a) \(\left(2x+\frac{1}{3}\right)^4\ge0\Rightarrow A\ge-1\)

Dấu \(=\)xảy ra khi \(2x+\frac{1}{3}=0\Leftrightarrow x=-\frac{1}{6}\).

b) \(\left(\frac{4}{9}x-\frac{2}{15}\right)^6\ge0\Rightarrow B\le3\)

Dấu \(=\)xảy ra khi \(\frac{4}{9}x-\frac{2}{15}=0\Leftrightarrow x=\frac{3}{10}\).

Tìm GTNN và GTLN mà

Dặt tử = A 

A^2 = \(x+\sqrt{x^2-y^2}+x-\sqrt{x^2-y^2}-2\sqrt{x^2-x^2+y^2}\)

= \(2x-2\sqrt{y^2}=2x-2y=2\left(x-y\right)\)

=> A  = \(\sqrt{2\left(x-y\right)}\)

Lấy tử chia mẫu là xong 

C lớn nhất khi (x-3)²+1 bé nhất 

=>x²-9 +1 bé nhất 

x²-8 bé nhất 

=>x² khác 8 và x²-8 bé nhất => x² -8=1

=>x²=9=>x=3

D lớn nhất khi |x-2|+2 bé nhất =>x-2 bé nhất=>x-2=0 =>x=2

2.

Áp dụng bất đẳng thức Cauchy - schwarz ( hay còn gọi là bất đẳng thức Cosi ):

\(\frac{x^2}{y+1}+\frac{y^2}{z+1}+\frac{z^2}{x+1}=\frac{\left(x+y+z\right)^2}{x+y+z+3}=\frac{9}{3+3}=\frac{9}{6}=\frac{3}{2}\)

Dấu "=" xảy ra khi x = y = z = 1

1: 

Áp dụng bất đẳng thức Cô si:

\(x\left(y+\frac{x}{1+y}\right)+y\left(z+\frac{y}{1+z}\right)+z\left(x+\frac{z}{1+x}\right)\)

\(=\left(x+y+z\right)\left[\left(y+\frac{x}{1+y}\right)+\left(z+\frac{y}{1+z}\right)+\left(x+\frac{z}{1+x}\right)\right]\)

\(=1\left[\left(x+y+z\right)+\left(\frac{x}{1+y}+\frac{y}{1+z}+\frac{z}{1+x}\right)\right]\)

\(=1\left[1+\left(\frac{x+y+z}{1+y+1+z+1+x}\right)\right]\)

\(=1\left[1+\left(\frac{1}{3+\left(x+y+z\right)}\right)\right]\)

\(=1\left[1+\frac{1}{4}\right]\)

\(=1+\frac{5}{4}=\frac{9}{4}\)

Dấu "=" xảy ra khi x = y = z = \(\frac{1}{3}\)

2. áp dạng bất đẳng thức cauchy - schwarz dạng engel

\(\frac{x^2}{y+1}+\frac{y^2}{z+1}+\frac{z^2}{x+1}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3}=\frac{3^2}{3+3}=\frac{9}{6}=\frac{3}{2}\)

dấu bằng xay ra khi x=y=z=1

lm bất đẳng thức cô si nhé!!! Thanks

\(\frac{1}{4}+\frac{1}{3}:\left(2x-1\right)=-5\)

\(\frac{1}{3}:\left(2x-1\right)=-5-\frac{1}{4}\)

\(\frac{1}{3}:\left(2x-1\right)=-\frac{20}{4}-\frac{1}{4}\)

\(\frac{1}{3}:\left(2x-1\right)=-\frac{21}{4}\)

\(\left(2x-1\right)=\frac{1}{3}:-\frac{21}{4}\)

\(\left(2x-1\right)=\frac{1}{3}.-\frac{4}{21}\)

\(\left(2x-1\right)=-\frac{4}{63}\)

2x= -4/63 + 1

2x = 59/63

x = 59/63 : 2

x = 59/126

1/3:(2.x-1)=-5-1/4

1/3:(2.x-1)=-21/4

2.x-1=1/3:-21/4

2.x-1=-4/63

2.x=-4/63+1

2.x=\(3\frac{59}{63}\)

x=\(3\frac{59}{63}\):2

x=\(1\frac{61}{63}\)

\(\frac{1}{4}+\frac{1}{3}\div\left(2x-1\right)=-5\)

\(\frac{1}{3}\div\left(2x-1\right)=-5-\frac{1}{4}\)

\(\frac{1}{3}\div\left(2x-1\right)=\frac{-20}{4}+\frac{-1}{4}\)

\(\frac{1}{3}\div\left(2x-1\right)=\frac{-21}{4}\)

\(2x-1=\frac{1}{3}\div\frac{-21}{4}\)

\(2x-1=\frac{1}{3}.\frac{-4}{21}\)

\(2x-1=\frac{-4}{63}\)

\(2x=\frac{-4}{63}+1\)

\(2x=\frac{-4}{63}+\frac{63}{63}\)

\(2x=\frac{59}{63}\)

\(x=\frac{59}{63}\div2\)

\(x=\frac{59}{63}.\frac{1}{2}\)

\(x=\frac{59}{126}\)

\(2\cdot\left(2x-6\right)+\left(x-1\right)=2\)

\(\Leftrightarrow4x-12+x-1-2=0\)

\(\Leftrightarrow5x-15=0\)

\(\Leftrightarrow5x=15\)

\(\Leftrightarrow x=3\)