Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(P=\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x-1-x+4}\)
\(=\dfrac{1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}-2}{3}=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b: P=1/4
=>\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\)
=>\(4\left(\sqrt{x}-2\right)=3\sqrt{x}\)
=>\(4\sqrt{x}-8-3\sqrt{x}=0\)
=>\(\sqrt{x}=8\)
=>x=64
c: Khi \(x=4+2\sqrt{3}\) thì \(P=\dfrac{\sqrt{4+2\sqrt{3}}-2}{3\cdot\sqrt{4+2\sqrt{3}}}\)
\(=\dfrac{\sqrt{3}+1-2}{3\left(\sqrt{3}+1\right)}=\dfrac{\sqrt{3}-1}{3\sqrt{3}+3}=\dfrac{2-\sqrt{3}}{3}\)
a: \(P=\dfrac{x+\sqrt{x}+1+11\sqrt{x}-11+34}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x+12\sqrt{x}+24}{\sqrt{x}+2}\)
b: Thay \(x=3-2\sqrt{2}\) vào P, ta được:
\(P=\dfrac{3-2\sqrt{2}+12\left(\sqrt{2}-1\right)+24}{\sqrt{2}-1+2}\)
\(=\dfrac{27-2\sqrt{2}+12\sqrt{2}-12}{\sqrt{2}+1}=5+5\sqrt{2}\)
1: Khi x=9 thì \(A=\dfrac{3+1}{3-1}=\dfrac{4}{2}=2\)
2:
a: \(P=\left(\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: \(2P=2\sqrt{x}+5\)
=>\(P=\sqrt{x}+\dfrac{5}{2}\)
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\dfrac{5}{2}=\dfrac{2\sqrt{x}+5}{2}\)
=>\(\sqrt{x}\left(2\sqrt{x}+5\right)=2\sqrt{x}+2\)
=>\(2x+3\sqrt{x}-2=0\)
=>\(\left(\sqrt{x}+2\right)\left(2\sqrt{x}-1\right)=0\)
=>\(2\sqrt{x}-1=0\)
=>x=1/4
Bạn có thể làm hộ mình câu c được không?Nếu được thì mình cảm ơn bạn nhiều!
2. \(P=x^2-x\sqrt{3}+1=\left(x^2-x\sqrt{3}+\frac{3}{4}\right)+\frac{1}{4}=\left(x-\frac{\sqrt{3}}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\)
Dấu '=' xảy ra khi \(x=\frac{\sqrt{3}}{2}\)
Vây \(P_{min}=\frac{1}{4}\)khi \(x=\frac{\sqrt{3}}{2}\)
3. \(Y=\frac{x}{\left(x+2011\right)^2}\le\frac{x}{4x.2011}=\frac{1}{8044}\)
Dấu '=' xảy ra khi \(x=2011\)
Vây \(Y_{max}=\frac{1}{8044}\)khi \(x=2011\)
4. \(Q=\frac{1}{x-\sqrt{x}+2}=\frac{1}{\left(x-\sqrt{x}+\frac{1}{4}\right)+\frac{7}{4}}=\frac{1}{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{7}{4}}\le\frac{4}{7}\)
Dấu '=' xảy ra khi \(x=\frac{1}{4}\)
Vậy \(Q_{max}=\frac{4}{7}\)khi \(x=\frac{1}{4}\)
