\(A=1-\sqrt{1-6x+9x^2}+\left(3x-1\right)^2\)
\(A=1-\sqrt{\left(3x-1\right)^2}+\left(3x-1\right)^2\)
\(A=1-\left(3x-1\right)+\left(3x-1\right)^2\)
\(A=1-3x+1+9x^2-6x+1\)
\(A=9x^2-9x+3\)
\(A=\left(3x\right)^2-2.3x.\frac{9}{6}+\frac{81}{36}-\frac{27}{36}\)
\(A=\left(3x-\frac{9}{6}\right)^2-\frac{27}{36}\)
\(A=\left(3x-\frac{9}{6}\right)^2-\frac{3}{4}\ge0\forall x\)
Dấu = xảy ra khi:
\(3x-\frac{9}{6}=0\Leftrightarrow3x=\frac{9}{6}\Leftrightarrow x=0,5\)
Vậy Amin = -3/4 tại x = 0,5
A=1-\(\sqrt{\left(3x-1\right)^2}\)+(3x-1)^2
A=1-/3x-1/+(3x-1)^2
đặt t=/3x-1/ với t>=0
khi đó A=t^2-t+1
A=t^2-t+1/4+3/4
A=(t-1/2)^2+3/4
khi đó A>=3/4
dấu bằng xảy ra khi t=1/2 hay x=1/2
Chúc bạn học tốt!
\(A=1-\sqrt{\left(3x-1\right)^2}+\left(3x-1\right)^2\)
\(A=\left(3x-1\right)^2-\left|3x-1\right|+1\)
+) Với \(x\ge\frac{1}{3}\)\(\Rightarrow\)\(A=\left(3x-1\right)^2-\left(3x-1\right)+\frac{1}{4}+\frac{3}{4}=\left(3x-\frac{3}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=\frac{1}{2}\) ( tm )
+) Với \(x< \frac{1}{3}\)\(\Rightarrow\)\(A=\left(3x-1\right)^2+\left(3x-1\right)+\frac{1}{4}+\frac{3}{4}=\left(3x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=\frac{1}{6}\) ( tm )
Vậy GTNN của \(A=\frac{3}{4}\) khi \(\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{6}\end{cases}}\)
A=\(1-\sqrt{1-6x+9x^2}+\left(3x-1\right)^2\)
\(=1-\sqrt{\left(3x-1\right)^2}+\left(3x-1\right)^2\)
\(=1-\left|3x-1\right|+\left(3x-1\right)^2\)
Đặt: \(t=\left|3x-1\right|\); \(t\ge0\)
Ta có: \(A=1-t+t^2=t^2-2.t.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(t-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
"=" xảy ra <=> \(t-\frac{1}{2}=0\Leftrightarrow t=\frac{1}{2}\)(tm)
Với \(t=\frac{1}{2}\) ta có: \(\left|3x-1\right|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}3x-1=\frac{1}{2}\\3x-1=-\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{6}\end{cases}}}\)(tm)
Vậy giá trị nhỏ nhất của A =3/4 khi x=1/2 hoặc x=1/6
