Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1. \(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}.\dfrac{4}{\sqrt{a}}\right)=\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.4=\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.4=\dfrac{-64\sqrt{a}}{a-4}\)Nếu nhân tu thứ 2 của phép tính là \(\sqrt{a}-\dfrac{4}{\sqrt{a}}\) thì kết quả của phép tính là -16 nha bạn
2.\(\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right).\left(1-\dfrac{1}{\sqrt{a}}\right)=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}.\dfrac{-\left(1-\sqrt{a}\right)}{\sqrt{a}}=\dfrac{-2\sqrt{a}}{\left(1+\sqrt{a}\right)\sqrt{a}}=\dfrac{-2}{1+\sqrt{a}}\)\(\left(a>0,a\ne1\right)\)
Câu 1 :
a ) \(\sqrt{0,36.100}=\sqrt{36}=6\)
b ) \(\sqrt[3]{-0,008}=\sqrt[3]{\left(-0,2\right)^3}=-0,2\)
c ) \(\sqrt{12}+6\sqrt{3}+\sqrt{27}=2\sqrt{3}+6\sqrt{3}+3\sqrt{3}=11\sqrt{3}\)
Câu 2 :
a ) \(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}=a-\sqrt{ab}+b\)
a)\(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right)\):\(\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a-1}\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a-1}\right)}\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{\left(a-1\right)\left(a-4\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{a-1-a+4}\)
\(=\dfrac{1}{\sqrt{a}}.\dfrac{\sqrt{a}-2}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\) ĐKXĐ: \(x>0\) \(a\ne4\) \(a\ne1\)
b) \(Q>0\)
\(\Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}>0\)
mà \(3\sqrt{a}>0\) (Kết hợp ĐKXĐ \(a>0\))
\(\Leftrightarrow\sqrt{a}-2>0\)
\(\Leftrightarrow\sqrt{a}>2\)
\(\Leftrightarrow a>4\) (Thỏa mãn ĐKXĐ)
Vậy \(a>4\) thì \(Q>0\)
____♫ Chúc bạn học tốt ♫____
1: \(A=\dfrac{a+1-2\sqrt{a}}{a+1}:\dfrac{a+1-2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a+1\right)}\)
\(=\sqrt{a}+1\)
2: Khi \(a=2010-2\sqrt{2009}\) thì \(A=\sqrt{2009}-1+1=\sqrt{2009}\)
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
a)
\(\dfrac{\left(\sqrt{x^2+4}-2\right)\left(\sqrt{x^2+4}-2\right)\left(x+\sqrt{x}+1\right)\sqrt{x-2\sqrt{x}+1}}{x\left(x\sqrt{x}-1\right)}\\=\dfrac{\left(\left(\sqrt{x^2+4}\right)^2-4\right)\left(\left(x+\sqrt{x}+1\right)\sqrt{\left(x-1\right)^2}\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{\left(x^2+4-4\right)\left(\left(x+\sqrt{x}+1\right)\left(x-1\right)\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{x^2\left(x^3-1\right)}{x\left(x\sqrt{x}-1\right)}=x^2\sqrt{x}\)
b)
\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{a}{\sqrt{a}}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\right)\left(\dfrac{a-4}{\sqrt{a}}\right)\\ =\dfrac{-8\sqrt{a}}{a-4}\cdot\dfrac{a-4}{\sqrt{a}}=-8\)
c)
\(\left(\dfrac{\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)}\right)\left(1-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}}{\sqrt{a}}-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\\ =\dfrac{2a+2}{a-1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(a+1\right)}{a+1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
d)
\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}^3+1\right)}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\\ =\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)+x+1\\ =x-\sqrt{x}-x-\sqrt{x}+x+1\\ =x-2\sqrt{x}+1\\ =\left(x-1\right)^2\)
A = \(\left(\dfrac{a-1}{\sqrt{a}-1}-2\right)\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}+1\right)=\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-2\right)\left(\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}+1\right)=\left(\sqrt{a}+1-2\right)\left(\sqrt{a}+1\right)=\left(\sqrt{a}-1\right)\left(\sqrt{a}-1\right)=a-1\)
\(B=\left(\dfrac{a\sqrt{a}-a}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}=\left(\dfrac{a\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{a-2}{a+2}=\left(\dfrac{a}{\sqrt{a}}-\dfrac{a-\sqrt{a}+1}{\sqrt{a}}\right)\cdot\dfrac{a-2}{a+2}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}=\dfrac{\left(\sqrt{a}-1\right)\left(a-2\right)}{\sqrt{a}\left(a+2\right)}\)
\(C=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{a}{a-1}\right):\left(\sqrt{a}-\dfrac{\sqrt{a}}{\sqrt{a}+1}\right)=\left(\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{a-1}-\dfrac{a}{a-1}\right):\left(\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)-\sqrt{a}}{\sqrt{a}+1}\right)=\dfrac{\sqrt{a}}{a-1}:\dfrac{a}{\sqrt{a}+1}=\dfrac{\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}+1}{a}=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\)
\(D=\dfrac{a+\sqrt{a}}{\sqrt{a}}+\dfrac{a+4}{\sqrt{a}+2}=\sqrt{a}+1+\dfrac{a+4}{\sqrt{a}+2}=\dfrac{\sqrt{a}\left(\sqrt{a}+2\right)+\sqrt{a}+2+a+4}{\sqrt{a}+2}=\dfrac{a+2\sqrt{a}+\sqrt{a}+2+a+4}{\sqrt{a}+2}=\dfrac{2a+3\sqrt{a}+6}{\sqrt{a}+2}\)
\(E=\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}+\dfrac{1-\sqrt{a}}{a+\sqrt{a}}\right)=\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)+1-\sqrt{a}}{\sqrt{a}\left(\sqrt{a}+1\right)}=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\cdot\dfrac{a-1+1-\sqrt{a}}{\sqrt{a}\left(\sqrt{a}+1\right)}=\dfrac{\left(\sqrt{a}-1\right)\cdot\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}\cdot\sqrt{a}}=\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}}\)
điều kiện xác định là : \(a>0;a\ne1\)
ta có : \(P=\left(\dfrac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\dfrac{\sqrt{a}-2}{a-1}\right)\dfrac{\left(\sqrt{a}-1\right)\left(a-1\right)}{\sqrt{a}}\)
\(P=\left(\dfrac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}-\dfrac{\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\)
\(P=\left(\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\)
\(P=\left(\dfrac{a-\sqrt{a}+2\sqrt{a}-2-\left(a+\sqrt{a}-2\sqrt{a}-2\right)}{\sqrt{a}+1}\right)\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
\(P=\dfrac{a-\sqrt{a}+2\sqrt{a}-2-a-\sqrt{a}+2\sqrt{a}+2}{\sqrt{a}+1}.\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
\(P=\dfrac{2\sqrt{a}}{\sqrt{a}+1}.\dfrac{\sqrt{a}-1}{\sqrt{a}}=\dfrac{2}{\sqrt{a}+1}.\sqrt{a}-1=\dfrac{2\left(\sqrt{a}-1\right)}{\sqrt{a}+1}\)
\(P=\dfrac{2\sqrt{a}-2}{\sqrt{a}+1}\) (biểu thức này luôn phụ thuộc vào biến) (đpcm)
không phụ thuộc vào biến mà bạn