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\(P=\frac{\sqrt{x}\left(\sqrt{x}-3\right)+2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{x-9}\) dk \(x\ge0;x\ne9\)
\(=\frac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}\)
\(=\frac{3\sqrt{x}-9}{x-9}=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3}{\sqrt{x}+3}\)
b)
\(P=\frac{1}{3}\Leftrightarrow\frac{3}{\sqrt{x}+3}=\frac{1}{3}\Leftrightarrow\sqrt{x}+3=9\Leftrightarrow\sqrt{x}=6\Leftrightarrow x=36\)
vay ......................................
nếu có sai bn thông cảm nha
ĐKXĐ: \(x\ge0;x\ne9\)
\(A=\left(\frac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\frac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)}\)
\(=\left(\frac{2x+6\sqrt{x}+x-3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right).\left(\frac{\sqrt{x}-3}{2\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}-1\right)}=\frac{3}{2\left(\sqrt{x}+3\right)}\)
Ta có: \(A=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right).\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\) ( ĐK: \(x\ne0,\)\(x\ne9,\)\(x\ge3\))
\(\Leftrightarrow A=\frac{\sqrt{x}.\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)}.\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{3\sqrt{x}-x+x+9}{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)}.\frac{2\sqrt{x}+4}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{3\sqrt{x}-9}{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)}.\frac{2\sqrt{x}+4}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{3\left(\sqrt{x}-3\right)}{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)}.\frac{2\sqrt{x}+4}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{3.\left(2\sqrt{x}+4\right)}{\left(9-x\right).\sqrt{x}}\)
\(\Leftrightarrow A=\frac{6\sqrt{x}+12}{9\sqrt{x}-x}\)
a) \(A=\frac{\sqrt{x}\left(\sqrt{x}-3\right)+2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(A=\frac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(A=\frac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3}{\sqrt{x}+3}\)
b) \(A=\frac{1}{3}=>\frac{3}{\sqrt{x}+3}=\frac{1}{3}\)
\(=>\sqrt{x}+3=9\)
\(=>\sqrt{x}=6=>x=36\)
c) \(A\)\(lớn\)\(nhất\)\(< =>\frac{3}{\sqrt{x}+3}lớn\)\(nhất\)
\(=>\sqrt{x}+3\)\(nhỏ\)\(nhất\)
\(Mà\)\(\sqrt{x}+3>=3
\)
\(Do\)\(đó\)\(\sqrt{x}+3=3=>x=0\)
ĐK: x > 0
a) Rút gọn M
M = \(\frac{\sqrt{x}}{x+\sqrt{x}}:\left(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
= \(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}:\left(\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
= \(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}:\left(\frac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
b) \(\frac{1}{M}=\frac{x+\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\frac{1}{\sqrt{x}}+1\ge2+1=3\)
=> M \(\le\)1/3
=> GTLN của M =1/ 3 khi \(\sqrt{x}=\frac{1}{\sqrt{x}}\Leftrightarrow x=1\) thỏa mãn
Vậy max M = 1/3 tại x = 1
1. Trục căn thức ở mẫu:
\(A=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+....+\frac{1}{\sqrt{2001}+\sqrt{2005}}+\frac{1}{\sqrt{2005}+\sqrt{2009}}\)
=\(\frac{\sqrt{5}-1}{4}+\frac{\sqrt{9}-\sqrt{5}}{4}+\frac{\sqrt{13}-\sqrt{9}}{4}+....+\frac{\sqrt{2005}-\sqrt{2001}}{4}+\frac{\sqrt{2009}-\sqrt{2005}}{4}\)
\(=\frac{\sqrt{2009}-1}{4}\)
2/ \(x=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)
=> \(x^3=\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)^3\)
\(=3+2\sqrt{2}+3-2\sqrt{2}+3\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right).\sqrt[3]{3+2\sqrt{2}}.\sqrt[3]{3-2\sqrt{2}}\)
\(=6+3x\)
=> \(x^3-3x=6\)
=> \(B=x^3-3x+2000=6+2000=2006\)
\(A=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+\frac{\sqrt{9}-\sqrt{13}}{9-13}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)
\(A=\frac{1-\sqrt{5}+\sqrt{5}-\sqrt{9}+\sqrt{9}-\sqrt{13}+...+\sqrt{2001}-\sqrt{2005}}{-4}\)
\(A=\frac{1-\sqrt{2005}}{-4}=\frac{\sqrt{2005}-1}{4}\)
ĐK: \(x\ge0;x\ne9\)
\(A=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}+\frac{3x+9}{x-9}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)+2\sqrt{x}\left(\sqrt{x}-3\right)+3x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-3\sqrt{x}+2x-6\sqrt{x}+3x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{-9x+9}{x-9}\)