Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
đkxđ là \(x\ne1;x>0\)
\(Q=\frac{\sqrt{x}\left(\left(\sqrt{x}\right)^3-1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(Q=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-2\sqrt{x}-1+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(Q=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)
gtnn \(x-\sqrt{x}+1=x-\frac{1}{2}.2.\sqrt{x}+\frac{1}{4}+\frac{3}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
gtnn 3/4
ý c bạn tự làm nha mk chịu
C=\(\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right).\left(\sqrt{x}-1\right)}\right).\frac{\sqrt{x}+1}{\sqrt{x}}\)
C=\(\frac{\left(\sqrt{x}+2\right).\left(x-1\right)-\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
C=\(\frac{x\sqrt{x}-\sqrt{x}+2x-2-\left(x-1\right)}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
C=\(\frac{x-1+x\sqrt{x}-\sqrt{x}}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
C=\(\frac{\left(x-1\right).\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
C=\(\frac{1}{\sqrt{x}}=\frac{\sqrt{x}}{x}\)
\(x=\frac{\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5-1}\right)^2}}{\sqrt{20}}=\frac{2\sqrt{5}}{\sqrt{20}}=1\)
=>P=(1+1-1)2016=1
\(A=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)\(A=\dfrac{2x-6\sqrt{x}+x+\sqrt{x+}3\sqrt{x}+3+3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)\(A=\dfrac{3x-13\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
ĐK x >0
\(PT\Leftrightarrow2x+2\sqrt{x^2-\frac{1}{x^4}}=\frac{4}{x^2}.\)
\(\Leftrightarrow2\sqrt{x^2-\frac{1}{x^4}}=\frac{4}{x^2}-2x\)
\(\Leftrightarrow x^2-\frac{1}{x^4}=\frac{4}{x^4}-\frac{4}{x}+x^2\)(chia cả 2 vế cho 2)
\(\Leftrightarrow\frac{5}{x^4}-\frac{4}{x}=0\Leftrightarrow5-4x^3=0\Leftrightarrow4x^3=5\)
\(\Leftrightarrow x^3=\frac{5}{4}\Leftrightarrow x=\sqrt[3]{\frac{5}{4}}\)
Vậy................................
\(\frac{\sqrt{x}+2}{\sqrt{x}-5}=\frac{\sqrt{x}-5+7}{\sqrt{x}-5}=1+\frac{7}{\sqrt{x}-5}\)
\(=\frac{\left(\sqrt{x}-5\right)+7}{\sqrt{x}-5}=1+\frac{7}{\sqrt{x}-5}\)
Em nghĩ thế thôi chứ sai đúng em ko biết đâu nha