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NT
0
7 tháng 8 2016
diều kiện x >= 0
P=\(\left(\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right).\frac{4\sqrt{x}}{3}\)
= \(\frac{x+2-x+\sqrt{x}-1}{x\sqrt{x}+1}.\frac{4\sqrt{x}}{3}\)
=\(\frac{\sqrt{x}+1}{x\sqrt{x}+1}.\frac{4\sqrt{x}}{3}\)=\(\frac{4\sqrt{x}}{3x-3\sqrt{x}+3}\)
P=8/9
<=> \(\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\frac{8}{9}\)
<=> \(3\sqrt{x}=2x-2\sqrt{x}+1\)
<=> \(2x-5\sqrt{x}+2=0\)
<=> \(\left[\begin{array}{nghiempt}x=4\\x=\frac{1}{4}\end{array}\right.\)
vậy x=4 hoặc x=1/4 thì p=8/9
7 tháng 8 2016
a) \(P=\left(\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right)\cdot\frac{4\sqrt{x}}{3}\left(ĐK:x\ge0;x\ne-1\right)\)
\(=\left[\frac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}+1}\right]\cdot\frac{4\sqrt{x}}{3}\)
\(=\frac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{4\sqrt{x}}{3}\)
\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{4\sqrt{x}}{3}\)
\(=\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
b) Để P=8/9
\(\Leftrightarrow\)\(\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\frac{8}{9}\)
\(\Leftrightarrow24\left(x-\sqrt{x}+1\right)=36\sqrt{x}\)
\(\Leftrightarrow24x-24\sqrt{x}+24-36\sqrt{x}=0\)
\(\Leftrightarrow24x-60\sqrt{x}+24=0\)
\(\Leftrightarrow12\left(2x-5\sqrt{x}+2\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{x}\right)-\left(4\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}-1\right)-2\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2\sqrt{x}-1=0\\\sqrt{x}-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{x}=\frac{1}{2}\\\sqrt{x}=2\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{4}\left(tm\right)\\x=4\left(tm\right)\end{array}\right.\)
AV
0
22 tháng 10 2016
a/ Bạn tự tìm ĐKXĐ
\(A=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{x}\left(\sqrt{y}+1\right)}{1-\sqrt{xy}}+1\right):\left(1-\frac{\sqrt{x}\left(\sqrt{y}+1\right)}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\right)\)
Xét
- \(=\frac{\left(\sqrt{x}+1\right)\left(1-\sqrt{xy}\right)+\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)+\left(1+\sqrt{xy}\right)\left(1-\sqrt{xy}\right)}{\left(1+\sqrt{xy}\right)\left(1-\sqrt{xy}\right)}\)
\(=\frac{\sqrt{x}-x\sqrt{y}+1-\sqrt{xy}+xy+\sqrt{xy}+x\sqrt{y}+\sqrt{x}+1-xy}{\left(1+\sqrt{xy}\right)\left(1-\sqrt{xy}\right)}\)
\(=\frac{2\sqrt{x}+2}{\left(1+\sqrt{xy}\right)\left(1-\sqrt{xy}\right)}\)
- \(1-\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\)
\(=\frac{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)-\left(\sqrt{xy}+\sqrt{x}\right)\left(\sqrt{xy}+1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}\)
\(=\frac{xy-1-xy-\sqrt{xy}-x\sqrt{y}-\sqrt{x}-x\sqrt{y}+\sqrt{x}-\sqrt{xy}+1}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}\)
\(=\frac{-2\sqrt{xy}-2x\sqrt{y}}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}=\frac{-2\sqrt{xy}\left(\sqrt{x}+1\right)}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}\)
\(\Rightarrow A=\frac{2\left(\sqrt{x}+1\right)}{\left(1+\sqrt{xy}\right)\left(1-\sqrt{xy}\right)}:\frac{2\sqrt{xy}\left(\sqrt{x}+1\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}=\frac{1}{\sqrt{xy}}\)
b/ Áp dụng BĐT \(\left(a+b\right)^2\ge4ab\) với \(a=\frac{1}{\sqrt{x}},b=\frac{1}{\sqrt{y}}\) được :
\(A=\frac{1}{\sqrt{x}.\sqrt{y}}\le\frac{1}{4}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)^2=\frac{1}{4}.6^2=9\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\sqrt{x}=\sqrt{y}\\\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=6\end{cases}}\Leftrightarrow x=y=\frac{1}{9}\)
Vậy ........................................................
22 tháng 4 2020
" m " ở đâu vậy bạn ,sửa đề câu b) : Tìm x để P =\(A-9\sqrt{x}\)
Bài giải
a) ĐKXĐ: \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
A = \(\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
= \(\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right).\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
= \(\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
= \(\frac{\sqrt{x}-1}{\sqrt{x}}\)
Vậy A = \(\frac{\sqrt{x}-1}{\sqrt{x}}\)với x > 0 ; x \(\ne1\)
b) P = A - \(9\sqrt{x}=\frac{\sqrt{x}-1}{\sqrt{x}}-9\sqrt{x}=1-\left(\frac{1}{\sqrt{x}}+9\sqrt{x}\right)\)
Áp dụng BĐT Côsi : \(\frac{1}{\sqrt{x}}+9\sqrt{x}\ge2.3=6\)
Dấu "=" xảy ra khi \(\frac{1}{\sqrt{x}}=9\sqrt{x}\Leftrightarrow1=9x\Leftrightarrow x=\frac{1}{9}\)
=> P \(\ge-5\).Vậy Max P = -5 khi x = \(\frac{1}{9}\)
ĐKXĐ: \(x\ge0\). Ta có:
\(P=\frac{\sqrt{x}-1}{\sqrt{x}}-9\sqrt{x}=1-\frac{1}{\sqrt{x}}-9\sqrt{x}=1-\left(\frac{1}{\sqrt{x}}+9\sqrt{x}\right)\)
Để P đạt GTLN thì \(\frac{1}{\sqrt{x}}+9\sqrt{x}\) đạt GTNN. Áp dụng BĐT Cô-si ta có:
\(\frac{1}{\sqrt{x}}+9\sqrt{x}\ge2\sqrt{\frac{1}{\sqrt{x}}.9\sqrt{x}}=6\Rightarrow P\le1-6=-5\)
Xảy ra đẳng thức khi và chỉ khi \(\frac{1}{\sqrt{x}}=9\sqrt{x}\Leftrightarrow9x=1\Leftrightarrow x=\frac{1}{9}\) (thỏa mãn)
Vậy max P = -5 khi và chỉ khi x = 1/9
\(P=1-\frac{1}{\sqrt{x}}-9\sqrt{x}=1-\left(\frac{1}{\sqrt{x}}+9\sqrt{x}\right)\le1-2\sqrt{\frac{1}{\sqrt{x}}\cdot9\sqrt{x}}=1-6=-5\)
Vậy MAx P = -5 tại x = 1/9