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a) \(A=\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=2\sqrt{7}-3\sqrt{7}+\dfrac{\sqrt{7}\left(\sqrt{7}+1\right)}{\sqrt{7}}-\left|\sqrt{7}+1\right|\)
\(=-\sqrt{7}+\sqrt{7}+1-\sqrt{7}-1=-\sqrt{7}\)
\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-3+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}=\dfrac{2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}\)
\(=\dfrac{8}{\sqrt{x}-3}\)
b) \(A>B\Rightarrow-\sqrt{7}>\dfrac{8}{\sqrt{x}-3}\Rightarrow\dfrac{8}{\sqrt{x}-3}+\sqrt{7}< 0\)
\(\Rightarrow\dfrac{\sqrt{7x}+8-3\sqrt{7}}{\sqrt{x}-3}< 0\)
Ta có: \(\left\{{}\begin{matrix}8=\sqrt{64}\\3\sqrt{7}=\sqrt{63}\end{matrix}\right.\Rightarrow8-3\sqrt{7}>0\Rightarrow8-3\sqrt{7}+\sqrt{7x}>0\)
\(\Rightarrow\sqrt{x}-3< 0\Rightarrow\sqrt{x}< 3\Rightarrow x< 9\Rightarrow0< x< 9\)
\(a,A=\sqrt{27}+\frac{2}{\sqrt{3}-2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
\(=3\sqrt{3}+\frac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\left(\sqrt{3}-1\right)\)
\(=3\sqrt{3}+\frac{2\sqrt{3}+4}{3-4}-\sqrt{3}+1\)
\(=3\sqrt{3}-2\sqrt{3}-4-\sqrt{3}+1\)
\(=-3\)
\(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
b, Ta có \(B< A\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}< -3\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}+3< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-1+3\sqrt{x}}{\sqrt{x}}< 0\)
\(\Leftrightarrow\frac{4\sqrt{x}-1}{\sqrt{x}}< 0\)
\(\Leftrightarrow4\sqrt{x}-1< 0\left(Do\sqrt{x}>0\right)\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{4}\)
\(\Leftrightarrow0< x< \frac{1}{2}\)(Kết hợp ĐKXĐ)
Vậy ...
a) Biểu thức A :
\(A=\sqrt{28}-\sqrt{63}+\frac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\sqrt{7}.\sqrt{4}-\sqrt{7}.\sqrt{9}+\frac{\sqrt{7}\left(\sqrt{7}+1\right)}{\sqrt{7}}-\left|\sqrt{7}+1\right|\)
\(=\sqrt{7}.2-\sqrt{7}.3+\sqrt{7}+1-\sqrt{7}-1\)(do \(\sqrt{7};1>0\))
\(=-\sqrt{7}\)
Biểu thức B :
ĐKXĐ : \(x\ge0;x\ne9\)
Ta có : \(B=\left(\frac{1}{\sqrt{x}+3}+\frac{1}{\sqrt{x}-3}\right).\frac{4\sqrt{x}+12}{\sqrt{x}}\)
\(=\frac{\sqrt{x}-3+\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}\)
\(=\frac{8}{\sqrt{x}-3}\)
a, \(A=\sqrt{28}-\sqrt{63}+\frac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=2\sqrt{7}-3\sqrt{7}+\sqrt{7}+1-\sqrt{7}-1=-\sqrt{7}\)
\(B=\left(\frac{1}{\sqrt{x}+3}+\frac{1}{\sqrt{x}-3}\right)\frac{4\sqrt{x}+12}{\sqrt{x}}\)ĐK : \(x>0;x\ne9\)
\(=\left(\frac{\sqrt{x}-3+\sqrt{x}+3}{x-9}\right)\frac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}=\frac{8\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-3\right)}=\frac{8}{\sqrt{x}-3}\)
b, Ta có : \(A>B\Rightarrow-\sqrt{7}>\frac{8}{\sqrt{x}-3}\Rightarrow-\sqrt{7}>\frac{8}{\sqrt{x}-3}\)
tự giải bft này nhé