dfrac{3+3\sqrt{7}-6-7+2\sqrt{7}}{\sqrt{7}-2}\)
\(=\dfrac{5\sqrt{7}-10}{\sqrt{7}-2}\)
\(=\dfrac{5\left(\sqrt{7}-2\right)}{\sqrt{7}-2}\)
\(=5\)
Vậy \(A=5\)
\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{x-\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x-1}\left(dkxd:x\ge0,x\ne1\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left(\dfrac{x-1}{\sqrt{x}+1}\right)\)
\(=\dfrac{\sqrt{x}.\sqrt{x}-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\dfrac{x-\sqrt{x}}{x-\sqrt{x}}.\left(\sqrt{x}-1\right)\)
\(=\sqrt{x}-1\)
Vậy \(B=\sqrt{x}-1\)
\(b,\) Để \(B< A\) thì \(\sqrt{x}-1< 5\)
\(\Leftrightarrow\sqrt{x}< 6\)
\(\Leftrightarrow x< 36\)
AT
10 tháng 7 2021
a) \(B=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\left(x\ge0,x\ne1\right)\)
\(=\left(\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{x+\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x-1}{x+\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{x+\sqrt{x}+1}{x-1}=\dfrac{1}{x-1}\)
14 tháng 8 2023
a: ĐKXĐ: a>=0; b>=0; ab<>0; a<>1\(M=\dfrac{3\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-1\right)}\)
\(=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{1}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-1\right)}\)
\(=\dfrac{a-2\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\cdot\dfrac{1}{a-1}=\dfrac{1}{a-1}\)
b: M nguyên khi a-1 thuộc {1;-1}
=>a thuộc {2;0}
RH
17 tháng 12 2023
a) ĐKXD: \(\left\{{}\begin{matrix}a>0\\a\ne1\\a\ne4\end{matrix}\right.\)
b) Với \(a>0;a\ne1;a\ne4\), ta có:
\(B=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\\ =\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ =\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ =\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
c)\(B\le\dfrac{1}{3}\rightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}\le\dfrac{1}{3}\rightarrow\dfrac{-2}{\sqrt{a}}\le0\) (đúng với mọi a thoả ĐKXĐ).
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
18 tháng 12 2023
a, ĐKXĐ:
\(\left\{{}\begin{matrix}\left|a\right|>1^2\\\left|a\right|>0\\\left|a\right|>2^2\end{matrix}\right.\Leftrightarrow a>4\)
b,
\(B=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\\ B=\dfrac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\left[\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)\right]}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ B=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\left(a-1\right)-\left(a-4\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ B=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}\\ B=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
\(c,B\le\dfrac{1}{3}\\ \Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}\le\dfrac{1}{3}\\ \Leftrightarrow3\left(\sqrt{a}-2\right)\le3\sqrt{a}\\ \Leftrightarrow\sqrt{a}-2\le\sqrt{a}\\ \Leftrightarrow\sqrt{a}-\sqrt{a}\le2\\ \Leftrightarrow0\le2\left(luôn.đúng\right)\)
Vậy: Với a>4 thì \(B\le\dfrac{1}{3}\)
8 tháng 2 2021
a) ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\b>0\\a\ne b\end{matrix}\right.\)
P = \(\dfrac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}.\left[\left(\dfrac{a+\sqrt{ab}+b-3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right):\dfrac{a-b}{a+\sqrt{ab}+b}\right]\)= \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}.\left[\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}.\dfrac{a+\sqrt{ab}+b}{a-b}\right]\)
= \(\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}.\dfrac{\sqrt{a}-\sqrt{b}}{a-b}\)
= \(\dfrac{1}{a-\sqrt{ab}+b}\)
b) có a = 16 và b = 4 (thoả mãn ĐKXĐ)
Thay a = 16, b =4 vào P có:
P = \(\dfrac{1}{16-\sqrt{16.4}+4}\)= \(\dfrac{1}{12}\)
Vậy tại a =16, b = 4 thì P = \(\dfrac{1}{12}\)
a) ĐKXĐ: \(a>1;a\ne-1\)
\(B=\left(\dfrac{3}{\sqrt{1+a}}+\dfrac{\sqrt{1-a}.\sqrt{1+a}}{\sqrt{1+a}}\right):\dfrac{3+\sqrt{1-a^2}}{\sqrt{1-a^2}}\)
\(\Leftrightarrow B=\dfrac{3+\sqrt{1-a}.\sqrt{1+a}}{\sqrt{1+a}}.\dfrac{\sqrt{1+a}.\sqrt{1-a}}{3+\sqrt{1+a}.\sqrt{1-a}}\)
\(\Leftrightarrow B=\sqrt{1-a}\)
b) Thay a=\(\dfrac{\sqrt{3}}{2+\sqrt{3}}\) vào B ta được:
\(B=\sqrt{1-\dfrac{\sqrt{3}}{2+\sqrt{3}}}\)
\(\Leftrightarrow B\) \(=\sqrt{\dfrac{2+\sqrt{3}-\sqrt{3}}{2+\sqrt{3}}}\)
\(\Leftrightarrow B\) \(=\sqrt{\dfrac{2}{2+\sqrt{3}}}\)
\(\Leftrightarrow B\)\(=\sqrt{\dfrac{4}{4+2\sqrt{3}}}\) \(\Leftrightarrow B\) \(=\dfrac{\sqrt{4}}{\sqrt{3+2\sqrt{3}+1}}\)
\(\Leftrightarrow B=\dfrac{2}{\sqrt{\left(\sqrt{3}+1\right)^2}}\) \(\Leftrightarrow B=\dfrac{2}{\sqrt{3}+1}=\dfrac{2.\left(\sqrt{3}-1\right)}{3-1}=\sqrt{3}-1\)
c) Có \(\sqrt{B}>B\) \(\Leftrightarrow\sqrt{\sqrt{1-a}}>\sqrt{1-a}\)
\(\Leftrightarrow\sqrt{1-a}>1-a\)
\(\Leftrightarrow\sqrt{1-a}-\left(1-a\right)>0\)
\(\Leftrightarrow\sqrt{1-a}.\left(1-\sqrt{1-a}\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{1-a}>0\\1-\sqrt{1-a}>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{1-a}< 0\\1-\sqrt{1-a}< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a< 1\\a>0\end{matrix}\right.\\\left\{{}\begin{matrix}a>1\\a< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}0< a< 1\\a>1;a< 0\end{matrix}\right.\)