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Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
\(P=\frac{2}{\sqrt{x}-1}+\frac{2\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}+\frac{x-10\sqrt{x}+3}{\sqrt{x^3}-1}\)
\(=\frac{2\left(x+\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+x-10\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{2x+2\sqrt{x}+2+2x-2+x-10\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{5x-8\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(5\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{5\sqrt{x}-3}{x+\sqrt{x}+1}\)
Với \(x\ge0;x\ne1\), ta có:
\(P=\frac{2}{\sqrt{x}-1}+\frac{2.\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}+\frac{x-10\sqrt{x}+3}{\sqrt{x^3}-1}\)
\(P=\frac{2.\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}+\frac{2.\left(\sqrt{x}+1\right).\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}+\frac{x-10\sqrt{x}+3}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{2x+2\sqrt{x}+2+2.\left(x-1\right)+x-10\sqrt{x}+3}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{3x-8\sqrt{x}+5+2x-2}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{5x-\sqrt{8x}+3}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{5x-5\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{\left(\sqrt{x}-1\right).\left(5\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}=\frac{5\sqrt{x}-3}{x+\sqrt{x}+1}\)
Vậy với \(x\ge0;x\ne1\) ta có: \(P=\frac{5\sqrt{x}-3}{x+\sqrt{x}+1}\)
a) \(\sqrt{12}-3\sqrt{75}+0,5\sqrt{\left(-6\right)^2\cdot3}\)
\(=2\sqrt{3}-15\sqrt{3}+0,5\sqrt{108}\)
\(=-13\sqrt{3}+3\sqrt{3}\)
\(=-10\sqrt{3}\)
b) \(3\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}-\sqrt{4+2\sqrt{3}}\)
\(=3\left|\sqrt{2}-\sqrt{3}\right|-\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=3\left(\sqrt{3}-\sqrt{2}\right)-\left|\sqrt{3}+1\right|\)
\(=3\sqrt{3}-3\sqrt{2}-\sqrt{3}-1\)
\(=2\sqrt{3}-3\sqrt{2}-1\)
c) \(\left(\frac{2x+1}{x\sqrt{x}-1}-\frac{\sqrt{x}}{x+\sqrt{x}+1}\right)\div\frac{1}{x-2\sqrt{x}+1}\)
\(=\frac{2x+1-\left(\sqrt{x}-1\right)\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\div\frac{1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)^2\)
\(=\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)^2\)
\(=\sqrt{x}-1\)
Câu 3 :
\(ĐKXĐ:x>0\)
\(P=\left(\frac{2}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+2}\right):\frac{2\sqrt{x}}{x+2\sqrt{x}}\)
\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}\cdot\frac{x+2\sqrt{x}}{2\sqrt{x}}\)
\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{2\sqrt{x}}\)
b) Để P = 3
\(\Leftrightarrow\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}=3\)
\(\Leftrightarrow2\sqrt{x}+4+x=6\sqrt{x}\)
\(\Leftrightarrow x-4\sqrt{x}+4=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\)
\(\Leftrightarrow\sqrt{x}-2=0\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\)(tm)
Vậy để \(P=3\Leftrightarrow x=4\)
Câu 1 : Hình như sai đề !! Mik sửa :
\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(A=\left(\frac{x}{x\sqrt{x}-4\sqrt{x}}-\frac{6}{3\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right):\left(\frac{x-4+10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{6}{\sqrt{x}+2}\)
\(\Leftrightarrow A=\frac{-6\left(\sqrt{x}+2\right)}{6\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow A=-\frac{1}{\sqrt{x}-2}\)
b) Để A < 2
\(\Leftrightarrow-\frac{1}{\sqrt{x}-2}< 2\)
\(\Leftrightarrow-1< 2\sqrt{x}-4\)
\(\Leftrightarrow2\sqrt{x}>3\)
\(\Leftrightarrow\sqrt{x}>1,5\)
\(\Leftrightarrow x>2,25\)
Vậy để \(A< 2\Leftrightarrow x>2,25\)
Lời giải:
a)
\(A=\frac{\sqrt{3}-1+\sqrt{3}+1}{(\sqrt{3}+1)(\sqrt{3}-1)}+2-\sqrt{3}=\frac{2\sqrt{3}}{3-1}+2-\sqrt{3}=\sqrt{3}+2-\sqrt{3}=2\)
b)
\(B=\left(\frac{1}{\sqrt{x}(\sqrt{x}-1)}+\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}\right):\frac{\sqrt{x}}{(\sqrt{x}-1)^2}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}.(\sqrt{x}-1)}.\frac{(\sqrt{x}-1)^2}{\sqrt{x}}=\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{x}=\frac{x-1}{x}\)
Bài làm
Rút gọn
\(\left(\frac{1}{\sqrt{x}-1}-\frac{\sqrt{x}}{1-x}\right)\cdot\frac{x-\sqrt{x}}{2\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right)\cdot\frac{\sqrt{x}(\sqrt{x}-1)}{2\sqrt{x}+1}\)
\(=\left(\frac{\sqrt{x}+1}{(\sqrt{x}-1)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\frac{\sqrt{x}(\sqrt{x}-1)}{2\sqrt{x}+1}\)
\(=\frac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}+1}\)
\(=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Tính:
\(\frac{3-\sqrt{3}}{\sqrt{3}+2}+\frac{\sqrt{3}}{\sqrt{3}-2}+\frac{21}{\sqrt{3}}\)
\(=\frac{3-\sqrt{3}}{\sqrt{3}+2}+\frac{\sqrt{3}}{\sqrt{3}-2}+\frac{7\sqrt{3}\cdot\sqrt{3}}{\sqrt{3}}\)
\(=\frac{3-\sqrt{3}}{\sqrt{3}+2}+\frac{\sqrt{3}}{\sqrt{3}-2}+7\sqrt{3}\)
\(=\frac{\left(3-\sqrt{3}\right)\left(\sqrt{3}-2\right)}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}+\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+7\sqrt{3}\)
\(=\frac{3\sqrt{3}-3-6+2\sqrt{3}}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}+\frac{3+2\sqrt{3}}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+7\sqrt{3}\)
\(=\frac{3\sqrt{3}-3-6+2\sqrt{3}+3+2\sqrt{3}}{3-4}+7\sqrt{3}\)
\(=\frac{7\sqrt{3}-6}{-1}+7\sqrt{3}\)
\(=6-7\sqrt{3}+7\sqrt{3}\)
\(=6\)
Bài làm
\(\sqrt{42-10\sqrt{17}}+\sqrt{\left(\sqrt{17}-\sqrt{16}\right)^2}\)
\(=\sqrt{42-10\sqrt{17}}+\left|\sqrt{17}-\sqrt{16}\right|\)
\(=\sqrt{25-10\sqrt{17}+17}+\sqrt{17}-\sqrt{16}\)
\(=\sqrt{\left(5-\sqrt{17}\right)^2}+\sqrt{17}-\sqrt{16}\)
\(=\left|5-\sqrt{17}\right|+\sqrt{17}-\sqrt{16}\)
\(=5-\sqrt{17}+\sqrt{17}-\sqrt{16}\)
\(=5-4\)
\(=1\)