a) \(M=x^2-3x+10\)

\(M=x^2-2\cdot\dfrac{3}{2}\cdot x+\dfrac{9}{4}+\dfrac{31}{4}\)

\(M=\left(x^2-2\cdot\dfrac{3}{2}\cdot x+\dfrac{9}{4}\right)+\dfrac{31}{4}\)

\(M=\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}\)

Mà: \(\left(x-\dfrac{3}{2}\right)^2\ge0\) nên: \(M=\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}\ge\dfrac{31}{4}\)

Dấu "=" xảy ra 

\(\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}=\dfrac{31}{4}\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=0\)

\(\Leftrightarrow x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)

Vậy: \(M_{min}=\dfrac{31}{4}\) với \(x=\dfrac{3}{2}\)

b) \(N=2x^2+5y^2+4xy+8x-4y-100\)

\(N=x^2+x^2+4y^2+y^2+4xy+8x-4y-120+16+4\)

\(N=\left(x^2+4xy+4y^2\right)+\left(x^2+8x+16\right)+\left(y^2-4y+4\right)-120\)

\(N=\left(x+2y\right)^2+\left(x+4\right)^2+\left(y-2\right)^2-120\)

Mà:

\(\left\{{}\begin{matrix}\left(x+2y\right)^2\ge0\\\left(x+4\right)^2\ge0\\\left(y-2\right)^2\ge0\end{matrix}\right.\) nên \(N=\left(x+2y\right)^2+\left(x+4\right)^2+\left(y-2\right)^2-120\ge120\)

Dấu "=" xảy ra:

\(\left\{{}\begin{matrix}\left(x+2y\right)^2=0\\\left(x+4\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-4+2y=0\\x=-4\\y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=-4\\y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)

Vậy: \(N_{min}=120\) khi \(\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)

a

\(M=x^2-3x+10=x^2-2.\dfrac{3}{2}.x+\dfrac{9}{4}+\dfrac{31}{4}\\ =\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}\ge\dfrac{31}{4}\)

Min M \(=\dfrac{31}{4}\) khi và chỉ khi \(x=\dfrac{3}{2}\)

b

\(N=2x^2+5y^2+4xy+8x-4y-100\\ =x^2+8x+16+y^2-4y+4+x^2+4xy+4y^2-120\\ =\left(x+4\right)^2+\left(y-2\right)^2+\left(x+2y\right)^2-120\ge-120\)

Min N \(=-120\) khi và chỉ khi \(x=-4\) và \(y=2\)

\(A=\dfrac{6x-2}{3x^2+1}\\ \Leftrightarrow3Ax^2+A=6x-2\\ \Leftrightarrow3Ax^2-6x+A+2=0\)

Coi đây là PT bậc 2 ẩn x, PT có nghiệm 

\(\Leftrightarrow\Delta'=9-3\left(A+2\right)\ge0\\ \Leftrightarrow3-3A\ge0\\ \Leftrightarrow A\le1\)

Vậy A chỉ có max, không có min

\(A_{max}=1\Leftrightarrow3x^2+1=6x-2\Leftrightarrow3\left(x-1\right)^2=0\Leftrightarrow x=1\)

Cần cm : \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)

\(\Leftrightarrow\left(\left|a\right|+\left|b\right|\right)^2\ge\left(\left|a+b\right|\right)^2\Leftrightarrow a^2+2\left|ab\right|+b^2\ge a^2+2ab+b^2\)

\(\Leftrightarrow\left|ab\right|\ge ab\) (luôn đúng; dấu "=" xảy ra \(\Leftrightarrow ab\ge0\))

Áp dụng ta có :

\(A=\left|x+3\right|+5\left|6x+1\right|+\left|x-1\right|+3=\left(\left|x+3\right|+\left|1-x\right|\right)+5\left|6x+1\right|+3\)

\(\ge\left|x+3+1-x\right|+5\left|6x+1\right|+3=5\left|6x+1\right|+7\ge7\) có GTNN là 7

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+3\right)\left(1-x\right)\ge0\\\left|6x+1\right|=0\end{cases}\Rightarrow x=-\frac{1}{6}\left(TM\right)}\)

vẬY \(D_{min}=7\) khi \(x=-\frac{1}{6}\)

a) Do \(\left|1+2x\right|\ge0\Rightarrow\dfrac{-1}{4}\left|1+2x\right|\le0\)

\(\Rightarrow A=2,25-\dfrac{1}{4}\left|1+2x\right|\le2,25\)

\(maxA=2,25\Leftrightarrow x=-\dfrac{1}{2}\)

b) Do \(\left|2x-3\right|\ge0\Rightarrow3+\dfrac{1}{2}\left|2x-3\right|\ge3\)

\(\Rightarrow B=\dfrac{1}{3+\dfrac{1}{2}\left|2x-3\right|}\le\dfrac{1}{3}\)

\(maxB=\dfrac{1}{3}\Leftrightarrow x=\dfrac{3}{2}\)

mình ghi nhầm đề bài là Tìm giá trị lớn nhất nhé

\(C=3\left|x-2\right|+\left|3x+1\right|=\left|3x-6\right|+\left|3x+1\right|=\left|6-3x\right|+\left|3x+1\right|\)

\(\ge\left|6-3x+3x+1\right|=7\)

Dấu "=" xảy ra khi \(-\frac{1}{3}\le x\le2\)

b) \(\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)\)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(=\left(x^2+5x\right)^2-36\ge-36\)

Vậy GTNN của bt là -36\(\Leftrightarrow x^2+5x=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

a) \(3x^2-6x-1=3\left(x^2-2x-\frac{1}{3}\right)\)

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

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

Vậy GTNN của bt là - 4\(\Leftrightarrow x=1\)