Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
mk làm luôn
a)\(A=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}-1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right).\)
=\(\frac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}-1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}\)
=\(\frac{\left(3x+3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right).3}\)
=\(\frac{3x+3\sqrt{x}-1}{9\sqrt{x}-3}\)
=
a/ \(A=\frac{\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}}{1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}}\)
\(A=\frac{\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}-1\right)-\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\frac{8\sqrt{x}}{9x-1}}{1-\frac{3\sqrt{x}+1-3}{3\sqrt{x}+1}}\)
\(A=\frac{\frac{3x-4\sqrt{x}+1-3\sqrt{x}-1}{\left(3\sqrt{x}\right)^2-1}-\frac{8\sqrt{x}}{9x-1}}{1-1-\frac{3}{3\sqrt{x}+1}}\)
\(A=\frac{\frac{3x-7\sqrt{x}}{9x-1}-\frac{8\sqrt{x}}{9x-1}}{-\frac{3}{3\sqrt{x}+1}}\)
\(A=\frac{3x-7\sqrt{x}-8\sqrt{x}}{9x-1}\left(\frac{-\left(3\sqrt{x}+1\right)}{3}\right)\)
\(A=\frac{3x-15\sqrt{x}}{9x-1}\left(\frac{-3\sqrt{x}-1}{3}\right)\)
\(A=\frac{3\left(x-3\sqrt{x}\right)}{9x-1}\left(\frac{-3\sqrt{x}-1}{3}\right)\)
\(A=\frac{\left(x-3\sqrt{x}\right)\left(-3\sqrt{x}-1\right)}{9x-1}\)
\(A=\frac{3x\sqrt{x}+8x+3\sqrt{x}}{9x-1}\)
\(A=\frac{3x\sqrt{x}}{9x-1}+\frac{8x}{9x-1}+\frac{3\sqrt{x}}{9x-1}\)
\(A=\frac{x\sqrt{x}}{x-\frac{1}{3}}+\frac{8x}{9x-1}+\frac{\sqrt{x}}{x-\frac{1}{3}}\)
\(A=\frac{\sqrt{x}\left(x-1\right)}{x-\frac{1}{3}}+\frac{\frac{8}{3}x}{x-\frac{1}{3}}\)
\(A=\frac{\sqrt{x}\left(x-1\right)+\frac{8}{3}x}{x-\frac{1}{3}}\)
c/\(P=\frac{\frac{2\left(\sqrt{x}-1\right)}{x\sqrt{x}-1}}{1-\frac{x+2}{x+\sqrt{x}+1}}\)\(\Leftrightarrow P=\frac{2\left(\sqrt{x}-1\right)}{x\sqrt{x}-1}:\frac{\sqrt{x}-1}{x+\sqrt{x}+1}\)
\(\Leftrightarrow\frac{2\left(x+\sqrt{x}+1\right)}{x\sqrt{x}-1}\)
Xét P-1 ta có \(\frac{2x+2\sqrt[]{x}+2-x\sqrt{x}+1}{x\sqrt{x}-1}=\frac{2x+2\sqrt{x}-x\sqrt{x}+3}{x\sqrt{x}-1}\)
với x<1 thì tử dương, mẫu âm, với x>1 thì tử âm và mẫu dương
Từ đó ta luuon có P-1\(\le0\RightarrowĐPCM\)
a/\(\Leftrightarrow x=\frac{5-\sqrt{5}}{1-\sqrt{5}}+\frac{5+\sqrt{5}}{1+\sqrt{5}}-\frac{25-5}{1-5}-1\)
\(\Leftrightarrow x=0+5-1\Leftrightarrow x=4\)
Thay vào B đc \(B=\frac{4+2}{4+2+1}=\frac{6}{7}\)
b/
\(A=\left(\sqrt{x}+2\right):\left(\frac{x+8}{x\sqrt{x}+8}+\frac{\sqrt{x}}{x-2\sqrt{x}+4}-\frac{1}{2+\sqrt{x}}\right)\)
\(=\left(\sqrt{x}+2\right):\left(\frac{x+8+\sqrt{x}\left(\sqrt{x}+2\right)-\left(x-2\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right)\)
\(=\left(\sqrt{x}+2\right):\left(\frac{x+8+x+2\sqrt{x}-x+2\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right)\)
\(=\left(\sqrt{x}+2\right):\left(\frac{x+4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right)\)
\(=\left(\sqrt{x}+2\right):\left[\frac{\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right]\)
\(=\left(\sqrt{x}+2\right):\frac{\sqrt{x}+2}{x-2\sqrt{x}+4}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}{\sqrt{x}+2}\)
\(=x-2\sqrt{x}+4\)
=.= hok tốt!!
Với \(x>0,x\ne1\)
Ta có:\(A=\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\left(\frac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}+2}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\left(\frac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{\left(\sqrt{x}-1\right).\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}}\)
b) lấy kết quả rút gọn của câu A ta được
\(P=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}< 1.=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}-1< 0\)
\(P=\frac{x+\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}=\frac{x+2}{\sqrt{x}-1}\)
đề bài cho x>=0 ta suy ra luôn
\(x+2>0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow x< 1\)
vậy x <1 thì P < 1
\(P=\left(\frac{x+1+\sqrt{x}}{x+1}\right):\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\sqrt{x}\left(x+1\right)-\left(x+1\right)}\right).\)
\(P=\left(\frac{x+1+\sqrt{x}}{x+1}\right):\left(\frac{1}{\sqrt{x-1}}-\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(P=\left(\frac{x+1+\sqrt{x}}{x+1}\right):\left(\frac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(P=\frac{\left(x+\sqrt{x}+1\right)}{\left(x+1\right)}:\frac{\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}-1\right)}=\frac{\left(x+\sqrt{x}+1\right)}{\left(x+1\right)}.\frac{\left(x+1\right)}{\sqrt{x}-1}\)
\(P=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}\)