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\(A=\left[1:\left(1-\frac{\sqrt{a}}{1+\sqrt{a}}\right)\right]\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}-a+\sqrt{a}-1}\right]\)
\(=\left[1:\left(\frac{1+\sqrt{a}-\sqrt{a}}{1+\sqrt{a}}\right)\right]\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\right]\)
\(=\left(1:\frac{1}{1+\sqrt{a}}\right).\frac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)
\(=\left(\sqrt{a}+1\right).\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)
\(=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{a+1}=\frac{a-1}{a+1}\)
\(A=\left[1:\left(\dfrac{1+\sqrt{a}-\sqrt{a}}{\sqrt{a}+1}\right)\right]\cdot\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\right)\)
\(=\dfrac{\sqrt{a}+1}{1}\cdot\dfrac{a+1-2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{\sqrt{a}+1}{1}\cdot\dfrac{\left(a+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)^2}\)
\(=\dfrac{\left(a+1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}-1}\)
\(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}=\dfrac{\left(1-a\sqrt{a}\right)\left(1+\sqrt{a}\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}=\dfrac{1+\sqrt{a}-a\sqrt{a}-a^2}{1-a}=\dfrac{\left(1-a\right)\left(\sqrt{a}+a+1\right)}{1-a}\)
=> \(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}=a+2\sqrt{a}+1=\left(\sqrt{a}+1\right)^2\)
Tương tự \(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}=\left(\sqrt{a}-1\right)^2\)
biểu thức trong dấu ngoặc vuông = \(\left[\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)\right]^2=\left(a-1\right)^2\)
\(E=\dfrac{1-a^2}{\left(a-1\right)^2}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\\\sqrt{y}=b\end{matrix}\right.\), ta có:
\(A=\left[\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\times\dfrac{2}{a+b}+\dfrac{1}{a^2}+\dfrac{1}{b^2}\right]\)\(\times\dfrac{a^3+ab^2+a^2b+b^3}{ab^3+a^3b}\)
\(=\left(\dfrac{b+a}{ab}\times\dfrac{2}{a+b}+\dfrac{b^2+a^2}{a^2b^2}\right)\)\(\times\dfrac{a^2\left(a+b\right)+b^2\left(a+b\right)}{ab\left(a^2+b^2\right)}\)
\(=\dfrac{2ab+b^2+a^2}{a^2b^2}\times\dfrac{\left(a+b\right)\left(a^2+b^2\right)}{ab\left(b^2+a^2\right)}\)
\(=\dfrac{\left(a+b\right)^3}{a^3b^3}\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^3}{\sqrt{\left(xy\right)^3}}\)
Mạn phép ko chép lại đề , mk làm luôn
a) \(D=\left[\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(1+\sqrt{ab}\right)+\left(\sqrt{a}+\sqrt{b}\right)\left(1-\sqrt{ab}\right)}{1-ab}\right]:\dfrac{a+b+2ab+1-ab}{1-ab}\)\(D=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(1+\sqrt{ab}+1-\sqrt{ab}\right)}{1-ab}.\dfrac{1-ab}{a+b+ab+1}\)
\(D=\dfrac{2\left(\sqrt{a}+\sqrt{b}\right)}{\left(b+1\right)\left(a+1\right)}\)
D=A/B
a)
B=1+(a+b+2ab)/(1-ab)=(a+b+ab)/(1-ab)
dk: a,b≥0; a.b≠1
1/B=(1-ab)/(a+b+ab)
A=√a+√b)[(1+√ab)+(1-√ab)]/(1-ab)=2(√a+√b)/(1-ab)
D=2(√a+√b)/[(a+1)(b+1)]
b)
a=2/(√3+2)=2(2-√3)/[(2+√3)(2-√3)]=2(2-√3)=(√3-1)^2
\(A=\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\)
\(=\left(\dfrac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+4\sqrt{a}\right)\cdot\dfrac{a-1}{\sqrt{a}}\)
\(=\left(\dfrac{2\cdot2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+4\sqrt{a}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\)
\(=\left(\dfrac{4\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+4\sqrt{a}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\)
\(=\dfrac{4\sqrt{a}+4\sqrt{a}\cdot\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\)
\(=4\sqrt{a}\cdot\left(1+\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\right)\cdot\dfrac{1}{\sqrt{a}}\)
\(=4\left(1+a-1\right)\)
\(=4a\)
Để \(\sqrt{a}>A\) thì \(\sqrt{a}>4a\)
\(\Leftrightarrow a>\sqrt{4a}\left(đk:a\ge0\right)\)
\(\Leftrightarrow a>2\sqrt{a}\)
\(\Leftrightarrow2\sqrt{a}< a\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{a}< a\left(đk:a\ge0\right)\\2\sqrt{a}< a\left(đk:a< 0\right)\end{matrix}\right.\)