Bài 8: Rút gọn biểu thức chứa căn bậc hai
Các câu hỏi tương tự
Chứng minh:
\(\left(\sqrt{a}-\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a}{\sqrt{ab}+b}-\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)=\sqrt{b}-\sqrt{a}\)
\(\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a}{\sqrt{ab}+b}-\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)=\sqrt{b}-\sqrt{a}\)
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. \(\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}\) + \(\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\) .\(\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
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Chứng minh:
\(\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}=-1\)
Chứng minh:
\(\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}=-1\)
1. Rút gọn
D = \(\frac{\sqrt{1+\frac{2\sqrt{2}}{3}}+\sqrt{1-\frac{2\sqrt{2}}{3}}}{\sqrt{1+\frac{2\sqrt{2}}{3}}-\sqrt{1-\frac{2\sqrt{2}}{3}}}\)
2. Chứng minh rằng:
\(\frac{a\sqrt{b}+b}{a-b}.\sqrt{\frac{ab+b^2-2\sqrt{ab^3}}{a\left(a+2\sqrt{b}\right)+b}}\left(\sqrt{a}+\sqrt{b}\right)=b\) với ( a > b > 0 )
Rút gọn:
A=\(\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}:\frac{2}{\sqrt{a}+\sqrt{b}}\)
B=\(\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{ab}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\left(a-b\right)\left(a\sqrt{a}+a\right)}\)
Rút gọn biểu thức:
\(a,\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(b,\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
Cho a>0, b>0, a≠b, rút gọn biểu thức:
A = \(\left(\frac{\sqrt{a}}{\sqrt{ab}-b}+\frac{\sqrt{b}}{\sqrt{ab}-a}\right):\frac{\sqrt{a}+\sqrt{b}}{a\sqrt{b}-b\sqrt{a}}\)