\(\hept{\begin{cases}x^3+16x=6x^2+9\\9y^2+32=y^2+31y\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^3-6x^2+16x-9=0\\9y^2-y^2-31y+32=0\end{cases}}\)

Đề sai sao ý 

đề bài đúng nhé, mak mk cũng lm đc rồi

a) \(x^3-4x^2-8x+8\)

\(=x^3+8-4x^2-8x\)

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

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

\(=\left(x+2\right)\left(x^2-6x+4\right)\)

\(=\left(x+2\right)\left(x^2-6x+9-5\right)\)

\(=\left(x+2\right)\left[\left(x-3\right)^2-5\right]\)

\(=\left(x+2\right)\left(x-3-\sqrt{5}\right)\left(x-3+\sqrt{5}\right)\)

b) \(1+6x-6x^2-x\)

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

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

Ta có: a+b=-5

<=>(a+b)²=25

<=>a²+2ab+b²=25

<=>a²+b²+12=25

<=>a²+b²=17

Ta có: \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

Thay a+b=-5,ab=6,a²+b²=17 vào biểu thức trên ta được:

\(-5\left(17-6\right)=-5.11=-55\)

ta có

\(a+b=5\)

=>\(\left(a+b\right)^2=25\)

=>\(a^2+2ab+b^2=25\)

=>\(a^2+2.6+b^2=25\)

=>\(a^2+12+b^2=25\)

=>\(a^2+b^2=17\)

ta có

\(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

thay \(a+b=-5;ab=6;a^2+b^2=17\) vào bt trên ,ta có

\(-5\left(17-6\right)=-5.11=-55\)

vậy \(a^3+b^3=-55\)

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

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

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...