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a, Ta có : \(x=\sqrt{3+2\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}=4\)
Thay x = 4 => \(\sqrt{x}=2\) vào B ta được :
\(B=\frac{2+5}{2-3}=-7\)
b, Ta có : Với \(x\ge0;x\ne9\)
\(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\frac{4\left(\sqrt{x}-3\right)+2x-\sqrt{x}-13-\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}\)
\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13-x-3\sqrt{x}}{x-9}=\frac{x-25}{x-9}\)
Lại có \(P=\frac{A}{B}\Rightarrow P=\frac{\frac{x-25}{x-9}}{\frac{\sqrt{x}+5}{\sqrt{x}-3}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)
ĐKXĐ: \(x\ne3;x\ne-3\)
Biểu thức = \(\frac{2\sqrt{x}.\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
=\(\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\) = \(\frac{3x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\) =\(\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
=\(\frac{3\sqrt{x}}{\sqrt{x}-3}\)
\(a,E=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\left(Đk:x\ge0;x\ne\pm1\right)\)(Đề như này mới đúng!)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{-\left(3x+9\sqrt{x}-2\sqrt{x}-6\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{2x-2\sqrt{x}+3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{7\sqrt{x}-2-5x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{5\sqrt{x}+2\sqrt{x}-2-5x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\left(5\sqrt{x}-5x\right)+\left(2\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)
Vậy...
\(b,\)Ta có:\(\frac{2-5\sqrt{x}}{\sqrt{x}+3}=\frac{-15+17-5\sqrt{x}}{\sqrt{x}+3}=\frac{\left(-15-5\sqrt{x}\right)+17}{\sqrt{x}+3}=\frac{-5\left(\sqrt{x}+3\right)+17}{\sqrt{x}+3}=-5+\frac{17}{\sqrt{x}+3}\)
Vì \(\sqrt{x}\ge0\forall x\Rightarrow\sqrt{x}+3\ge3\forall x\Rightarrow\frac{17}{\sqrt{x}+3}\le\frac{17}{3}\Rightarrow-5+\frac{17}{\sqrt{x}+3}\le\frac{2}{3}\Rightarrow E\le\frac{2}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow x=0\)
Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
a)ĐKXĐ : x > 0
P = \(\left(\frac{x-1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{\sqrt{x}\left(1+\sqrt{x}\right)}\right)\)
= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\frac{1}{\sqrt{x}}.\left(\sqrt{x}-1+\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\frac{\sqrt{x}-1}{\sqrt{x}}.\left(1-\frac{1}{\sqrt{x}+1}\right)\)
= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\frac{\left(\sqrt{x}-1\right).\sqrt{x}}{\sqrt{x}}\)
= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
Vậy P = \(\frac{\sqrt{x}+1}{\sqrt{x}}\)
b) x = \(\frac{2}{2+\sqrt{3}}=\frac{2\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\frac{2.\left(2-\sqrt{3}\right)}{4-3}=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{3}-1\)
=> P = \(\frac{\sqrt{x}+1}{\sqrt{x}}=\frac{\sqrt{3}-1+1}{\sqrt{3}-1}=\frac{\sqrt{3}}{\sqrt{3}-1}\)
= \(\frac{\sqrt{3}\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3+1}\right)}=\frac{3+\sqrt{3}}{3-1}=\frac{3+\sqrt{3}}{2}\)
c)\(P\sqrt{x}=6\sqrt{x}-3-\sqrt{x-4}\)
\(\Leftrightarrow\frac{\left(\sqrt{x}+1\right)\sqrt{x}}{\sqrt{x}}=6\sqrt{x}-3-\sqrt{x-4}\)
\(\Leftrightarrow\sqrt{x}+1=6\sqrt{x}-3-\sqrt{x-4}\)
\(\Leftrightarrow\sqrt{x-4}=5\sqrt{x-4}\)
Đặt \(\hept{\begin{cases}a=\sqrt{x}\\b=\sqrt{x-4}\end{cases}\Rightarrow a^2+b^2=x-\left(x-4\right)=4}\)
\(\Rightarrow\hept{\begin{cases}a^2-b^2=4\\b=5a-4\end{cases}\Rightarrow\hept{\begin{cases}a^2-\left(5a-4\right)^2=4\left(^∗\right)\\b=5a-4\end{cases}}}\)
Từ (*) <=> a2 -(25a2 -40a + 16 ) =4
<=> -24a2 + 40a - 20 = 0
=> \(\Delta'=-80< 0\)
=> PT vô nghiệm
=> ko tồn tại x thỏa mãn