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

Khi x=25 thì A=|5-1|=4

b: \(B=\dfrac{-x+\sqrt{x}+2+\sqrt{x}-2}{x-4}=\dfrac{-x+2\sqrt{x}}{x-4}=\dfrac{-\sqrt{x}}{\sqrt{x}+2}\)

1. a, => -12x+60+21-7x = 5

=> 81 - 19x = 5

=> 19x = 81 - 5 = 76

=> x = 76 : 19 = 4

Tk mk nha

Áp dụng bđt \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\forall x;y;z\ge0\) ta được :

\(B=\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\ge\frac{9}{3+\left(a+b+c\right)}=\frac{9}{3+3}=\frac{9}{6}=\frac{3}{2}\)

Dấu "=" xảy ra <=> \(a=b=c=1\)

Vậy GTNN của B là \(\frac{3}{2}\) tại \(a=b=c=1\)

dùng bđt 1/x+1/y+1/z >/ 9/(x+y+z) với x,y,z>0 

\(P=\dfrac{1}{abc}+\dfrac{1}{a^2+b^2+c^2}=\dfrac{a+b+c}{abc}+\dfrac{1}{a^2+b^2+c^2}\)

\(=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{a^2+b^2+c^2}\left(1\right)\)

\(\)\(\left\{{}\begin{matrix}a+b+c=1\\\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\ge\dfrac{9}{ab+bc+ac}\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow P\ge\dfrac{9}{ab+bc+ac}+\dfrac{1}{a^2+b^2+c^2}\)

\(=\dfrac{1}{2\left(ab+bc+ac\right)}+\dfrac{1}{a^2+b^2+c^2}+\dfrac{17}{2\left(ab+bc+ac\right)}\)

\(\Rightarrow P\ge\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{17}{2\left(ab+bc+ac\right)}\)

\(\Rightarrow P\ge9+\dfrac{17}{2\left(ab+bc+ac\right)}\)

mà \(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{1}{3}\)

\(\Rightarrow P\ge9+\dfrac{17}{2.\dfrac{1}{3}}=9+\dfrac{17.3}{2}=\dfrac{18+17.3}{2}=\dfrac{69}{2}\)

\(\Rightarrow Min\left(P\right)=\dfrac{69}{2}\)

\(\dfrac{a}{1+b^2}=a-\dfrac{ab^2}{1+b^2}\ge a-\dfrac{ab^2}{2b}=a-\dfrac{1}{2}ab\)

Tương tự: \(\dfrac{b}{1+c^2}\ge b-\dfrac{1}{2}bc\) ; \(\dfrac{c}{1+a^2}\ge c-\dfrac{1}{2}ca\)

Cộng vế:

\(P\ge a+b+c-\dfrac{1}{2}\left(ab+bc+ca\right)\ge a+b+c-\dfrac{1}{6}\left(a+b+c\right)^2=\dfrac{3}{2}\)

\(P_{min}=\dfrac{3}{2}\) khi \(a=b=c=1\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(P=\frac{1}{2a-a^2}+\frac{1}{2b-b^2}+\frac{1}{2c-c^2}\)

\(\ge\frac{\left(1+1+1\right)^2}{2\left(a+b+c\right)-\left(a^2+b^2+c^2\right)}\)

\(=\frac{9}{2-\left(a^2+b^2+c^2\right)}\ge\frac{9}{2-\frac{\left(a+b+c\right)^2}{3}}\)

\(=\frac{9}{2-\frac{1}{3}}=\frac{9}{\frac{5}{3}}=\frac{27}{5}\)

Xảy ra khi \(a=b=c=\frac{1}{3}\)

@Thắng Nguyễn
Nếu đề là min của \(\text{ }\frac{1}{2x}-x^2+\frac{1}{2y}-y^2+\frac{1}{2z}-z^2\) thì liệu giải đ.c không nhỉ? 
 

chắc k đâu vì đề là a,b,c mà you sửa là x,y,z sao làm :v