\(M=2x^2+3x+4\)

\(\Rightarrow2M=4x^2+6x+8=\left(4x^2+6x+\frac{9}{4}\right)+\frac{23}{4}=\left(2x+\frac{3}{2}\right)^2+\frac{23}{4}\)

Vì\(\left(2x+\frac{3}{2}\right)^2\ge0\Rightarrow M=\left(2x+\frac{3}{2}\right)^2+\frac{23}{4}\ge\frac{23}{4}\)

Dấu "=" xảy ra khi x=-3/4

Vậy Mmin=23/4 khi x=-3/4

\(M=2x^2+3x+4\)

\(2M=4x^2+6x+8\)

\(2M=\left(4x^2+6x+\frac{9}{4}\right)+\frac{23}{4}\)

\(2M=\left(2x+\frac{3}{2}\right)^2+\frac{23}{4}\)

Mà  \(\left(2x+\frac{3}{2}\right)^2\ge0\forall x\)

\(\Rightarrow2M\ge\frac{23}{4}\Leftrightarrow M\ge\frac{23}{8}\)

Dấu "=" xảy ra khi : 

\(2x+\frac{3}{2}=0\Leftrightarrow x=-\frac{3}{4}\)

Vậy ....

Ta có : 

\(x^2+4y^2-4x-4y+5=0\)

\(\Leftrightarrow\)\(\left(x^2-4x+4\right)+\left(4y^2-4y+1\right)=0\)

\(\Leftrightarrow\)\(\left[x^2-2.x.2+2^2\right]+\left[\left(2y\right)^2-2.2y.1+1^2\right]=0\)

\(\Leftrightarrow\)\(\left(x-2\right)^2+\left(2y-1\right)^2=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}\left(x-2\right)^2=0\\\left(2y-1\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-2=0\\2y-1=0\end{cases}}}\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x=2\\2y=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=\frac{1}{2}\end{cases}}}\)

Vậy \(x=2\) và \(y=\frac{1}{2}\)

Chúc bạn học tốt ~ 

        \(x^2+4y^2-4x-4y+5=0\)

\(\Leftrightarrow\)\(\left(x^2-4x+4\right)+\left(4y^2-4y+1\right)=0\)

\(\Leftrightarrow\)\(\left(x-2\right)^2+\left(2y-1\right)^2=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x-2=0\\2y-1=0\end{cases}}\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x=2\\y=\frac{1}{2}\end{cases}}\)

Vậy

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

\(\Leftrightarrow\)\(x^3+1-x^3-3x=1\)

\(\Leftrightarrow\)\(-3x=0\)

\(\Leftrightarrow\)\(x=0\)

Vậy...

\(A=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)

\(=x^3-8-x^3-3x^2-3x-1+3x^2-3\)

\(=-3x-11\)

(x+2)(x+3)-(x-2)(x+5)=0

=> x²+5x+6-x²-3x+10=0

=>2x+16=0 

 =>2x=-16

=>x=-8

a)  \(\left(x+6\right)^2-x\left(x+9\right)=0\)

\(\Leftrightarrow\)\(x^2+12x+36-x^2-9x=0\)

\(\Leftrightarrow\)\(3x+36=0\)

\(\Leftrightarrow\)\(x=-12\)

Vậy...

b) \(6x\left(2x+5\right)-\left(3x+4\right)\left(4x-3\right)=9\)

\(\Leftrightarrow\)\(12x^2+30x-12x^2-7x+12=9\)

\(\Leftrightarrow\)\(23x+12=9\)

\(\Leftrightarrow\)\(x=-\frac{3}{23}\)

Vậy

c) \(2x\left(8x+3\right)-\left(4x+1\right)=13\)

\(\Leftrightarrow\)\(16x^2+6x-4x-1=13\)

\(\Leftrightarrow\)\(16x^2+2x-14=0\)

\(\Leftrightarrow\)\(8x^2+x-7=0\)

\(\Leftrightarrow\)\(\left(x+1\right)\left(8x-7\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-1\\x=\frac{7}{8}\end{cases}}\)

Vậy

d) \(\left(x-4\right)^2-x\left(x+4\right)=0\)

\(\Leftrightarrow\)\(x^2-8x+16-x^2-4x=0\)

\(\Leftrightarrow\)\(-12x+16=0\)

\(\Leftrightarrow\)\(x=\frac{4}{3}\)

Vậy

e) \(\left(x-2\right)^2-\left(2x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\)\(x^2-4x+4-2x^2+x+6=0\)

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

\(\Leftrightarrow\)\(\left(2-x\right)\left(x+5\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)

Vậy