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1. \(x=\frac{1}{9}\) thỏa mãn đk: \(x\ge0;x\ne9\)
Thay \(x=\frac{1}{9}\) vào A ta có:
\(A=\frac{\sqrt{\frac{1}{9}}+1}{\sqrt{\frac{1}{9}}-3}=-\frac{1}{2}\)
2. \(B=...\)
\(B=\frac{3\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{4x+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{3x-9\sqrt{x}+x+3\sqrt{x}-4x-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
3. \(P=A:B=\frac{\sqrt{x}+1}{\sqrt{x}-3}:\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{\sqrt{x}+3}{-6}\)
Vì \(\sqrt{x}+3\ge3\forall x\)\(\Rightarrow\frac{\sqrt{x}+3}{-6}\le\frac{3}{-6}=-\frac{1}{2}\)
hay \(P\le-\frac{1}{2}\)
Dấu "=" xảy ra <=> x=0
ĐK: \(x-9\ne0\Rightarrow x\ne9\)
\(\sqrt{x}\ge0\Rightarrow x\ge0\)
\(x+\sqrt{x}-6\ne0\Rightarrow x+3\sqrt{x}-2\sqrt{x}-6\ne0\Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\ne0\)
\(\Rightarrow\sqrt{x}-2\ne0\Rightarrow\sqrt{x}\ne2\Rightarrow x\ne4\)
ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)
\(A=\left(\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{1}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\left(\frac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\left(\frac{1+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\frac{1+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4\sqrt{x}-12}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-3\right)}\)
2, Với \(x=\frac{25}{16}\)\(\Rightarrow\sqrt{x}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)
\(A=\frac{\frac{5}{4}\left(\frac{5}{4}-2\right)}{4\left(\frac{5}{4}-3\right)}=\frac{5}{4}.\left(-\frac{3}{4}\right):4\left(-\frac{7}{4}\right)=-\frac{15}{16}:-7=\frac{15}{112}\)
\(\orbr{\begin{cases}\orbr{\begin{cases}\\\end{cases}}\\\end{cases}}\)\(\orbr{\begin{cases}\orbr{\begin{cases}\sqrt{x}-2< 0\\\sqrt{x}-3>0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}< 2\\\sqrt{x}>3\end{cases}}\Rightarrow\orbr{\begin{cases}x< 4\\x>9\end{cases}}}\\\orbr{\begin{cases}\sqrt{x}-2>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}>2\\\sqrt{x}< 3\end{cases}\Rightarrow\orbr{\begin{cases}x>4\\x< 9\end{cases}}}}\end{cases}}\)
mọi ng ơi mk viết thiếu dấu ngoặc nha.thiếu ngoặc lownns nha. đóng ngoắc ở trước dấu chia
a) đk: \(\hept{\begin{cases}a\ge0\\a\ne16\end{cases}}\)
Ta có:
\(C=\frac{a}{a-16}-\frac{2}{\sqrt{a}-4}-\frac{2}{\sqrt{a}+4}\)
\(C=\frac{a-2\cdot\left(\sqrt{a}+4\right)-2\cdot\left(\sqrt{a}-4\right)}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}\)
\(C=\frac{a-2\sqrt{a}-8-2\sqrt{a}+8}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}\)
\(C=\frac{a-4\sqrt{a}}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}=\frac{\sqrt{a}}{\sqrt{a}+4}\)
b) Ta có: \(a=9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\)
\(\Rightarrow\sqrt{a}=\sqrt{5}-2\)
Khi đó: \(C=\frac{\sqrt{5}-2}{\sqrt{5}-2+4}=\frac{\sqrt{5}-2}{\sqrt{5}+2}=\frac{\left(\sqrt{5}-2\right)^2}{1}=9-4\sqrt{5}\)
\(C=\frac{a}{a-16}-\frac{2}{\sqrt{a}-4}-\frac{2}{\sqrt{a}+4}\)
a) ĐKXĐ : \(\hept{\begin{cases}a\ge0\\a\ne16\end{cases}}\)
\(=\frac{a}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}-\frac{2\left(\sqrt{a}+4\right)}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}-\frac{2\left(\sqrt{a}-4\right)}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}\)
\(=\frac{a-2\sqrt{a}-8-2\sqrt{a}+8}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}\)
\(=\frac{a-4\sqrt{a}}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}\)
\(=\frac{\sqrt{a}\left(\sqrt{a}-4\right)}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}=\frac{\sqrt{a}}{\sqrt{a}+4}\)
b) Với \(a=9-4\sqrt{5}\)( tmđk )
\(C=\frac{\sqrt{a}}{\sqrt{a}+4}=1-\frac{4}{\sqrt{a}+4}\)
\(C=1-\frac{4}{\sqrt{9-4\sqrt{5}}+4}\)
\(=1-\frac{4}{\sqrt{5-4\sqrt{5}+4}+4}\)
\(=1-\frac{4}{\sqrt{\left(\sqrt{5}-2\right)^2}+4}\)
\(=1-\frac{4}{\left|\sqrt{5}-2\right|+4}\)
\(=1-\frac{4}{\sqrt{5}-2+4}\)
\(=1-\frac{4}{\sqrt{5}+2}\)
\(=\frac{\sqrt{5}+2-4}{\sqrt{5}+2}\)
\(=\frac{\sqrt{5}-2}{\sqrt{5}+2}\)
\(=\frac{\left(\sqrt{5}-2\right)\left(\sqrt{5}-2\right)}{1}=9-4\sqrt{5}\)
a, \(\sqrt{\frac{9}{4}-\sqrt{2}}=\sqrt{\frac{9}{4}-\frac{4\sqrt{2}}{4}}=\sqrt{\frac{9-4\sqrt{2}}{4}}\)
\(=\sqrt{\frac{\left(2\sqrt{2}\right)^2-4\sqrt{2}+1}{4}}=\sqrt{\frac{\left(2\sqrt{2}-1\right)^2}{4}}=\frac{2\sqrt{2}-1}{2}\)
b, \(\sqrt{\frac{129}{16}+\sqrt{2}}=\sqrt{\frac{129+16\sqrt{2}}{16}}=\sqrt{\frac{\left(8\sqrt{2}\right)^2+16\sqrt{2}+1}{16}}\)
\(=\sqrt{\frac{\left(8\sqrt{2}+1\right)^2}{16}}=\frac{8\sqrt{2}+1}{4}\)