TA CÓ:

\(ab+\frac{1}{a^2}+\frac{1}{b^2}\ge ab+2\sqrt{\frac{1}{a^2}.\frac{1}{b^2}}=ab+\frac{2}{ab}\ge2\sqrt{ab.\frac{2}{ab}}=2\sqrt{2}\)

\(a,2x-1-3x\left(2x-1\right)=0\)

\(\Leftrightarrow2x-1-6x^2+3x=0\)

\(\Leftrightarrow5x-1-6x^2=0\)

\(\Leftrightarrow6x^2-5x+1=0\)

\(\Leftrightarrow6x^2-2x-3x+1=0\)

\(\Leftrightarrow2x\left(3x-1\right)-\left(3x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\3x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=1\\3x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{3}\end{cases}}\)

\(b,2x^2+4x=0\)

\(\Leftrightarrow2x\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x+4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)

\(A=4-x^2+2x\)

\(=-\left(x^2-2x+1\right)+5\)

\(=-\left(x-1\right)^2+5\le5\)

Dấu "= " ra khi x=1

\(B=\left(2x-1\right)^2+\left(x+2\right)^2\)

\(B=4x^2-4x+1+x^2+4x+4\)

\(B=5x^2+5\)

Ta có: \(5x^2\ge0\forall x\)

\(\Rightarrow5x^2+5\ge5\forall x\)

\(B=5\Leftrightarrow5x^2=0\Leftrightarrow x=0\)

Vậy \(B_{min}=5\Leftrightarrow x=0\)

Tham khảo nhé~

B\(=\left(2x-1\right)^2+\left(x+2\right)^2\)

\(=4x^2-4x+1+x^2+4x+4\)

\(=5x^2+5\ge5\)

Dấu "=" xảy ra khi x=0

Vậy ...

Giúp t ư,gắt thế :(

SLT:\(A=3\left(x^2+y^2\right)=x^2+y^2+2\left(x^2+y^2\right)\)

\(=x^2+y^2+2\left(xy+12\right)\)

\(=x^2+y^2+2xy+24=\left(x+y\right)^2+24\ge24\)

Nên GTNN của A là 24 đạt được khi\(x=-y\) hay \(2x^2=-x^2+12\Leftrightarrow3x^2=12\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)

\(x=2\Rightarrow y=-2;x=-2\Rightarrow y=2\)

cảm ơn, là giúp tui mak nếu hiểu là gíup tao cũng đc, cho thân thiện^^