Question

[Mai Xuân Lộc]

Giải giúp mình với nhé ^_^

 

Answer 1

[Hoàng Như Quỳnh]

\(11,ĐKXĐ:x\ge0;x\ne1\)

\(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)

\(B=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)

\(B=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{1+\sqrt{x}}\)

\(B=\frac{\sqrt{x}-1}{\sqrt{x}}\)

\(b,P=9\sqrt{x}-B=9\sqrt{x}-\frac{\sqrt{x}-1}{\sqrt{x}}\)

\(P=9\sqrt{x}-1+\frac{1}{\sqrt{x}}\)

\(9\sqrt{x}+\frac{1}{\sqrt{x}}\ge2\sqrt{9\sqrt{x}.\frac{1}{\sqrt{x}}}=2\sqrt{9}=6\)( cô-si)

\(P\ge6-1=5\)dấu "=" xảy ra khi và chỉ khi \(x=\frac{1}{9}\)

\(< =>MIN:P=5\)

Answer 2

[Hoàng Như Quỳnh]

\(12,P=\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(P=\left(\frac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(P=\frac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(P=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(P=\frac{\sqrt{x}+1}{\sqrt{x}}\)

\(b,P+\sqrt{x}=\frac{\sqrt{x}+1}{\sqrt{x}}+\sqrt{x}\)

xét\(\frac{\sqrt{x}+1}{\sqrt{x}}+\sqrt{x}-3\)

\(\frac{\sqrt{x}+1+x-3\sqrt{x}}{\sqrt{x}}\)

\(\frac{x-2\sqrt{x}+1}{\sqrt{x}}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\ge0\)

mà dấu "=" xảy ra khi và chỉ khi \(x=1\left(KTM\right)\)

vậy dấu "=" ko xảy ra \(\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)

\(< =>P+\sqrt{x}>3\)