Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Sửa đề: \(C=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(a,C=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\left(a>0;a\ne1;a\ne4\right)\\ C=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\\ b,C\ge\dfrac{1}{6}\Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}-\dfrac{1}{6}\ge0\Leftrightarrow\dfrac{\sqrt{a}-4}{6\sqrt{a}}\ge0\\ \Leftrightarrow\sqrt{a}-4\ge0\left(6\sqrt{a}>0\right)\\ \Leftrightarrow a\ge16\)
\(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\left(đk:a>0,a\ne1\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+2}=\dfrac{1}{\sqrt{a}}.\dfrac{\sqrt{a}-2}{1}=\dfrac{\sqrt{a}-2}{\sqrt{a}}\)
Để A nguyên
\(\Leftrightarrow A=\dfrac{\sqrt{a}-2}{\sqrt{a}}=1-\dfrac{2}{\sqrt{a}}\in Z\)
Do \(\sqrt{a}>0,\sqrt{a}\ne1\)
\(\Leftrightarrow\sqrt{a}\inƯ\left(2\right)=\left\{2\right\}\)
\(\Leftrightarrow a=4\)
a) \(A=\sqrt{9a}-\sqrt{16a}-\sqrt{49a}=3\sqrt{a}-4\sqrt{a}-7\sqrt{a}=-8\sqrt{a}\)
b) \(B=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}}-\left(\sqrt{3}+\sqrt{2}\right)\)
\(=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}}-\left(\sqrt{3}+\sqrt{2}\right)\)
\(=2+\sqrt{3}+\sqrt{2}+1-\sqrt{3}-\sqrt{2}=3\)
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 \(P_{max}=1\)
a) ĐK:\(x\ge0;x\ne9\)
\(P=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\dfrac{2\sqrt{x}-2-\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)
b)\(P=-\dfrac{3}{\sqrt{x}+3}\)
Có \(\sqrt{x}+3\ge3;\forall x\ge0\)
\(\Leftrightarrow-\dfrac{3}{\sqrt{x}+3}\ge-\dfrac{1}{3}\)
\(P_{min}=-\dfrac{1}{3}\Leftrightarrow x=0\)
a) Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
\(C=\left(\dfrac{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}+a\right)}{1+\sqrt{a}}-\sqrt{a}\right)\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)^2\left(\sqrt{a}+1\right)^2}+\sqrt{a}\)
\(=\left(1-2\sqrt{a}+a\right).\dfrac{1}{\left(\sqrt{a}-1\right)^2}+\sqrt{a}\)
\(=\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)^2}+\sqrt{a}=\sqrt{a}+1\)
Để \(C=3\Rightarrow\sqrt{a}+1=3\Rightarrow\sqrt{a}=2\Rightarrow a=4\)
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)
a/ \(B=(\dfrac{2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-6}{x-9}):(1+\dfrac{6}{x-9})\)
= \((\dfrac{2}{\sqrt{x}-3}+\dfrac{\sqrt{x}+6}{(\sqrt{x}-3)(\sqrt{x}+3)}):(\dfrac{x-9}{x-9}+\dfrac{6}{x-9})\)
=\((\dfrac{2(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}+\dfrac{\sqrt{x}-6}{(\sqrt{x}-3)(\sqrt{x}+3)}):(\dfrac{x-3}{x-9})\)
=\((\dfrac{2\sqrt{x}+6+\sqrt{x}-6}{(\sqrt{x}-3)(\sqrt{x}+3)}):(\dfrac{x-3}{x-9})\)
=\((\dfrac{2\sqrt{x}+6+\sqrt{x}-6}{x-9}).(\dfrac{x-9}{x-3})\)
= \(\dfrac{3\sqrt{x}}{x-3}\)
Vậy B=\(\dfrac{3\sqrt{x}}{x-3}\)
b/ Để B≥0 thì \(\dfrac{3\sqrt{x}}{x-3} \)≥0
\(<=>\begin{cases} x-3 không= 0\\ 3\sqrt{x}>/0 \end{cases} \)
<=> \(\begin{cases} x không= 3\\ x>/0 \end{cases} \)
Vậy để B≥0 thì x không = 3 và x≥0
a) Ta có: \(C=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(1+\dfrac{2}{\sqrt{x}-1}\right)\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
b) Để C<-1 thì C+1<0
\(\Leftrightarrow\dfrac{\sqrt{x}-1+\sqrt{x}}{\sqrt{x}}< 0\)
\(\Leftrightarrow2\sqrt{x}-1< 0\)
\(\Leftrightarrow x< \dfrac{1}{4}\)
Kết hợp ĐKXĐ, ta được: \(0< x< \dfrac{1}{4}\)
a) \(C=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\left(a>0.a\ne1\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}+1-\sqrt{a}-2}{\sqrt{a}-1}=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{-1}{\sqrt{a}-1}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\left(1-\sqrt{a}\right)=-\dfrac{1}{\sqrt{a}}\)
b) \(C=\dfrac{1}{4}\Rightarrow-\dfrac{1}{\sqrt{a}}=\dfrac{1}{4}\Rightarrow\sqrt{a}=-4\) (vô lý) \(\Rightarrow\) không có a thỏa đề