\(A=\dfrac{a-\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{a-1}{\sqrt{a}-1}\)
\(=\dfrac{a-1}{\sqrt{a}-1}=\sqrt{a}+1\)
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Bài 8: Rút gọn biểu thức chứa căn bậc hai
Các câu hỏi tương tự
Rút gọn biểu thức:
\(a,\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(b,\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
rút gọn A = \(\frac{\sqrt{a}\left(1-a\right)^2}{1+a}:\left[\left(\frac{1-\sqrt{a^3}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1+\sqrt{a^3}}{1+\sqrt{a}}-\sqrt{a}\right)\right]\)
Đề bài: Rút gọn biểu thức:
1. frac{sqrt{a^2+x^2}+sqrt{a^2-x^2}}{sqrt{a^2+x^2}-sqrt{a^2-x^2}}-sqrt{^{ }frac{a^4}{x^4}-1}
2. left(frac{1-asqrt{a}}{1-sqrt{a}}+sqrt{a}right) . left(frac{1+asqrt{a}}{1+sqrt{a}}-sqrt{a}right)
3. left(frac{3}{sqrt{1+x}}sqrt{1-x}right) : left(frac{3}{sqrt{1-x^2}}+1right)
4. left(sqrt{a}+frac{b-sqrt{ab}}{sqrt{a}+sqrt{b}}right) : left(frac{a}{sqrt{ab+b}}+frac{b}{sqrt{ab}-a}-frac{a+b}{sqrt{ab}}right)
5. frac{sqrt{a}+sqrt{b}-1}{a+sqrt{ab}} + frac{sqrt{a}-sqrt{b}}{2...
Đọc tiếp
Đề bài: Rút gọn biểu thức:
1. \(\frac{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}-\sqrt{a^2-x^2}}-\sqrt{^{ }\frac{a^4}{x^4}-1}\)
2. \(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\) . \(\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)
3. \(\left(\frac{3}{\sqrt{1+x}}\sqrt{1-x}\right)\) : \(\left(\frac{3}{\sqrt{1-x^2}}+1\right)\)
4. \(\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right)\) : \(\left(\frac{a}{\sqrt{ab+b}}+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)\)
5. \(\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}\) + \(\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\) .\(\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
Các bạn giúp tớ nhé, hứa sẽ tick, tớ cảm ơn!!!!
\(\left[\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a^2}}\right]\left[\sqrt{\frac{1}{a^2}-1}-\frac{1}{a}\right]\)
Chứng minh:
\(\left(\frac{2a+1}{\sqrt{a^3-1}}-\frac{\sqrt{a}}{a+\sqrt{a}+1}\right)\left(\frac{1+\sqrt{a^3}}{1+\sqrt{a}}-\sqrt{a}\right)=\sqrt{a}-1\)
cho biểu thức P = \(\left(\frac{\sqrt{a}-1}{3\sqrt{a}+\left(\sqrt{a}-1\right)^2}-\frac{1-3\sqrt{a}+a}{a\sqrt{a}-1}-\frac{1}{\sqrt{a}-1}\right):\frac{a+1}{1-\sqrt{a}}\)
rút gọn P
cho biểu thức E=\(\left(1+\frac{\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)\)
G=\(\left(\frac{a\sqrt{a}-3}{a-2\sqrt{a}-3}-\frac{2\sqrt{a-3}}{\sqrt{a}+1}+\frac{\sqrt{a}+3}{3-\sqrt{a}}\right)\left(\frac{a+8}{a-1}\right)\)
cho biểu thức P=\(\left[\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\right]:\left(\frac{1}{\sqrt{a}+1}+\frac{1}{\sqrt{a}-1}\right)\)
a/ Rút gọn P
b/ Tìm a để \(\frac{1}{P}-\frac{\sqrt{a}+1}{8}\ge1\)