ĐK: \(\left(x-2\right)\left(x^2+1\right)+2x\left(x-2\right)\ne0\)

\(\Leftrightarrow\left(x-2\right)\left(x+1\right)^2\ne0\Leftrightarrow x\ne-1;2\)

Ta có: \(A=\frac{x^2\left(x-2\right)+4\left(x-2\right)}{\left(x-2\right)\left(x^2+2x+1\right)}=\frac{x^2+4}{\left(x+1\right)^2}=\frac{t^2-2t+5}{t^2}\left(t=x+1\right)\)

\(=\frac{5}{t^2}-\frac{2}{t}+1=5\left(\frac{1}{t}-\frac{1}{5}\right)^2+\frac{4}{5}\ge\frac{4}{5}\)

Đẳng thức xảy ra khi t = 5 hay x=4

Vậy..

Vì \(\hept{\begin{cases}\left(2x+3\right)^2\ge0\\\left|x^2-\frac{9}{4}\right|\ge0\end{cases}}\)=> \(D\ge3\cdot0+2\cdot0+3,5=3,5\)

Dấu = xảy ra khi       \(x=-\frac{3}{2}\)

Ta có: 

\(D=3\left(2x+3\right)^2+2\left|x^2-\frac{9}{4}\right|+3,5\)

Mà: \(\left(2x+3\right)^2\ge0\)          với mọi x 

   \(\left|x^2-\frac{9}{4}\right|\ge0\)   với mọi x

\(\Rightarrow3\left(2x+3\right)^2+2\left|x^2-\frac{9}{4}\right|\ge0\)

\(\Rightarrow3\left(2x+3\right)^2+2\left|x^2-\frac{9}{4}\right|+3,5\ge3,5\)

Dấu "=" xảy ra \(\Leftrightarrow\) \(\hept{\begin{cases}2x+3=0\\x^2-\frac{9}{4}=0\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{3}{2}\\x=\pm\frac{3}{2}\end{cases}\Rightarrow}x=\pm\frac{3}{2}}\)

Vậy: GTNN của D bằng 3,5 khi x = \(\pm\)\(\frac{3}{2}\)

Bạn phuong là sai r, hợp của chúng phải là x=-3/2 mới đúng

1 M=\(x^2-4xy+4y^2-2x+4y+10\)

=\(\left(x^2-4xy+4y^2\right)+\left(-2x+4y\right)+10\)

\(=\left(x-2y\right)^2-2\left(x-2y\right)+10\)

\(=\left(x-2y\right)\left(x-2y-2\right)+10\)

vì \(\left(x-2y\right)\left(x-2y-2\right)\ge0\)

nên \(\left(x-2y\right)\left(x-2y-2\right)+10\ge10\)

\(\Rightarrow\)A\(\ge13\)

dấu "=" xảy ra khi (x-2y)(x-2y-2)=0

\(\left[{}\begin{matrix}x-2y=0\\x-2y-2=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}2y=x\\x-2y=2\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0;y=0\\x=2;y=1\end{matrix}\right.\)

vậy GTNN của M=10 khi x=0; y=0

x=2;y=1

\(D\left(x\right)=\frac{\left(x^2+2x+3\right)\left(x^2+2x+9\right)}{x^2+2x+1}\)

Đặt \(a=x^2+2x+1=\left(x+1\right)^2\ge0\)

\(\Rightarrow\)\(D\left(x\right)=\frac{\left(a+2\right)\left(a+8\right)}{a}=\frac{a^2+10a+16}{a}\)

Áp dụng BĐT AM-GM ta có:\(a^2+16\ge2\sqrt{a^2.16}=2.4a=8a\)

\(\Rightarrow D\left(x\right)\ge\frac{8a+10a}{a}=\frac{18a}{a}=18\)

Nên minD(x)=18 đạt được khi \(a=4\Leftrightarrow\left(x+1\right)^2=4\Rightarrow\orbr{\begin{cases}x+1=2\\x+1=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)

\(f\left(x\right)=\frac{2x^2-2x+3}{x^2-x+2}=\frac{2\left(x^2-x+2\right)-1}{x^2-x+2}=2-\frac{1}{x^2-x+2}=2-\frac{1}{\left(x-\frac{1}{2}\right)^2+\frac{7}{4}}\)

Ta thấy : \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}\forall x\)

\(\Leftrightarrow\frac{1}{\left(x-\frac{1}{2}\right)^2+\frac{7}{4}}\ge\frac{1}{\frac{7}{4}}=\frac{4}{7}\forall x\)

\(\Rightarrow f\left(x\right)=2-\frac{1}{\left(x-\frac{1}{2}\right)^2+\frac{7}{4}}\ge2-\frac{4}{7}=\frac{10}{7}\forall x\) có GTNN là \(\frac{10}{7}\)

Dấu "=" xảy ra <=> \(\left(x-\frac{1}{2}\right)^2=0\Leftrightarrow x=\frac{1}{2}\)

Vậy \(f\left(x\right)_{min}=\frac{10}{7}\) tại \(x=\frac{1}{2}\)

Sai rồi bạn!

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