K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

16 tháng 7 2018

a) \(B=\)\(\dfrac{\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}}{\dfrac{\sqrt{x}}{x+\sqrt{x}}}\) ĐKXĐ: x>0

=\(\dfrac{\dfrac{\sqrt{x}+1+\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}}{\dfrac{\sqrt{x}}{x+\sqrt{x}}}\)

\(=\dfrac{x+\sqrt{x}+1}{x+\sqrt{x}}:\dfrac{\sqrt{x}}{x+\sqrt{x}}\)

=\(\dfrac{x+\sqrt{x}+1}{x+\sqrt{x}}\times\dfrac{x+\sqrt{x}}{\sqrt{x}}\)

\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)

b)

Theo câu a ) ta có :

B=\(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)

Xét : \(x+\sqrt{x}+1=x+2.\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

=\(\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\) (với mọi x>0) (1)

Xét:

\(\sqrt{x}>0\) (2)

Từ (1) và (2) =>\(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}>0\) (ĐPCM)

c) B=\(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\) ( theo câu a)

=\(\dfrac{x}{\sqrt{x}}+\dfrac{1}{\sqrt{x}}+1\)

=\(\sqrt{x}+\dfrac{1}{\sqrt{x}}+1\)

Áp dụng BĐT cô si cho \(\sqrt{x}\)\(\dfrac{1}{\sqrt{x}}\)

Ta có : \(\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\dfrac{1}{\sqrt{x}}}\)

=2

Vậy :\(\sqrt{x}+\dfrac{1}{\sqrt{x}}+1\ge2+1\)

Hay\(\sqrt{x}+\dfrac{1}{\sqrt{x}}+1\ge3\)

Min B= 3 Dấu "=" xảy ra khi x=1

CHÚC BẠN HỌC TỐThiuhiu

16 tháng 7 2018

thanks nha

a: \(P=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

b:Sửa đề: 2A

2A=2căn x+5

=>(2căn x+2)/căn x=2căn x+5

=>2x+5căn x-2căn x-2=0

=>2x+3căn x-2=0

=>(căn x+2)(2căn x-1)=0

=>x=1/4

AH
Akai Haruma
Giáo viên
26 tháng 5 2023

Lời giải:
\(P=\left[\frac{\sqrt{x}+1}{(\sqrt{x}+1)(\sqrt{x}-1)}+\frac{x}{\sqrt{x}(\sqrt{x}-1)}\right]:\frac{\sqrt{x}+1}{\sqrt{x}}\)

\(=\left[\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}-1}\right].\frac{\sqrt{x}}{\sqrt{x}+1}\)

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

b. Áp dụng BĐT AM-GM

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

\(=(\sqrt{x}-1)+\frac{1}{\sqrt{x}-1}+2\geq 2\sqrt{(\sqrt{x}-1).\frac{1}{\sqrt{x}-1}}+2=2+2=4\)

Vậy $M_{\min}=4$ khi $\sqrt{x}-1=\frac{1}{\sqrt{x}-1}$

$\Rightarrow \sqrt{x}-1=0$

$\Leftrightarrow x=1$

16 tháng 11 2021

a: \(A=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)

a: \(A=\left(\dfrac{\sqrt{3}\left(x-\sqrt{3}\right)+3}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\right)\cdot\dfrac{x^2+3+x\sqrt{3}}{x\sqrt{3}}\)

\(=\dfrac{x\sqrt{3}}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\cdot\dfrac{x^2+x\sqrt{3}+3}{x\sqrt{3}}\)

\(=\dfrac{1}{x-\sqrt{3}}\)

b: \(B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\)

\(=x-\sqrt{x}-x-\sqrt{x}+x+1\)

\(=x-2\sqrt{x}+1\)

c: \(C=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{x\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}}\)

\(=\dfrac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)

\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)

a: \(B=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-4}\cdot\dfrac{\sqrt{x}+2}{x+16}=\dfrac{1}{\sqrt{x}-2}\)

b: Khi x=9 thì B=1/(3-2)=1