Cho bt : \(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}\)
Khi x thỏa mãn dkxd. Hãy tìm GTNN của bt B, với B = A(x - 1)
K
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21 tháng 2 2022
a: \(P=\left(\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}+1\right)}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{x+1+\sqrt{x}}{x+1}\)
\(=\dfrac{2\sqrt{x}+x+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{x+1}{x+\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
b: Thay \(x=9+2\sqrt{7}\) vào P, ta được:
\(P=\dfrac{\sqrt{9+2\sqrt{7}}+1}{9+2\sqrt{7}+\sqrt{9+2\sqrt{7}+1}}\simeq0,25\)
KD
28 tháng 1 2020
\(a,Đkxđ:\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
\(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x+1}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x^3}-1\right)}{x+\sqrt{x}+1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}=\sqrt{x}\left(\sqrt{x}-1\right)\)
\(=x-\sqrt{x}\)
\(b,P=x-\sqrt{x}=x-\sqrt{x}+\frac{1}{4}-\frac{1}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\)
Ta có: \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\forall x\ge0\)
\(\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\forall x\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow x=\frac{1}{4}\)
\(Min_P=-\frac{1}{4}\Leftrightarrow x=\frac{1}{4}\)
c, Đề thiếu không bạn?
N
11 tháng 9 2019
1a
\(A=\frac{3}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{a^4+b^4}{2}\ge\frac{6}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{\frac{\left(a^2+b^2\right)^2}{2}}{2}\)
\(\ge10+\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{4}=10+\frac{1}{16}=\frac{161}{16}\)
Dau '=' xay ra khi \(a=b=\frac{1}{2}\)
Vay \(A_{min}=\frac{161}{16}\)
N
11 tháng 9 2019
1b.\(B=\frac{1}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{a^8+b^8}{4}\ge\frac{2}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{\frac{\left(a^4+b^4\right)^2}{2}}{4}\)
\(\ge6+\frac{\left[\frac{\left(a^2+b^2\right)^2}{2}\right]^2}{8}\ge6+\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{32}=6+\frac{1}{128}=\frac{769}{128}\)
Dau '=' xay ra khi \(a=b=\frac{1}{2}\)
Vay \(B_{min}=\frac{769}{128}\)khi \(a=b=\frac{1}{2}\)
16 tháng 10 2017
Câu 1) a) ĐKXĐ \(x\ge0,\)\(x\ne4\)A=\(\frac{x+2\sqrt{x}-4}{2\left(x-4\right)}\)b) Mình chưa làm được Câu 2) a) ĐKXĐ \(x>0,\)\(x\ne4\)A=\(\frac{\sqrt{x}-1}{\sqrt{x}}\)b) Để a<\(\frac{1}{2}\)\(\Rightarrow\)\(\frac{\sqrt{x}-1}{\sqrt{x}}< \frac{1}{2}\)\(\Rightarrow x< 1\)\(\Rightarrow0< x< 1\)thỏa mãn bài toán c) Ta có A=\(\frac{\sqrt{x}-1}{\sqrt{x}}=1-\frac{1}{\sqrt{x}}\), để A \(\in Z\)\(\Rightarrow\sqrt{x}\inƯ\left(1\right)\), \(\Rightarrow x=1\)( thỏa mãn ĐK)
8 tháng 7 2019
\(ĐKXĐ:x\ge0\)
Đề sai???
Sửa lại
\(a,P=\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}-1}{x-\sqrt{x}+1}\)
\(=\frac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\frac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{x+2-x+\sqrt{x}-1+x-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}}{x-\sqrt{x}+1}\)
22 tháng 1 2017
c) \(\frac{1}{P}=1+\frac{x}{\sqrt{x}+1}\)\(=1+\frac{x-1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}+1}\)
\(=1+\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\frac{1}{\sqrt{x}+1}\)
\(=1+\sqrt{x}-1+\frac{1}{\sqrt{x}+1}\)
\(=-1+\sqrt{x}+1+\frac{1}{\sqrt{x}+1}\)\(\ge-1+2\sqrt{\left(\sqrt{x}+1\right)\left(\frac{1}{\sqrt{x}+1}\right)}=1\)
Dau "=" xay ra khi x = 0
\(ĐKXĐ:\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(B=A\left(x-1\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}\right)\left(x-1\right)\)
\(=\sqrt{x}\left(\sqrt{x}+1\right)-2\left(\sqrt{x}-1\right)-2\)
\(=x+\sqrt{x}-2\sqrt{x}+2-2\)
\(=x-\sqrt{x}\)
\(=x-2.\frac{1}{2}\sqrt{x}+\frac{1}{4}-\frac{1}{4}\)
\(=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\)
\(\ge-\frac{1}{4}\)
Dấu "=" xảy ra khi \(\sqrt{x}-\frac{1}{2}=0\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\left(tm\right)\)
Vậy \(Min_B=-\frac{1}{4}\) khi \(x=\frac{1}{4}\)