\(A=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(5-3x\right)^2}\)

\(A=3x-1+5-3x=4\)

\(A\)có giá trị ko phụ thuộc vào biến x

\(A=\sqrt{1-6x+9x^2}+\sqrt{9x^2-12x+4}\)

\(=\sqrt{\left(1-3x\right)^2}+\sqrt{\left(3x-2\right)^2}\)

\(=\left|1-3x\right|+\left|3x-2\right|\)

\(\ge\left|1-3x+3x-2\right|=\left|-1\right|=1\)

Dấu "=" xảy ra \(\Leftrightarrow\left(1-3x\right)\left(3x-2\right)\ge0\Leftrightarrow\frac{1}{3}\le x\le\frac{2}{3}\)

Vậy \(A_{min}=1\) tại \(\frac{1}{3}\le x\le\frac{2}{3}\)

Xin lỗi cậu tớ mới học lớp 7 thôi

a) \(A=\sqrt{4x^2+4x+2}=\sqrt{4x^2+4x+1+1}=\sqrt{\left(2x+1\right)^2+1}\)

Vì \(\left(2x+1\right)^2\ge0\forall x\)\(\Rightarrow\left(2x+1\right)^2+1\ge1\forall x\)

\(\Rightarrow A\ge\sqrt{1}=1\)

Dấu " = " xảy ra \(\Leftrightarrow2x+1=0\)\(\Leftrightarrow2x=-1\)\(\Leftrightarrow x=\frac{-1}{2}\)

Vậy \(minA=1\Leftrightarrow x=\frac{-1}{2}\)

b) \(B=\sqrt{2x^2-4x+5+1}=\sqrt{2x^2-4x+2+3+1}=\sqrt{2\left(x^2-2x+1\right)+4}\)

\(=\sqrt{2\left(x-1\right)^2+4}\)

Vì \(\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow2\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow2\left(x-1\right)^2+4\ge4\forall x\)

\(\Rightarrow B\ge\sqrt{4}=2\)

Dấu " = " xảy ra \(\Leftrightarrow x-1=0\)\(\Leftrightarrow x=1\)

Vậy \(minB=2\Leftrightarrow x=1\)

Mơn bạn nha

\(A=1-|1-3x|+|3x-1|^2\)

\(=\left(|3x-1|-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

\(\Rightarrow minA=\frac{3}{4}\Leftrightarrow x=\frac{1}{2}\)hoặc \(x=\frac{1}{6}\)

\(Y=\sqrt{\left(3x+2\right)^2+7}\ge\sqrt{0+7}=\sqrt{7}\)

\(Y_{Min}=\sqrt{7}\Leftrightarrow3x+2=0\Leftrightarrow x=-\frac{2}{3}\)

a) A = x² + 12x + 39

= ( x² + 12x + 36 ) + 3

= ( x + 6 )² + 3 ≥ 3 ∀ x

Đẳng thức xảy ra ⇔ x + 6 = 0 => x = -6

=> MinA = 3 ⇔ x = -6

B = 9x² - 12x 

= 9( x² - 4/3x + 4/9 ) - 4

= 9( x - 2/3 )² - 4 ≥ -4 ∀ x

Đẳng thức xảy ra ⇔ x - 2/3 = 0 => x = 2/3

=> MinB = -4 ⇔ x = 2/3

b) C = 4x - x² + 1

= -( x² - 4x + 4 ) + 5

= -( x - 2 )² + 5 ≤ 5 ∀ x

Đẳng thức xảy ra ⇔ x - 2 = 0 => x = 2

=> MaxC = 5 ⇔ x = 2

D = -4x² + 4x - 3

= -( 4x² - 4x + 1 ) - 2

= -( 2x - 1 )² - 2 ≤ -2 ∀ x

Đẳng thức xảy ra ⇔ 2x - 1 = 0 => x = 1/2

=> MaxD = -2 ⇔ x = 1/2

Ta có A = x² + 12x + 39 = (x² + 12x + 36) + 3 = (x + 6)² + 3 \(\ge\)3

Dấu "=" xảy ra <=> x + 6 = 0

=> x = -6

Vậy Min A = 3 <=> x = -6

Ta có B = 9x² - 12x = [(3x)² - 12x + 4] - 4 =(3x - 2)² - 4 \(\ge\)-4

Dấu "=" xảy ra <=> 3x - 2 =0

=> x = 2/3

Vậy Min B = -4 <=> x = 2/3

b) Ta có C = 4x - x² + 1 = -(x² - 4x - 1) = -(x² - 4x + 4) + 5 = -(x - 2)² + 5 \(\le\)5

Dấu "=" xảy ra <=> x - 2 = 0

=> x = 2

Vậy Max C = 5 <=> x = 2

Ta có D = -4x² + 4x - 3 = -(4x² - 4x + 1) - 2 = -(2x - 1)² - 2 \(\le\)-2

Dấu "=" xảy ra <=> 2x - 1 = 0

=> x = 0,5

Vậy Max D = -2 <=> x = 0,5

\(\frac{4x^2+9x+18\sqrt{x}+9}{4x\sqrt{x}+4\sqrt{x}}+\frac{4x\sqrt{x}+4\sqrt{x}}{4x^2+9x+18\sqrt{x}+9}-2=\frac{\left(-4x\sqrt{x}+4x^2+9x+22\sqrt{x}+9\right)^2}{\left(4x^2+9x+18\sqrt{x}+9\right)\left(4x\sqrt{x}+4\sqrt{x}\right)}\ge0\)

Đặt \(M=\frac{4x^2+9x+18\sqrt{x}+9}{4x\sqrt{x}+4x}\left(x>0\right)\Rightarrow M>0\)

Đặt \(y=\sqrt{x}>0\)ta có \(M=\frac{4x^2+9x+18\sqrt{x}+9}{4x\sqrt{x}+4x}=\frac{4y^4+9y^2+18y+9}{4y^3+4y^2}\)\(=\frac{3\left(4y^3+4y^2\right)+\left(4y^2-12y^3-3y^2+18y+9\right)}{4y^3+4y^2}=3+\frac{\left(2y^2-3y-3\right)^2}{4y^3+4y^2}\ge3\)

\(y>0\Rightarrow\hept{\begin{cases}4y^3+4y^2>0\\\left(2y^2-3y-3\right)^2\ge0\end{cases}\Rightarrow\frac{\left(2y-3y-3\right)^2}{4y^3+4y^2}\ge0}\)

Đẳng thức xảy ra \(\Leftrightarrow2y^2-3y-3=0\Leftrightarrow y=\frac{3+\sqrt{33}}{4}\left(y>0\right)\)

\(\Rightarrow x=\left(\frac{3+\sqrt{33}}{4}\right)^2=\frac{21+3\sqrt{33}}{8}\)

Khi đó \(A=M+\frac{1}{M}=\frac{8M}{9}+\left(\frac{M}{9}+\frac{1}{M}\right)\ge\frac{8\cdot3}{9}+2\sqrt{\frac{M}{9}\cdot\frac{1}{M}}=\frac{8}{3}+\frac{2}{3}=\frac{10}{3}\)

Đẳng thức xảy ra \(\Leftrightarrow\hept{\begin{cases}M=3\\\frac{M}{9}=\frac{1}{M}\end{cases}\Leftrightarrow M=3\Leftrightarrow x=\frac{21+3\sqrt{33}}{8}}\)

Vậy \(A_{min}=\frac{10}{3}\Leftrightarrow x=\frac{21+3\sqrt{33}}{8}\)