Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) DK de P xác dinh : \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
b) \(P=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{1-x}+\frac{\left(\sqrt{x}-2\right)^2+3\sqrt{x}-x}{1-\sqrt{x}}\)
\(=\frac{\sqrt{x}}{1-\sqrt{x}}+\frac{-\sqrt{x}+4}{1-\sqrt{x}}\)
\(=\frac{4}{1-\sqrt{x}}\)
c) de P > o thì \(1-\sqrt{x}>0\Rightarrow\sqrt{x}< 1\Rightarrow0< x< 1\)
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
\(A=\left(\frac{2x+1}{\sqrt{x^3}-1}-\frac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\frac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\) \(ĐKXĐ:x\ge0;x\ne1\)
\(A=\left(\frac{2x+1}{\sqrt{x^3}-1}-\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x^3}-1}\right).\left[\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{1+\sqrt{x}}-\sqrt{x}\right]\)
\(A=\left(\frac{2x+1-x+\sqrt{x}}{\sqrt{x^3}-1}\right)\left(x-\sqrt{x}+1-\sqrt{x}\right)\)
\(A=\left(\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(x-2\sqrt{x}+1\right)\)
\(A=\frac{1}{\sqrt{x}-1}\left(\sqrt{x}-1\right)^2\)
\(A=\sqrt{x}-1\)
vậy \(A=\sqrt{x}-1\)
b) Theo câu a ta có : \(A=\sqrt{x}-1\)với \(ĐKXĐ:x\ge0;x\ne1\)
Theo bài ra \(A=3\Leftrightarrow\sqrt{x}-1=3\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\) ( TM ĐKXĐ \(x\ge0;x\ne1\))
vậy \(x=16\)thì \(A=3\)
a) \(A=\left(\sqrt{6}+\sqrt{10}\right).\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=2\sqrt{2}\)
\(B=\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}+1\)
\(=\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+1\)
\(=\frac{4}{x-4}+1\)
\(=\frac{4}{x-4}+\frac{x-4}{x-4}=\frac{x}{x-4}\)
Câu 3 :
\(ĐKXĐ:x>0\)
\(P=\left(\frac{2}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+2}\right):\frac{2\sqrt{x}}{x+2\sqrt{x}}\)
\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}\cdot\frac{x+2\sqrt{x}}{2\sqrt{x}}\)
\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{2\sqrt{x}}\)
b) Để P = 3
\(\Leftrightarrow\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}=3\)
\(\Leftrightarrow2\sqrt{x}+4+x=6\sqrt{x}\)
\(\Leftrightarrow x-4\sqrt{x}+4=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\)
\(\Leftrightarrow\sqrt{x}-2=0\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\)(tm)
Vậy để \(P=3\Leftrightarrow x=4\)
Câu 1 : Hình như sai đề !! Mik sửa :
\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(A=\left(\frac{x}{x\sqrt{x}-4\sqrt{x}}-\frac{6}{3\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right):\left(\frac{x-4+10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{6}{\sqrt{x}+2}\)
\(\Leftrightarrow A=\frac{-6\left(\sqrt{x}+2\right)}{6\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow A=-\frac{1}{\sqrt{x}-2}\)
b) Để A < 2
\(\Leftrightarrow-\frac{1}{\sqrt{x}-2}< 2\)
\(\Leftrightarrow-1< 2\sqrt{x}-4\)
\(\Leftrightarrow2\sqrt{x}>3\)
\(\Leftrightarrow\sqrt{x}>1,5\)
\(\Leftrightarrow x>2,25\)
Vậy để \(A< 2\Leftrightarrow x>2,25\)
B=\(\left(\frac{2x+1-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)\(\left(\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{1+\sqrt{x}}-\sqrt{x}\right)\)ĐK :\(x>0;x\ne1\)
B=\(\frac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\left(x-\sqrt{x}+1-\sqrt{x}\right)\)
B=\(\frac{\left(x+\sqrt{x}+1\right)\left(x-2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
B=\(\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\)
B=\(\sqrt{x}-1\)
b, Để B=3 =>\(\sqrt{x}-1=3\)
<=>\(\sqrt{x}=4\)
<=> x=16 (nhận)
Vậy x =16 thì B=3