a.

\(y'=x^2+2\left(m^2-1\right)x+2m-3\)

\(y''=2x+2\left(m^2-1\right)\)

Hàm đạt cực đại tại \(x=2\) khi: \(\left\{{}\begin{matrix}y'\left(2\right)=0\\y''\left(2\right)< 0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}4+4\left(m^2-1\right)+2m-3=0\\4+2\left(m^2-1\right)< 0\end{matrix}\right.\)

Do \(2m^2+2>0\) ;\(\forall m\) nên ko tồn tại m thỏa mãn yêu cầu đề bài

b.

\(y'=x^2+2mx+3\)

\(y''=2x+2m\)

Hàm đạt cực đại tại \(x=-3\) khi: \(\left\{{}\begin{matrix}9-6m+3=0\\-6+2m< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m=2\\m< 3\end{matrix}\right.\)

\(\Rightarrow m=2\)

1.a) có: \(|x-\frac{3}{2}|,|x+1|,\left|x-2\right|\ge0\Rightarrow4x\ge0\Rightarrow x\ge0\)

\(x\ge0\Rightarrow x-\frac{3}{2}\ge\frac{-3}{2}\Rightarrow\left|x-\frac{3}{2}\right|\ge\left|\frac{-3}{2}\right|=\frac{3}{2}\Rightarrow\left|x-\frac{3}{2}\right|=x-\frac{3}{2}\)

cmtt: \(|x-2|=x-2\)

\(\Rightarrow3x-\frac{3}{2}+1-2=4x\)

\(\Rightarrow3x-\frac{5}{2}=4x\)

\(\Rightarrow x=\frac{-5}{2}\left(ko,t/m\right)\)

\(3,\\ a,=a^2+2a+1-a^2+2a-1-3a^2+3=-3a^2+4a+3\\ b,=\left(m^3-m+1-m^2+3\right)^2=\left(m^3-m^2-m+4\right)^2\\ 4,\\ a,\Leftrightarrow25x^2+10x+1-25x^2+9=3\\ \Leftrightarrow10x=-7\Leftrightarrow x=-\dfrac{7}{10}\\ b,\Leftrightarrow-9x^2+30x-25+9x^2+18x+9=30\\ \Leftrightarrow48x=46\Leftrightarrow x=\dfrac{23}{24}\\ c,\Leftrightarrow x^2+8x+16-x^2+1=16\\ \Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)

\(y'=mx^2-2\left(m-1\right)x+3\left(m-2\right)\)

\(y'\ge0\) ; \(\forall x\ge2\)

\(\Leftrightarrow mx^2-2\left(m-1\right)x+3\left(m-2\right)\ge0\) ; \(\forall x\ge2\)

\(\Leftrightarrow mx^2-2mx+3m\ge6-x\)

\(\Leftrightarrow m\left(x^2-2x+3\right)\ge6-x\)

\(\Leftrightarrow m\ge\dfrac{6-x}{x^2-2x+3}\)

\(\Rightarrow m\ge\max\limits_{x\ge2}\dfrac{6-x}{x^2-2x+3}=\dfrac{4}{3}\)

Vậy \(m\ge\dfrac{4}{3}\)

sao ra được 4/3 ạ?

Bài 1: 

c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)

Ta có: \(\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1}\)

\(\Leftrightarrow\dfrac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\dfrac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\dfrac{6x+8}{\left(4x-1\right)\left(4x+1\right)}\)

Suy ra: \(-12x-3=8x-2-6x-8\)

\(\Leftrightarrow-12x-3-2x+10=0\)

\(\Leftrightarrow-14x+7=0\)

\(\Leftrightarrow-14x=-7\)

\(\Leftrightarrow x=\dfrac{1}{2}\)(nhận)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

1: Khi m=3 thì hệ phương trình (1) trở thành:

\(\left\{{}\begin{matrix}3x-2y=-1\\2x+3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{13}\\y=\dfrac{5}{13}\end{matrix}\right.\)

2: Khi x=-1/2 và y=2/3 vào hệ phương trình, ta được:

\(\left\{{}\begin{matrix}2\cdot\dfrac{-1}{2}+3\cdot\dfrac{2}{3}=1\\-\dfrac{1}{2}m-\dfrac{4}{3}=-1\end{matrix}\right.\Leftrightarrow m\cdot\dfrac{-1}{2}=\dfrac{1}{3}\)

hay m=-2/3

M = \(\frac{1}{3}\)+ \(\frac{1}{6}\)+ \(\frac{1}{10}\)+ \(\frac{1}{15}\)( Mẫu chung: 60 )
M = \(\frac{20}{60}\)+ \(\frac{10}{60}\)+ \(\frac{6}{60}\)+ \(\frac{4}{60}\)
M = \(\frac{40}{60}\)
M = \(\frac{2}{3}\)

M = 1/3 + 1/6 + 1/10 + 1/15

M= 1/3+ 6 + 10 + 15

M = 1/34