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a/ \(P=\left[1-\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\left[\frac{3-\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}-2}{\sqrt{x}+3}-\frac{9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)
\(=\left(1-\frac{\sqrt{x}}{\sqrt{x}+3}\right):\left[\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)+\left(\sqrt{x}-2\right)^2-9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)
\(=\left(\frac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}+3}\right):\left[\frac{9-x+x-4\sqrt{x}+4-9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)
\(=\frac{3}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{13-4\sqrt{x}-9x}\)
\(=\frac{3\sqrt{x}-6}{13-4\sqrt{x}-9x}\)
b/ \(P=1\Rightarrow\frac{3\sqrt{x}-6}{13-4\sqrt{x}-9x}=1\Rightarrow3\sqrt{x}-6=13-4\sqrt{x}-9x\)
\(\Rightarrow9x+7\sqrt{x}-19=0\)
Mình k biết mình sai chỗ nào nữa, bạn xem giúp mình với
P/s: Không chắc lắm nha!
\(a,B=\frac{10\sqrt{x}+12+\sqrt{x}\left(\sqrt{x}-3\right)-\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x+6\sqrt{x}+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+3}{\sqrt{x}-3}\)
\(b,C=\frac{x-1}{\sqrt{x}-3}:\frac{\sqrt{x}+3}{\sqrt{x}-3}=\frac{x-1}{\sqrt{x}+3}\)
Vì\(\hept{\begin{cases}x\ge0\\\sqrt{x}+3>0\end{cases}\Rightarrow}x-1\ge-1\)
\(\Rightarrow C_{min}=-1\Leftrightarrow x=0\)
Vậy................
Với x = 0 thì C = -1/3 chứ có phải là -1 đâu .
b)
Ta có: \(C=\frac{x-1}{\sqrt{x}+3}=\sqrt{x}-3+\frac{8}{\sqrt{x}+3}=\left(\sqrt{x}+3+\frac{9}{\sqrt{x}+3}\right)-6-\frac{1}{\sqrt{x}+3}\)
\(\ge2\sqrt{\left(\sqrt{x}+3\right).\frac{9}{\sqrt{x}+3}}-6-\frac{1}{3}=-\frac{1}{3}\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}\sqrt{x}+3=\frac{9}{\sqrt{x}+3}\\x=0\end{cases}}\Leftrightarrow x=0\)
Vậy min C = -1/3 tại x =0
a. \(A=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(x+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{\sqrt{x}+3}\)
. \(x=2.\left(4+\sqrt{15}\right).\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4-\sqrt{15}}\)
\(\Rightarrow x=\left(\sqrt{5}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\sqrt{2}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)^2.\left(\sqrt{5}-\sqrt{3}\right)^3\)\(=4\left(\sqrt{5}-\sqrt{3}\right)\)
Thay \(x=4\left(\sqrt{5}-\sqrt{3}\right)\Rightarrow A=\frac{3}{\sqrt{4\left(\sqrt{5}-\sqrt{3}\right)}+3}\)
\(=\frac{3}{2\sqrt{\left(\sqrt{5}-\sqrt{3}\right)}+3}\)
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
a) \(A=\left(\frac{1}{\sqrt{x}+3}-\frac{4}{9-x}\right).\frac{2\sqrt{x}-6}{\sqrt{x}+1}\)
\(A=\left[\frac{\sqrt{x}-3}{x-9}+\frac{4}{x-9}\right].\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}\)
\(A=\frac{\sqrt{x}-3+4}{x-9}.\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}\)
\(A=\frac{\sqrt{x}+1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}\)
\(A=\frac{2}{\sqrt{x}+3}\)
vậy \(A=\frac{2}{\sqrt{x}+3}\)