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Bài 1 :
a )\(A=\frac{3-\sqrt{3}}{\sqrt{3}-1}+\frac{\sqrt{35}-\sqrt{15}}{\sqrt{5}}-\sqrt{28}\)
\(A=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\frac{\sqrt{5}\left(\sqrt{7}-\sqrt{3}\right)}{\sqrt{5}}-\sqrt{28}\)
\(A=\sqrt{3}+\sqrt{7}-\sqrt{3}-\sqrt{28}\)
\(A=\sqrt{7}-\sqrt{28}\)
\(A=\sqrt{7}-2\sqrt{7}=-\sqrt{7}\)
Vậy \(A=-\sqrt{7}\)
b)\(B=\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\frac{\sqrt{a}+\sqrt{b}}{a-b}\left(a,b>0;a\ne b\right)\)
\(B=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}:\frac{\sqrt{a}+\sqrt{b}}{a-b}\)
\(B=\left(\sqrt{a}+\sqrt{b}\right).\frac{a-b}{\sqrt{a}+\sqrt{b}}\)
\(B=a-b\)
Vậy \(B=a-b\left(a,b>0;a\ne b\right)\)
_Minh ngụy_
Bài 2 :
a )\(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\left(x>0\right)\)
\(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
Vậy \(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\left(x>0\right)\)
b) \(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\left(x>0\right)\)
Ta có : \(B>0\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}>0\)
Vì : \(\sqrt{x}\ge0\forall x\Rightarrow\)để \(B>O\)cần \(\sqrt{x}-1>0\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)( thỏa mãn \(x>0\))
Vậy \(x>1\)thì \(B>0\)
_Minh ngụy_
\(a,A=\sqrt{27}+\frac{2}{\sqrt{3}-2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
\(=3\sqrt{3}+\frac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\left(\sqrt{3}-1\right)\)
\(=3\sqrt{3}+\frac{2\sqrt{3}+4}{3-4}-\sqrt{3}+1\)
\(=3\sqrt{3}-2\sqrt{3}-4-\sqrt{3}+1\)
\(=-3\)
\(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
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) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(A=\left(\sqrt{27}+3\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(\Leftrightarrow A=3\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(\Leftrightarrow A=3\left(5-3\right)\)
\(\Leftrightarrow A=6\)
\(B=\left(\frac{1}{1-\sqrt{x}}-\frac{1}{1+\sqrt{x}}\right):\left(\frac{\sqrt{x}}{1-\sqrt{x}}\right)\)
\(\Leftrightarrow B=\frac{1+\sqrt{x}-1+\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\cdot\frac{1-\sqrt{x}}{\sqrt{x}}\)
\(\Leftrightarrow B=\frac{2\sqrt{x}\left(1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\sqrt{x}}\)
\(\Leftrightarrow B=\frac{2}{1+\sqrt{x}}\)
b) Để A = 6B
\(\Leftrightarrow6=6.\frac{2}{1+\sqrt{x}}\)
\(\Leftrightarrow\frac{2}{1+\sqrt{x}}=1\)
\(\Leftrightarrow1+\sqrt{x}=2\)
\(\Leftrightarrow\sqrt{x}=1\)
\(\Leftrightarrow x=1\left(ktm\right)\)
Vậy để \(A=6B\Leftrightarrow x\in\varnothing\)
Ps: Em không chắc lắm đâu ạ, thử sức em mới lớp 8 :)) Sợ nhất mấy cái căn căn này ...
a) Ta có :
\(A=\left(\sqrt{27}+3\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(3\sqrt{3}+3\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=3\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=3\left[\left(\sqrt{5}\right)^2-\left(\sqrt{3}\right)^2\right]\)
\(=3\left(5-3\right)=3\cdot2=6\)
Vậy : \(A=6\)
Ta có : \(B=\left(\frac{1}{1-\sqrt{x}}-\frac{1}{1+\sqrt{x}}\right):\left(\frac{\sqrt{x}}{1-\sqrt{x}}\right)\)
\(=\frac{1+\sqrt{x}-1+\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\cdot\frac{1-\sqrt{x}}{\sqrt{x}}\)
\(=\frac{2\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\cdot\frac{1-\sqrt{x}}{\sqrt{x}}\)
\(=\frac{2}{1+\sqrt{x}}\)
b) Để \(A=6B\)
\(\Leftrightarrow6=6\cdot\frac{2}{1+\sqrt{x}}\)
\(\Leftrightarrow\frac{2}{1+\sqrt{x}}=1\)
\(\Leftrightarrow\frac{2-1-\sqrt{x}}{1+\sqrt{x}}=0\)
\(\Rightarrow\sqrt{x}=1\)
\(\Leftrightarrow x=1\) ( không thỏa mãn ĐKXĐ của biểu thức B )
Vậy : không có giá trị nào thỏa mãn đề bài.