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 \(\sqrt{x}\)và \(\dfrac{1}{\sqrt{x}}\)

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 TỐT

\(a,\)\(a\left(a+b\right)+b\left(a-b\right)\)

\(=\left(a+b\right)\left(a-b\right)\)

\(=a^2-b^2\)

\(b,\) \(a^{n+1}\left(a+b\right)-b\left(a^{n+1}+b^{n-1}\right)\)

=\(a.a^{n+1}+b.a^{n+1}-b.a^{n+1}+b.b^{n-1}\)

=\(a^n-b^n\)

Cho \(x^3+y^3+z^3=1\)

\(\dfrac{x^2}{\sqrt{1-x^2}}+\dfrac{y^2}{\sqrt{1-y^2}}+\dfrac{z^2}{\sqrt{1-z^2}}\)\(\ge2\)

\(Đkxđ:x\ge3+\sqrt{12}\) \(\sqrt{x^2+6x-3}=x-3\)

\(\Leftrightarrow\left(\sqrt{x^2+6x-3}\right)^2=\left(x-3\right)^2\)

\(\Leftrightarrow x^2+6x-3=x^2-6x+9\)

\(\Leftrightarrow x^2+6x-3-x^2+6x-9=0\)

\(\Leftrightarrow12x-12=0\)

\(\Leftrightarrow12x=12\)

\(\Leftrightarrow x=1\)

C=(2-x)(x+4)

=\(2x-x^2+8-4x\)

=\(-\left(x^2-2x+1\right)+9\)\(\le9\)

MaxC=9 Dấu "=" xảy ra khi x=1

Áp dụng BĐT cô si cho:

!)\(\dfrac{3}{x}+\dfrac{9}{y}\)\(\ge2\sqrt{\dfrac{3}{x}.\dfrac{9}{y}}\ge2\sqrt{\dfrac{3.9}{xy}}=2\sqrt{\dfrac{27}{3}}=6\)

!!) Tương tự ta có:

\(3x+y\ge2\sqrt{3xy}\ge6\)

Vậy: K=\(\dfrac{3}{x}+\dfrac{9}{y}-\dfrac{26}{3x+y}\)\(\ge6-\dfrac{26}{6}=\dfrac{5}{3}\)

Min K=\(\dfrac{5}{3}\) Dấu "=' xảy ra khi y=1 và x=3

Cậu cần gải nk k ạ