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\(A=2x^2-6x-\sqrt{7}\)
\(=2\left(x^2-3x-\sqrt{\frac{7}{2}}\right)\)
\(=2\left(x^2-3x+\frac{9}{4}-\frac{9+2\sqrt{7}}{4}\right)\)
\(=2\left[\left(x-\frac{3}{2}\right)^2-\frac{9+2\sqrt{7}}{4}\right]\)
\(=2\left(x-\frac{3}{2}\right)^2-\frac{9+2\sqrt{7}}{2}\)
Vì \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-\frac{3}{2}\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-\frac{3}{2}\right)^2-\frac{9+2\sqrt{7}}{2}\ge-\frac{9+2\sqrt{7}}{2}\)
Vậy \(Min_A=\frac{-9+2\sqrt{7}}{2}\Leftrightarrow x=\frac{3}{2}\)
a) \(4x^2+12x+10=\left(2x+3\right)^2+1\ge1\)
Dấu "="\(\Leftrightarrow x=-2\)
b) \(B=\left(3x-1\right)^2+4\ge4\)
Dấu "="\(\Leftrightarrow x=\frac{1}{3}\)
a, \(A=4x^2+12x+10\)
\(=\left(2x+1\right)^2+1\ge1\forall x\)
Dấu"=" xảy ra<=> \(\left(2x+1\right)^2=0\)
\(\Leftrightarrow x=\frac{-1}{2}\)
\(b,B=9x^2-6x+5\)
\(=\left(3x-1\right)^2+4\ge4\forall x\)
Dấu"=" xảy ra<=> \(\left(3x-1\right)^2=0\)
\(\Leftrightarrow x=\frac{1}{3}\)
= \(\left(9x^2+12xy+4y^2\right)+\left(x^2+6x+9\right)+2017\)
\(=\left(3x+2y\right)^2+\left(x+3\right)^2+2017\ge2017\)
=> \(MinP=2017\Leftrightarrow\left\{{}\begin{matrix}2y=-3x\\x=-3\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}x=-3\\y=\dfrac{9}{2}\end{matrix}\right.\)
Ô cho mình hỏi \(Min\) là gì ạ lớp 9 rồi mà chưa học bao giờ.
a)\(A=4x^2+4x+11\)
\(=4x^2+4x+1+10\)
\(=\left(2x+1\right)^2+10\ge10\)
Dấu = khi \(x=\frac{-1}{2}\)
Vậy MinA=10 khi \(x=\frac{-1}{2}\)
b)\(B=3x^2-6x+1\)
\(=3x^2-6x+3-2\)
\(=3\left(x^2-2x+1\right)-2\)
\(=3\left(x-1\right)^2-2\ge-2\)
Dấu = khi \(x=1\)
Vậy MinB=-2 khi \(x=1\)
c)\(C=x^2-2x+y^2-4y+6\)
\(=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+1\)
\(=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\)
Dấu = khi \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
Vậy MinC=1 khi \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)