giúp em bài này với ạ
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
=3987654277777777773790123539876542777777777737901235
theo mình là vậy.
ĐK: \(x\ge0,x\ne1\).
\(A=\frac{x\sqrt{x}+1}{x-1}-\frac{x-1}{\sqrt{x}+1}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(=\frac{x-\sqrt{x}+1}{\sqrt{x}-1}-\left(\sqrt{x}-1\right)\)
\(=\frac{x-\sqrt{x}+1-\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\)
\(=\frac{x-\sqrt{x}+1-\left(x-2\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-1}\)
Với \(x=\frac{9}{4}\):
\(A=\frac{\sqrt{\frac{9}{4}}}{\sqrt{\frac{9}{4}}-1}=\frac{\frac{3}{2}}{\frac{3}{2}-1}=3\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}< 1\Leftrightarrow\frac{\sqrt{x}}{\sqrt{x}-1}-1< 0\Leftrightarrow\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}< 0\)
\(\Leftrightarrow\frac{1}{\sqrt{x}-1}< 0\)
\(\Leftrightarrow\sqrt{x}-1< 0\)
\(\Leftrightarrow0\le x< 1\).
\(\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(=\frac{\left(x\sqrt{y}-y\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\frac{\sqrt{xy}.x+xy-xy-\sqrt{xy}.y}{x-y}\)
\(=\frac{\sqrt{xy}.x-\sqrt{xy}.y}{x-y}\)
\(\frac{=\sqrt{xy}.\left(x-y\right)}{x-y}\)
\(=\sqrt{xy}\)
a, \(\sqrt{16x}=8\)ĐK : x >= 0
\(\Leftrightarrow4\sqrt{x}=8\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)
c, \(\sqrt{9\left(x-1\right)}=21\Leftrightarrow3\sqrt{x-1}=21\Leftrightarrow\sqrt{x-1}=7\)ĐK : x >= 1
\(\Leftrightarrow x-1=49\Leftrightarrow x=50\)
ĐK: \(x\ge0,x\ne1\).
\(P=\left(\frac{3}{x-1}+\frac{1}{\sqrt{x}+1}\right)\div\frac{1}{\sqrt{x}+1}\)
\(=\left(\frac{3}{x-1}+\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\left(\sqrt{x}+1\right)\)
\(=\frac{3+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\left(\sqrt{x}+1\right)\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}-1}\)
\(P=\frac{5}{4}\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}-1}=\frac{5}{4}\)(\(x\ge0,x\ne1\))
\(\Rightarrow4\left(\sqrt{x}+2\right)=5\left(\sqrt{x}-1\right)\)
\(\Leftrightarrow\sqrt{x}=13\)
\(\Leftrightarrow x=169\)(tm)
\(M=\frac{x+12}{\sqrt{x}-1}.\frac{1}{P}=\frac{x+12}{\sqrt{x}-1}.\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{x+12}{\sqrt{x}+2}=\frac{x-4}{\sqrt{x}+2}+\frac{16}{\sqrt{x}+2}\)
\(=\sqrt{x}-2+\frac{16}{\sqrt{x}+2}=\sqrt{x}+2+\frac{16}{\sqrt{x}+2}-4\ge2\sqrt{\left(\sqrt{x}+2\right)\frac{16}{\sqrt{x}+2}}-4=4\)
Dấu \(=\)khi \(\sqrt{x}+2=\frac{16}{\sqrt{x}+2}\Leftrightarrow x=4\)(tm)