Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1) \(VT=\frac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}=VP\)(ĐPCM)
2) \(VT=\text{[}\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a+b-\sqrt{ab}\right)}{\left(\sqrt{a}+\sqrt{b}\right)}-\sqrt{ab}\text{]}.\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}\)
\(=\frac{\left(a+b-\sqrt{ab}-\sqrt{ab}\right)\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}\)\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}=\frac{\left(a-b\right)^2}{\left(a-b\right)^2}=1=VP\)(ĐPCM)
4) \(VT=\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a=VP\)(ĐPCM)
a: \(=\dfrac{\sqrt{a}-1}{\sqrt{a}\left(a-\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{1}\)
\(=a-1\)
b: \(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\left(\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}+\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\dfrac{\sqrt{ab}+b+\sqrt{ab}-b}{\sqrt{a}\left(a-b\right)}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{1}{\sqrt{a}}\)
c: \(=\dfrac{a\sqrt{b}+b}{a-b}\cdot\sqrt{\dfrac{ab+b^2-2b\sqrt{ab}}{a^2+2a\sqrt{b}+b}}\cdot\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\sqrt{\dfrac{b\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(a+\sqrt{b}\right)^2}}\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a+\sqrt{b}}=b\)
3)\(...=\left[\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}\right].\frac{1-xy}{x+xy}\)
= \(\frac{\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}}{1-xy}.\frac{1-xy}{x\left(1+y\right)}\)= \(\frac{2\sqrt{x}+2y\sqrt{x}}{x\left(1+y\right)}=\frac{2\sqrt{x}\left(1+y\right)}{x\left(1+y\right)}=\frac{2}{\sqrt{x}}\)
a: \(=4a-4\sqrt{10a}-9\sqrt{10a}=4a-13\sqrt{10a}\)
b: \(=\sqrt{x}\left(4-\sqrt{2}\right)\cdot\sqrt{x}\left(1-\sqrt{2}\right)\)
\(=x\cdot\left(4-4\sqrt{2}-\sqrt{2}+2\right)\)
\(=\left(6-5\sqrt{2}\right)x\)
c: \(=\dfrac{2}{2x-1}\cdot x\sqrt{5}\cdot\left(2x-1\right)=2x\sqrt{5}\)
a) \(\sqrt{\dfrac{x-2\sqrt{x+1}}{x+2\sqrt{x+1}}}\) = \(\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}\) = \(\dfrac{\sqrt{x-1}}{\sqrt{x+1}}\)
b) \(\dfrac{x-1}{\sqrt{y}-1}\)\(\sqrt{\dfrac{y-2\sqrt{y+1}}{\left(x-1\right)^4}}\)
= \(\dfrac{x-1}{\sqrt{y}-1}\) \(\sqrt{\dfrac{\left(y-1\right)^4}{\left(x-1\right)^4}}\)
= \(\dfrac{x-1}{\sqrt{y}-1}\)\(\dfrac{\left(\sqrt{y}-1\right)^4}{\left(x-1\right)^2}\)
= \(\dfrac{\sqrt{y-1}}{x-1}\)
Chúc bạn học tốt :3
= \(\left[\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\right].\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)
= \(\dfrac{b-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}.\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)
= b-a
a)
\(P=\left(\dfrac{b-a}{\sqrt{b}-\sqrt{a}}-\dfrac{a\sqrt{a}-b\sqrt{b}}{a-b}\right):\dfrac{\left(\sqrt{b}-\sqrt{a}\right)^2+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\)
\(=\left[\sqrt{b}+\sqrt{a}-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right]:\dfrac{b-\sqrt{ab}+a}{\sqrt{a}+\sqrt{b}}\)
\(=\left(\sqrt{b}+\sqrt{a}-\dfrac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\right).\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-a-\sqrt{ab}-b}{\sqrt{a}+\sqrt{b}}.\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\)
\(=\dfrac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}.\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\)\(=\dfrac{\sqrt{ab}}{a-\sqrt{ab}+b}\)
b) \(P=\dfrac{\sqrt{ab}}{a-\sqrt{ab}+b}=\dfrac{\sqrt{ab}}{\left(\sqrt{a}-\dfrac{1}{2}\sqrt{b}\right)^2+\dfrac{3}{4}b}\)
Vì \(\left(\sqrt{a}-\dfrac{1}{2}\sqrt{b}\right)^2+\dfrac{3}{4}b>0;\forall a\ge0;b\ge0;a\ne b\)
\(\sqrt{ab}\ge0\)\(\forall a\ge0;b\ge0\)
\(\Rightarrow P=\dfrac{\sqrt{ab}}{\left(\sqrt{a}-\dfrac{1}{2}\sqrt{b}\right)^2+\dfrac{3}{4}b}\ge0\)
Vậy...
cảm ơn tất cả mọi người