rút gọn pt
\(\left(\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right)\sqrt{\dfrac{1}{b}-\dfrac{1}{a}}\)
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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 = \(\dfrac{1}{12}\)
7 tháng 8 2023
\(I=\dfrac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left[\left(\dfrac{a+\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right)\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right]\)
\(=\dfrac{a+2\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left(\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\cdot\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-b\right)}\)
\(=\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)^2\cdot\left(a-\sqrt{ab}+b\right)}\)
Khi a=16 và b=4 thì \(I=\dfrac{16+4+4\cdot\sqrt{16\cdot4}}{\left(4-2\right)^2\cdot\left(16-\sqrt{16\cdot4}+4\right)}=\dfrac{20+4\cdot8}{4\cdot12}\)
\(=\dfrac{20+32}{48}=\dfrac{52}{48}=\dfrac{13}{12}\)
26 tháng 8 2023
\(B=\left(\dfrac{1}{\sqrt[]{a}-1}-\dfrac{1}{\sqrt[]{a}}\right):\left(\dfrac{\sqrt[]{a}+1}{\sqrt[]{a}-2}-\dfrac{\sqrt[]{a}+2}{\sqrt[]{a}-1}\right)\left(1\right)\)
a) B xác định khi và chỉ khi :
\(\left\{{}\begin{matrix}a\ge0\\\sqrt[]{a}\ne0\\\sqrt[]{a}-1\ne0\\\sqrt[]{a}-2\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a>0\\a\ne1\\a\ne4\end{matrix}\right.\)
b) \(\left(1\right)\Leftrightarrow B=\left(\dfrac{\sqrt[]{a}-\left(\sqrt[]{a}-1\right)}{\sqrt[]{a}\left(\sqrt[]{a}-1\right)}\right):\left(\dfrac{\left(\sqrt[]{a}+1\right)\left(\sqrt[]{a}-1\right)-\left(\sqrt[]{a}+2\right)\left(\sqrt[]{a}-2\right)}{\left(\sqrt[]{a}-1\right)\left(\sqrt[]{a}-2\right)}\right)\)
\(\Leftrightarrow B=\left(\dfrac{1}{\sqrt[]{a}\left(\sqrt[]{a}-1\right)}\right):\left(\dfrac{a-1-\left(a-4\right)}{\left(\sqrt[]{a}-1\right)\left(\sqrt[]{a}-2\right)}\right)\)
\(\Leftrightarrow B=\left(\dfrac{1}{\sqrt[]{a}\left(\sqrt[]{a}-1\right)}\right):\left(\dfrac{3}{\left(\sqrt[]{a}-1\right)\left(\sqrt[]{a}-2\right)}\right)\)
\(\Leftrightarrow B=\left(\dfrac{1}{\sqrt[]{a}\left(\sqrt[]{a}-1\right)}\right).\left(\dfrac{\left(\sqrt[]{a}-1\right)\left(\sqrt[]{a}-2\right)}{3}\right)\)
\(\Leftrightarrow B=\dfrac{\sqrt[]{a}-2}{3\sqrt[]{a}}\)
30 tháng 1
\(A=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{1}{a-b}\left(\dfrac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\right)\)
\(=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{1}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\left(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\right)\)
\(=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{a+2\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
\(=\dfrac{2\sqrt{b}-\sqrt{a}-\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\dfrac{-\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
=-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}
4 tháng 7 2021
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 \(\sqrt{B}>B\) \(\Leftrightarrow\sqrt{\sqrt{1-a}}>\sqrt{1-a}\)
\(\Leftrightarrow\sqrt{1-a}>1-a\)
\(\Leftrightarrow\sqrt{1-a}-\left(1-a\right)>0\)
\(\Leftrightarrow\sqrt{1-a}.\left(1-\sqrt{1-a}\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{1-a}>0\\1-\sqrt{1-a}>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{1-a}< 0\\1-\sqrt{1-a}< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a< 1\\a>0\end{matrix}\right.\\\left\{{}\begin{matrix}a>1\\a< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}0< a< 1\\a>1;a< 0\end{matrix}\right.\)
\(=\dfrac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b}{a-b}\cdot\dfrac{\sqrt{ab}\cdot\sqrt{a-b}}{ab}\)
\(=\dfrac{4ab}{ab}\cdot\dfrac{1}{\sqrt{a-b}}=\dfrac{4}{\sqrt{a-b}}\)