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Bài 2:
a: \(A=2\sqrt{7}-1+\left(\sqrt{7}+4\right)\)
\(=2\sqrt{7}-1+\sqrt{7}+4=3\sqrt{7}+3\)
b: \(B=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(=\sqrt{x-1}+1+1-\sqrt{x-1}=2\)
\(A=\sqrt{x-2+2\sqrt{x-3}}+\sqrt{x+6+6\sqrt{x-3}}\\ A=\sqrt{x-3+2\sqrt{x-3}+1}+\sqrt{x-3+2.3.\sqrt{x-3}+9}\\ A=\sqrt{\left(\sqrt{x-3}+1\right)^2}+\sqrt{\left(\sqrt{x-3}+3\right)^2}\\ A=\left|\sqrt{x-3}+1\right|+\left|\sqrt{x-3}+3\right|\\ A=\sqrt{x-3}+1+\sqrt{x-3}+3\\ A=2\sqrt{x-3}+4\)
\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(A=\)\(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}.\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\)\(\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)\(=\frac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)\(=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
\(A=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
\(\Rightarrow\frac{-5\sqrt{x}+2}{\sqrt{x}+3}=-\frac{1}{7}\Rightarrow-7\left(-5\sqrt{x}+2\right)=\sqrt{x}+3\)
\(\Rightarrow35\sqrt{x}-14=\sqrt{x}+3\)
\(\Rightarrow34\sqrt{x}=17\)
\(\Rightarrow\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\left(tm\right)\)
Vậy với \(x=\frac{1}{4}\)thì \(A=-\frac{1}{7}\)
\(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)+3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6+3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-3x+11\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)\(=\frac{-3x+3\sqrt{x}+8\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{-3\sqrt{x}\left(\sqrt{x}-1\right)+8\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(8-3\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\frac{8-3\sqrt{x}}{\sqrt{x}-3}\)
a: \(P=\dfrac{1-\sqrt{x}}{\left(\sqrt{x}+2\right)\cdot\sqrt{x}}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{1-\sqrt{x}}=\dfrac{\sqrt{x}+2}{\sqrt{x}}\)
b: Để P=4/3 thì 4 căn x=3 căn x+6
=>x=36
a) A có nghĩa\(\Leftrightarrow x-y\ne0\Leftrightarrow x\ne y\)
b) \(A=\frac{x+y-2\sqrt{xy}}{x-y}=\frac{\left(\sqrt{x-\sqrt{y}}\right)^2}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
1)
a)
\(\sqrt{11-6\sqrt{2}}=\sqrt{2-2.3.\sqrt{2}+9}=\left|\sqrt{2}-3\right|=3-\sqrt{2}\)
\(A=3-\sqrt{2}+3+\sqrt{2}=6\)
b)
\(B^2=24+2\sqrt{12^2-4.11}=24+2\sqrt{100}=24+20=44\)
\(B=\sqrt{44}=2\sqrt{11}\)