Xác định giá trị của biểu thức:
\(A=\left(a+1\right)^{-1}+\left(b+1\right)^{-1}\) với \(a=\left(2+\sqrt{3}\right)^{-1},b=\left(2-\sqrt{3}\right)^{-1}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,ĐK:x>0;x\ne9\\ b,A=\dfrac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\\ A=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{2}{\sqrt{x}+3}\\ c,A>\dfrac{2}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}-\dfrac{2}{5}>0\\ \Leftrightarrow\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{5}>0\\ \Leftrightarrow\dfrac{2-\sqrt{x}}{5\left(\sqrt{x}+3\right)}>0\\ \Leftrightarrow2-\sqrt{x}>0\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}< 2\Leftrightarrow0< x< 4\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1}{\sqrt{x}}\)
b: Thay \(x=3+2\sqrt{2}\) vào P, ta được:
\(P=\dfrac{2\sqrt{2}+2}{\sqrt{2}+1}=2\)
a) Vì khi a>0 và \(a\notin\left\{4;1\right\}\) thì \(\left\{{}\begin{matrix}\sqrt{a}-1\ne0\\\sqrt{a}\ne0\\\sqrt{a}-2\ne0\end{matrix}\right.\)
nên Q xác định
b) Ta có: \(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
Để Q dương thì \(\sqrt{a}-2>0\)
\(\Leftrightarrow a>4\)
Kết hợp ĐKXĐ, ta được: a>4
\(1,\\ a,ĐK:\left\{{}\begin{matrix}x\ge0\\x+5\ge0\end{matrix}\right.\Leftrightarrow x\ge0\\ b,Sửa:B=\left(\sqrt{3}-1\right)^2+\dfrac{24-2\sqrt{3}}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+\dfrac{2\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+2\sqrt{3}=4\\ 3,\\ =\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1+\sqrt{x}}\right]\cdot\dfrac{\sqrt{x}-3+2-2\sqrt{x}}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\left(1-\sqrt{x}\right)\cdot\dfrac{-\sqrt{x}-1}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\dfrac{-\sqrt{x}-1}{\sqrt{x}-3}-2=\dfrac{-\sqrt{x}-1-2\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{-3\sqrt{x}+5}{\sqrt{x}-3}\)
a) Để \(\sqrt{\left|x\right|-1}\) xác định
<=> \(\left|x\right|\ge1\)
<=> \(\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\)
b) Để \(\sqrt{-\left|x+5\right|}\) xác định
<=> \(-\left|x+5\right|\ge0\)
Mà \(\left|x+5\right|\ge0\left(\forall x\right)\)
<=> x + 5 = 0 <=> x = -5
c) Để \(\sqrt{\left|x-1\right|-3}\) xác định
<=> \(\left|x-1\right|\ge3\)
<=> \(\left[{}\begin{matrix}x-1\ge3< =>x\ge4\\x-1\le-3< =>x\le-2\end{matrix}\right.\)
`a)đk:|x|-1>=0`
`<=>|x|>=1`
`<=>` \(\left[ \begin{array}{l}x \ge 1\\x\le -1\end{array} \right.\)
`b)đk:-|x+5|>=0`
`<=>|x+5|<=0`
Mà `|x+5|>=0`
`<=>|x+5|=0`
`<=>x=-5`
`c)đk:|x-1|-3>=0`
`|x-1|>=3`
`<=>` \(\left[ \begin{array}{l}x-1 \ge 3\\x-1 \le -3\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x \ge 4\\x \le -2\end{array} \right.\)
\(A=\dfrac{1}{a+1}+\dfrac{1}{b+1}=\dfrac{a+b+2}{\left(a+1\right)\left(b+1\right)}\)
\(=\dfrac{\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}+2}{\left(\dfrac{1}{2+\sqrt{3}}+1\right).\left(\dfrac{1}{2-\sqrt{3}}+1\right)}\)
\(=\dfrac{\dfrac{2-\sqrt{3}+2+\sqrt{3}+2\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}{\dfrac{3+\sqrt{3}}{2+\sqrt{3}}.\dfrac{3-\sqrt{3}}{2-\sqrt{3}}}=\dfrac{6}{6}=1\)
P/s: ( Nếu sai chỗ nào ns tui vs nha chứ nhiều số quá rối luôn )