Giúp mk vs
\(\left(\dfrac{a+\sqrt{a}}{1+\sqrt{a}}+\dfrac{a}{1-\sqrt{a}}\right)\times\left(\dfrac{a}{2\sqrt{a}}-\dfrac{1}{2}\right)\)
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16 tháng 9 2021
\(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\left(đk:a>0,a\ne1\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+2}=\dfrac{1}{\sqrt{a}}.\dfrac{\sqrt{a}-2}{1}=\dfrac{\sqrt{a}-2}{\sqrt{a}}\)
Để A nguyên
\(\Leftrightarrow A=\dfrac{\sqrt{a}-2}{\sqrt{a}}=1-\dfrac{2}{\sqrt{a}}\in Z\)
Do \(\sqrt{a}>0,\sqrt{a}\ne1\)
\(\Leftrightarrow\sqrt{a}\inƯ\left(2\right)=\left\{2\right\}\)
\(\Leftrightarrow a=4\)
25 tháng 2 2022
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(\Rightarrow A=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}}\)
28 tháng 4 2021
a) Ta có: \(A=\sqrt{23+6\sqrt{10}}-\sqrt{23-6\sqrt{10}}\)
\(=\sqrt{18+2\cdot3\sqrt{2}\cdot\sqrt{5}+5}-\sqrt{18-2\cdot3\sqrt{2}\cdot\sqrt{5}+5}\)
\(=\sqrt{\left(3\sqrt{2}+\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{2}-\sqrt{5}\right)^2}\)
\(=3\sqrt{2}+\sqrt{5}-3\sqrt{2}+\sqrt{5}\)
\(=2\sqrt{5}\)
28 tháng 4 2021
b) Ta có: \(B=\left(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}+1\right)\left(\dfrac{2-\sqrt{2}}{\sqrt{2}-1}-1\right)\)
\(=\left(\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}+1\right)\left(\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}-1\right)\)
\(=\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)\)
=2-1=2
AT
31 tháng 7 2021
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
5 tháng 10 2021
Sửa đề: \(C=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(a,C=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\left(a>0;a\ne1;a\ne4\right)\\ C=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\\ b,C\ge\dfrac{1}{6}\Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}-\dfrac{1}{6}\ge0\Leftrightarrow\dfrac{\sqrt{a}-4}{6\sqrt{a}}\ge0\\ \Leftrightarrow\sqrt{a}-4\ge0\left(6\sqrt{a}>0\right)\\ \Leftrightarrow a\ge16\)
6 tháng 7 2021
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 \(Max=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{4}\)
AH
Akai Haruma
Giáo viên
6 tháng 7 2021
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)
8 tháng 10 2023
ĐKXĐ: \(\left\{{}\begin{matrix}a>=0\\a< >1\end{matrix}\right.\)
\(\left(\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right)\cdot\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{\sqrt{a}-1}\right)\)
\(=\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{a\sqrt{a}}{a-1}\right)\cdot\dfrac{\sqrt{a}-1+\sqrt{a}+1}{a-1}\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2-a\sqrt{a}}{a-1}\cdot\dfrac{2\sqrt{a}}{a-1}\)
\(=\dfrac{2\sqrt{a}\left(a+2\sqrt{a}+1-a\sqrt{a}\right)}{\left(a-1\right)^2}\)
AH
Akai Haruma
Giáo viên
26 tháng 8 2021
Lời giải:
ĐKXĐ: $x>0$
a. \(P=\frac{x-1}{\sqrt{x}}:\left[\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}+1)}+\frac{1-\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}\right]\)
\(=\frac{x-1}{\sqrt{x}}:\frac{x-1+1-\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}=\frac{x-1}{\sqrt{x}}:\frac{\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}+1)}=\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{(\sqrt{x}+1)^2}{\sqrt{x}}\)
b.
\(x=\frac{4}{4+2\sqrt{3}}=(\frac{2}{\sqrt{3}+1})^2\Rightarrow \sqrt{x}=\frac{2}{\sqrt{3}+1}\)
\(P=\frac{(\frac{2}{\sqrt{3}+1}+1)^2}{\frac{2}{\sqrt{3}+1}}=\frac{3+3\sqrt{3}}{2}\)
26 tháng 8 2021
a: Ta có: \(P=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\dfrac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}}\)
ĐKXĐ: \(a>0 \) \(a\ne 1\)
=\((\sqrt{a} +\frac{a}{1-\sqrt{a} } )(\frac{\sqrt{a} }{2}-\frac{1}{2} )\)
=\(\frac{\sqrt{a} -a+a}{1-\sqrt{a} }\frac{\sqrt{a}-1 }{2}\)
=\(\frac{-\sqrt{a} }{2} \)
\(\left(\dfrac{a+\sqrt{a}}{1+\sqrt{a}}+\dfrac{a}{1-\sqrt{a}}\right).\left(\dfrac{a}{2\sqrt{a}}-\dfrac{1}{2}\right)\)
\(ĐKXĐ:a>0;a\ne1\)
\(=\left(\sqrt{a}+\dfrac{a}{1-\sqrt{a}}\right).\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2}\right)\)
\(=\dfrac{\sqrt{a}-a+a}{1-\sqrt{a}}.\dfrac{\sqrt{a}-1}{2}\)
\(=\dfrac{-\sqrt{a}}{2}\)