Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1,ĐKXĐ:x\ge0;x\ne4\)
\(A=\left(1+\frac{2}{\sqrt{x}}\right)\left(\frac{\sqrt{x}-2+\sqrt{x}+2-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
\(A=\left(1+\frac{2}{\sqrt{x}}\right)\left(\frac{2\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
\(A=\left(1+\frac{2}{\sqrt{x}}\right)\left(\frac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
\(A=\left(\frac{\sqrt{x}+2}{\sqrt{x}}\right)\left(\frac{2}{\sqrt{x}+2}\right)\)
\(A=\frac{2}{\sqrt{x}}\)
\(2,A>\frac{1}{2}\)
\(\Leftrightarrow\frac{2}{\sqrt{x}}>\frac{1}{2}\)
\(\Leftrightarrow\frac{2}{\sqrt{x}}-\frac{1}{2}>0\)
\(\Leftrightarrow\frac{4}{2\sqrt{x}}-\frac{\sqrt{x}}{2\sqrt{x}}>0\)
\(\Leftrightarrow\frac{4-\sqrt{x}}{2\sqrt{x}}>0\)
Do \(\sqrt{x}>0\Rightarrow2\sqrt{x}>0\)
\(\Rightarrow4-\sqrt{x}>0\)
\(\Leftrightarrow-\sqrt{x}>-4\)
\(\Leftrightarrow\sqrt{x}< 4\)
\(\Leftrightarrow x< 16\)
Kết hợp với ĐKXĐ thì \(0\le x< 16\)và \(x\ne4\)
\(3,A=-2\sqrt{x}+5\)
\(\Leftrightarrow\frac{2}{\sqrt{x}}=-2\sqrt{x}+5\)
\(\Leftrightarrow\sqrt{x}\left(-2\sqrt{x}+5\right)=2\)
\(\Leftrightarrow-2x+5\sqrt{x}-2=0\)
\(\Leftrightarrow-2x+2.5\sqrt{x}+2.5\sqrt{x}-2=0\)
\(\Leftrightarrow\left(-2x+2.5\sqrt{x}\right)+\left(2.5\sqrt{x}-2\right)=0\)
Đến đây thì mình chịu
Bạn tự giải nốt nhé
HỌC TỐT
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 \(A< 2\Leftrightarrow x>2,25\)
\(B=\frac{2+\sqrt{x}}{x-4\sqrt{x}+4}:\left(\frac{\sqrt{x}+2}{\sqrt{x}}+\frac{1}{\sqrt{x}-2}+\frac{6-x}{x+2\sqrt{x}}\right)\)
\(B=\frac{2+\sqrt{x}}{\left(\sqrt{x}-2\right)^2}:\left(\frac{\sqrt{x}+2}{\sqrt{x}}+\frac{1}{\sqrt{x}-2}+\frac{6-x}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)\)
\(B=\frac{2+\sqrt{x}}{\left(\sqrt{x}-2\right)^2}:\left(\frac{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)+\sqrt{x}\left(\sqrt{x}+2\right)+\left(6-x\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)
\(B=\frac{2+\sqrt{x}}{\left(\sqrt{x}-2\right)^2}:\left(\frac{x\sqrt{x}-8+x+2\sqrt{x}+6\sqrt{x}-12-x\sqrt{x}+2x}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)
\(B=\frac{2+\sqrt{x}}{\left(\sqrt{x}-2\right)^2}:\left(\frac{3x+8\sqrt{x}-20}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)
\(B=\frac{\sqrt{x}\left(2+\sqrt{x}\right)^2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)^2\left(3x+8\sqrt{x}-20\right)}\)
\(B=\frac{\sqrt{x}\left(2+\sqrt{x}\right)^2}{\left(\sqrt{x}-2\right)\left(3x+8\sqrt{x}-20\right)}\)
tới đây mình bí rồi cậu làm giúp mình đi
mại dzo
a, \(\left(\frac{1}{x+2\sqrt{x}}-\frac{1}{\sqrt{x}+2}\right):\frac{1-\sqrt{x}}{x+4\sqrt{x}+4}\)ĐK : x >= 0 ; \(x\ne1\)
\(=\left(\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right).\frac{\left(\sqrt{x}+2\right)^2}{1-\sqrt{x}}=\frac{\sqrt{x}+2}{\sqrt{x}}\)
b, \(F=\frac{5}{2}\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}}=\frac{5}{2}\Rightarrow2\sqrt{x}+4=5\sqrt{x}\Leftrightarrow3\sqrt{x}=4\Leftrightarrow x=\frac{16}{9}\)
ĐK : x > 0 , x khác 1
\(bthuc=\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\left(\sqrt{x}+2\right)^2}{1-\sqrt{x}}=\frac{\sqrt{x}+2}{\sqrt{x}}\)
Để bthuc = 5/2 thì \(\frac{\sqrt{x}+2}{\sqrt{x}}=\frac{5}{2}\Rightarrow2\sqrt{x}+4=5\sqrt{x}\Leftrightarrow3\sqrt{x}=4\Leftrightarrow x=\frac{16}{9}\left(tm\right)\)