Cho biểu thức \(B=\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{ab}}\right):\left(1+\frac{a+b+2ab}{1-ab}\right)\)
Rút gọn biểu thức:
A= \(\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{3\sqrt{ab}}{a\sqrt{a}+b\sqrt{b}}\right).\left[\left(\frac{1}{\sqrt{a}-\sqrt{b}}-\frac{3\sqrt{ab}}{a\sqrt{a}-b\sqrt{b}}\right):\frac{a-b}{a+\sqrt{ab}+b}\right]\)
\(\frac{\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right)}{\left(1+\frac{a+b+2ab}{1-ab}\right)}\)
\(\frac{\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right)}{\left(1+\frac{a+b+2ab}{1-ab}\right)}\)
Tử số: \(\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right)=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(1+\sqrt{ab}\right)+\left(\sqrt{a}-\sqrt{b}\right)\left(1-\sqrt{ab}\right)}{\left(1-\sqrt{ab}\right)\left(1+\sqrt{ab}\right)}\)
\(=\frac{\sqrt{a}+\sqrt{b}+a\sqrt{b}+b\sqrt{a}+\sqrt{a}-\sqrt{b}-a\sqrt{b}+b\sqrt{a}}{1-ab}\)
\(=\frac{2\sqrt{a}+2\sqrt{a}b}{1-ab}=\frac{2\sqrt{a}\left(1+b\right)}{1-ab}\)
Mẫu số: \(1+\frac{a+b+2ab}{1-ab}=\frac{1-ab+a+b+2ab}{1-ab}=\frac{1+ab+a+b}{1-ab}=\frac{\left(a+b\right)\left(b+1\right)}{1-ab}\)
Do đó:
\(\frac{\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right)}{\left(1+\frac{a+b+2ab}{1-ab}\right)}=\frac{\frac{2\sqrt{a}\left(1+b\right)}{1-ab}}{\frac{\left(a+1\right)\left(b+1\right)}{1-ab}}=\frac{2\sqrt{a}}{a+1}\)
Chứng minh các đẳng thức sau:
a) \(\left(1-a^2\right):\left[\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1
+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\right]+1=\frac{2}{1-a}\)
b) \(\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a}{\sqrt{ab}+b}
+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)=\sqrt{b}-\sqrt{a}\)
c) \(\frac{\sqrt{a}+\sqrt{b}-1}{a
+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a
+\sqrt{ab}}\right)=\frac{\sqrt{a}}{a}\)
d) \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2=1\)
Chứng minh các đẳng thức sau:
a) \(\left(1-a^2\right):\left(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\right)+1=\frac{2}{1-a}\)
b) \(\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a}{\sqrt{ab}+b}+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)=\sqrt{b}-\sqrt{a}\)
c) \(\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}.\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a+\sqrt{ab}}\right)=\frac{\sqrt{a}}{a}\)
Rút gọn biểu thức:
A= \(\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{3\sqrt{ab}}{a\sqrt{a}+b\sqrt{b}}\right).\left[\left(\frac{1}{\sqrt{a}-\sqrt{b}}-\frac{3\sqrt{ab}}{a\sqrt{a}-b\sqrt{b}}\right):\frac{a-b}{a+\sqrt{ab}+b}\right]\)
\(A=\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\right)\left[\left(\frac{1}{\sqrt{a}-\sqrt{b}}-\frac{3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right):\frac{a-b}{a+\sqrt{ab}+b}\right]\)
\(A=\left[\frac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\right].\left[\frac{a+b+\sqrt{ab}-3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}.\frac{a+\sqrt{ab}+b}{a-b}\right]\)
\(A=\left[\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\right].\left[\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right]\)
\(A=\frac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}.\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{1}{a-\sqrt{ab}+b}\)
Điều kiện : a, b\(\ge0\)
Rút gọn \(\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right):\left(1+\frac{a+b+2ab}{1-ab}\right)\)
mình thật sự muốn lm cho bạn nhưng nhìu việc quá,,,chỗ nào ko làm đc bạn hỏi mình,,,,mình làm cho
B=\(\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}+\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}-1\right):\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}-\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}+-\right)\)Rút gọn biểu thức
cho biểu thức
P=\(\left(\frac{\sqrt{a}}{\sqrt{ab}}+\frac{\sqrt{ab}+\sqrt{a}}{1-\sqrt{ab}}+1\right):\left(1+\frac{\sqrt{ab}+\sqrt{a}}{1-\sqrt{ab}}-\frac{\sqrt{a}+1}{\sqrt{ab}+1}\right)\\ \\ \\ \)
a) Rút gọn biểu thức
b) Cho \(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}=6\).Tìm giá trị lớn nhất của P
\(\)Cho biểu thức
\(B=\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{3\sqrt{ab}}{a\sqrt{a}+b\sqrt{b}}\right)\left(\left(\frac{1}{\sqrt{a}-\sqrt{b}}-\frac{3\sqrt{ab}}{a\sqrt{a}-b\sqrt{b}}\right):\frac{a-b}{a+\sqrt{ab}+b}\right)\)
a, Rút gọn B
b, Tính B khi a=16, b=4