f = | 3/4x - 2/5 | - 3

=> f >= 3

Dấu "=" xảy ra khi :

\(\frac{3}{4x}-\frac{2}{5}=0\)

\(\frac{3}{4x}=\frac{2}{5}\)

\(x=\frac{15}{8}\)

Vậy, gtnn của f là 3 khi x = 15/8

Ta có \(F=-3+\left|\frac{3}{4}x-\frac{2}{5}\right|=\left|\frac{3}{4}x-\frac{2}{5}\right|-3\)

Ta thấy \(\left|\frac{3}{4}x-\frac{2}{5}\right|\ge0\)với mọi x suy ra \(\left|\frac{3}{4}x-\frac{2}{5}\right|-3\ge-3\)

Khi đó \(F\ge-3\)

Do đó giá trị nhỏ nhất của F là -3 khi và chỉ khi \(\frac{3}{4}x-\frac{2}{5}=0\Rightarrow\frac{3}{4}x=\frac{2}{5}\Rightarrow x=\frac{8}{15}\)

Vậy.....

1) \(f\left(x\right)=6x^2-15x+4\)

\(\Rightarrow f\left(x\right)=6\left(x^2-\dfrac{5}{3}x\right)+4\)

\(\Rightarrow f\left(x\right)=6\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}-\dfrac{25}{36}\right)+4\)

\(\Rightarrow f\left(x\right)=6\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}\right)+4-\dfrac{25}{6}\)

\(\Rightarrow f\left(x\right)=6\left(x-\dfrac{5}{6}\right)^2-\dfrac{1}{6}\ge-\dfrac{1}{6}\left(6\left(x-\dfrac{5}{6}\right)^2\ge0,\forall x\right)\)

\(\Rightarrow GTNN\left(f\left(x\right)\right)=-\dfrac{1}{6}\left(tạix=\dfrac{5}{6}\right)\)

2) \(f\left(x\right)=4x^2-13x+5\)

\(\Rightarrow f\left(x\right)=4\left(x^2-\dfrac{13}{4}x\right)+5\)

\(\Rightarrow f\left(x\right)=4\left(x^2-\dfrac{13}{4}x+\dfrac{169}{64}-\dfrac{169}{64}\right)+5\)

\(\Rightarrow f\left(x\right)=4\left(x^2-\dfrac{13}{4}x+\dfrac{169}{64}\right)+5-\dfrac{169}{16}\)

\(\Rightarrow f\left(x\right)=4\left(x-\dfrac{13}{8}\right)^2-\dfrac{89}{16}\ge-\dfrac{89}{16}\left(4\left(x-\dfrac{13}{8}\right)^2\ge0,\forall x\right)\)

\(\Rightarrow GTNN\left(f\left(x\right)\right)=-\dfrac{89}{16}\left(tạix=\dfrac{13}{8}\right)\)

\(F=2x^2+8xy+11y^2-4x-2y+18\)

\(=2\left[x^2+2x\left(2y-1\right)+\left(2y-1\right)^2\right]+3\left(y^2+2y+1\right)+13\)

\(=2\left(x+2y-1\right)^2+3\left(y+1\right)^2+13\ge13\)

\(minF=13\Leftrightarrow\) \(\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)

A=4x²-4x+3

<=> A=4x²-4x+1+2

<=> A=(2x-1)²+2

Vì (2x-1)²\(\ge0\)nên \(\left(2x-1\right)^2+2\ge2\)

Vậy Min_A=2 khi x=\(\frac{1}{2}\)

A=

Ta có:A=4x²-4x+3

A=

A=+2

Vì 0

=>A=(2x−1)²+2≥2

Dấu"=" xảy ra khi:2x-1=0=>x=1/2

Vậy GTNN của A=2<=>x=1/2

\(A=\left(4x^2-4x+1\right)+2=\left(2x-1\right)^2+2\ge2\)

\(A_{min}=2\) khi \(x=\dfrac{1}{2}\)

\(4x^2-4x+3=\left(4x^2-4x+1\right)+2=\left(2x-1\right)^2+2\ge2\)

Dấu bằng xảy ra khi \(x=\dfrac{1}{2}\)

Vậy GTNN của 4x² - 4x + 3 là 2 khi \(x=\dfrac{1}{2}\)

\(\left|2x-1\right|+3\ge3\Leftrightarrow\dfrac{3+\left|2x-1\right|}{14}\ge\dfrac{3}{14}\)

Dấu \("="\Leftrightarrow2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)

\(\dfrac{-4x^2+4x}{15}=\dfrac{-4x^2+4x-1+1}{15}=\dfrac{-\left(2x-1\right)^2+1}{15}\)

Ta có \(-\left(2x-1\right)^2+1\le1\Leftrightarrow\dfrac{-\left(2x-1\right)^2+1}{15}\le\dfrac{1}{15}\)

Dấu \("="\Leftrightarrow2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)