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Ta có: \(\dfrac{3\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}+1\right)}\)
\(=\dfrac{3\left(\sqrt{5}-1\right)}{6+2\sqrt{5}}\)
\(=\dfrac{3\left(\sqrt{5}-1\right)\left(6-2\sqrt{5}\right)}{\left(6-2\sqrt{5}\right)\left(6+2\sqrt{5}\right)}\)
\(=\dfrac{3\left(6\sqrt{5}-10-6+2\sqrt{5}\right)}{16}\)
\(=\dfrac{3\left(8\sqrt{5}-16\right)}{16}\)
\(=\dfrac{3\cdot\left(\sqrt{5}-2\right)}{2}\)
\(\left(2+\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right)\cdot\left(2-\dfrac{5+\sqrt{5}}{\sqrt{5}+1}\right)\\ =\left(2+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\cdot\left(2-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)\\ =\left(2+\sqrt{5}\right)\cdot\left(2-\sqrt{5}\right)\\ =2^2-\sqrt{5}^2=4-5=-1\)
Ta có: \(\left(2+\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right)\cdot\left(2-\dfrac{5+\sqrt{5}}{\sqrt{5}+1}\right)\)
\(=\left(2+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\cdot\left(2-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)\)
\(=\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)\)
=4-5=-1
a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có : \(P=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(x-1\right)^2}{2}\)
\(=\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{x-2\sqrt{x}+\sqrt{x}-2-x-2\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{-2\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)}=-\sqrt{x}\left(\sqrt{x}-1\right)\)
b, Ta có : \(P=-x+\sqrt{x}=-x+\dfrac{2.\sqrt{x}.1}{2}-\dfrac{1}{4}+\dfrac{1}{4}\)
\(=-\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
Vậy \(Max=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{4}\)
Lời giải:
ĐKXĐ: $x\geq 0; x\neq 1$
a.
\(A=\frac{(\sqrt{x}-2)(x-1)}{2}-\frac{(\sqrt{x}+2)(1-x)^2}{2(x+2\sqrt{x}+1)}=\frac{(\sqrt{x}-2)(x-1)}{2}-\frac{(\sqrt{x}+2)(\sqrt{x}-1)^2(\sqrt{x}+1)^2}{2(\sqrt{x}+1)^2}\)
\(=\frac{(\sqrt{x}-2)(x-1)}{2}-\frac{(\sqrt{x}+2)(\sqrt{x}-1)^2}{2}=\frac{2\sqrt{x}-2x}{2}=\sqrt{x}-x\)
b.
$\sqrt{x}-x=\frac{1}{4}-(x-\sqrt{x}+\frac{1}{4})$
$=\frac{1}{4}-(\sqrt{x}-\frac{1}{2})^2$
$\leq \frac{1}{4}$
Vậy GTLN của biểu thức là $\frac{1}{4}$. Giá trị này đạt tại $\sqrt{x}-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}$ (thỏa đkxđ)
\(\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\left(2-\sqrt{5}\right)-\left(\sqrt{5}-1\right)\)
\(=2-\sqrt{5}-\sqrt{5}+1\)
\(=3-2\sqrt{5}\)
\(\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}=|2-\sqrt{5}|-|\sqrt{5}-1|.\)
\(=\sqrt{5}-2-\sqrt{5}+1\)(Vì \(2=\sqrt{4}< \sqrt{5};1=\sqrt{1}< \sqrt{5}\))
\(=-1\)
\(\sqrt{5+2\sqrt{6}}+\sqrt{10-4\sqrt{6}}=\sqrt{2+2.\sqrt{2}\sqrt{3}+3}+\sqrt{4-2.2.\sqrt{6}+6}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{6}\right)^2}\)
\(=|\sqrt{2}+\sqrt{3}|+|2-\sqrt{6}|\)
\(=\sqrt{2}+\sqrt{3}+\sqrt{6}-2\)( Vì \(\sqrt{6}>\sqrt{4}=2\))
\(\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6+1}\right)\left(5-2\sqrt{2}-\sqrt{3}\right)\)
\(=\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\sqrt{6+1}\left(5-2\sqrt{2}-\sqrt{3}\right)\)
\(=\sqrt{5}\left(1+\sqrt{2}\right)\left(1+\sqrt{3}\right)\left(5-2\sqrt{2}-\sqrt{3}\right)\)
\(=\sqrt{5}\left(\sqrt{6}+\sqrt{2}+\sqrt{3}+1\right)\left(5-2\sqrt{2}-\sqrt{3}\right)\)
\(=\sqrt{5}\left(2\sqrt{6}-2\right)\)
\(=2\sqrt{30}-2\sqrt{5}\)
a) Ta có: \(P=\left(\dfrac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}\right):\left(1+\dfrac{x+2xy+y}{1-xy}\right)\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}:\dfrac{1-xy+x+2xy+y}{1-xy}\)
\(=\dfrac{2\sqrt{x}\left(y+1\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}\cdot\dfrac{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}{x+xy+y+1}\)
\(=\dfrac{2\sqrt{x}\left(y+1\right)}{\left(y+1\right)\left(x+1\right)}=\dfrac{2\sqrt{x}}{x+1}\)
Đk:\(xy\ne1;x\ge0;y\ge0\)
\(P=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}:\dfrac{1-xy+x+y+2xy}{1-xy}\)
\(=\dfrac{\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}:\dfrac{1+x+y+xy}{1-xy}\)
\(=\dfrac{2\sqrt{x}+2y\sqrt{x}}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}:\dfrac{\left(1+x\right)\left(1+y\right)}{1-xy}\)\(=\dfrac{2\sqrt{x}\left(1+y\right)}{1-xy}.\dfrac{1-xy}{\left(1+x\right)\left(1+y\right)}=\dfrac{2\sqrt{x}}{1+x}\)
b) Áp dụng AM-GM có:
\(1+x\ge2\sqrt{x}\Leftrightarrow\)\(\dfrac{2\sqrt{x}}{1+x}\le1\)
Dấu "=" xảy ra khi x=1 (tm)
Vậy \(P_{max}=1\)
thôi không cần nữa các bạn ạ ! mình giải đc r đáp án là cosh(0) +1 nhé!