Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lời giải:
a.
\(A=\frac{2(\sqrt{x}-4)-3(\sqrt{x}+4)}{(\sqrt{x}-4)(\sqrt{x}+4)}+\frac{2\sqrt{x}+16}{(\sqrt{x}-4)(\sqrt{x}+4)}=\frac{-\sqrt{x}-20}{(\sqrt{x}-4)(\sqrt{x}+4)}+\frac{2\sqrt{x}+16}{(\sqrt{x}-4)(\sqrt{x}+4)}\\ =\frac{\sqrt{x}-4}{(\sqrt{x}-4)(\sqrt{x}+4)}=\frac{1}{\sqrt{x}+4}\)
b. Khi $x=4-2\sqrt{3}=(\sqrt{3}-1)^2\Rightarrow \sqrt{x}=\sqrt{3}-1$
$A=\frac{1}{\sqrt{3}-1+4}=\frac{1}{\sqrt{3}+3}$
\(=>A=\left(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left[\dfrac{\sqrt{x}+1-2}{x-1}\right]\)
\(=>A=\dfrac{\sqrt{x}+1}{\sqrt{x}}.\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}}\)
b,\(=>\dfrac{1}{\sqrt{x}}=\dfrac{1}{2}=>\sqrt{x}=2=>x=\sqrt{2}\left(tm\right)\)
\(A=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\text{x > 0, x ≠ 1}\)
\(A=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{x-1-x+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\) \(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)^2}\)
ĐKXĐ: \(x>0;x\ne1\)
\(A=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{2\left(\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)}-\dfrac{2-x}{x\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{x+2\sqrt{x}}{x\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{\left(x+2\sqrt{x}\right).x.\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x+2\sqrt{x}\right)}=\dfrac{x}{\sqrt{x}-1}\)
b.
\(x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}+1\)
\(\Rightarrow A=\dfrac{4+2\sqrt{3}}{\sqrt{3}+1-1}=\dfrac{4+2\sqrt{3}}{\sqrt{3}}=\dfrac{6+4\sqrt{3}}{3}\)
c.
Để \(\sqrt{A}\) xác định \(\Rightarrow\sqrt{x}-1>0\Rightarrow x>1\)
Ta có:
\(\sqrt{A}=\sqrt{\dfrac{x}{\sqrt{x}-1}}=\sqrt{\dfrac{x}{\sqrt{x}-1}-4+4}=\sqrt{\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}+4}\ge\sqrt{4}=2\)
Dấu "=" xảy ra khi \(\sqrt{x}-2=0\Rightarrow x=4\)
a.
\(A=\left(1-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{1}{\sqrt{a}-1}+\dfrac{1}{\sqrt{a}+1}\right)\)
\(=\left(\dfrac{1-\sqrt{a}}{\sqrt{a}}\right)\left(\dfrac{\sqrt{a}-1+\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\dfrac{1-\sqrt{a}}{\sqrt{a}}.\dfrac{2\sqrt{a}}{a-1}=\dfrac{2\left(1-\sqrt{a}\right)}{a-1}=\dfrac{-2\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\dfrac{-2}{\sqrt{a}+1}\)
b.
\(a-2\sqrt{2}\rightarrow\sqrt{a}=\sqrt{2}-1\)
\(=2-2\sqrt{2}+1\)
=\(\left(\sqrt{2}-1\right)^2\)
\(\rightarrow A=\dfrac{-2}{\sqrt{2}-1+1}=\dfrac{-1}{\sqrt{2}}=\sqrt{2}\)
=>\(A=\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right).\left(\dfrac{\sqrt{a}+1+\sqrt{a}-1}{a-1}\right)\left(a>0,a\ne1\right)\)
\(=\dfrac{\sqrt{a}-1}{\sqrt{a}}.\dfrac{2\sqrt{a}}{a-1}=\dfrac{2}{\sqrt{a}+1}\)
b, \(a=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\) thế vào A
\(=>A=\dfrac{2}{\sqrt{\left(\sqrt{2}-1\right) ^2}+1}=\dfrac{2}{\sqrt{2}}\)
a) Ta có: \(B=\left(\dfrac{x+3\sqrt{x}-3}{x-16}-\dfrac{1}{\sqrt{x}+4}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-4}\)
\(=\left(\dfrac{x+3\sqrt{x}-3-\sqrt{x}+4}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-4}\)
\(=\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\cdot\dfrac{\sqrt{x}-4}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+4}\)
a) \(ĐKXĐ:\left\{{}\begin{matrix}x>0\\x\ne1\\x\ne4\end{matrix}\right.\)
\(\Leftrightarrow B=\dfrac{\sqrt{x}-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow B=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)
\(\Leftrightarrow B=\dfrac{2-\sqrt{x}}{3\sqrt{x}}\)
b) \(x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}+1\) (*)
Thay (*) vào B , ta được : \(B=\dfrac{2-\sqrt{3}-1}{3\sqrt{3}+3}=\dfrac{-\sqrt{3}+1}{3\sqrt{3}+3}\)
a, \(A=\dfrac{\sqrt{x}-1}{x^2-x}:\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right)=\dfrac{\sqrt{x}-1}{x\left(\sqrt{x}\pm1\right)}:\left(\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{1}{x\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1}=1\)
b, Cho A = 1 rồi còn gì, hay đề lỗi bạn ?
\(x=4+2\sqrt{3}=\sqrt{3}^2+2\sqrt{3}+1=\left(\sqrt{3}+1\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}+1\right|=\sqrt{3}+1\)
xem là bài mình làm có sai đâu ko nhé nếu rút gọn ra kq khác thì thay bên trên vào nhé