mà đây là môn toán mà

\(a,\left(x+2\right)^{10}+\left(x+2\right)^8=0\\ \Leftrightarrow\left(x+2\right)^8\left[\left(x+2\right)^2+1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)^8=0\\\left(x+2\right)^2+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+2=0\\\left(x+2\right)^2=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\end{matrix}\right.\\ b,\left(x+3\right)^{10}-\left(x+3\right)^8=0\\ \Leftrightarrow\left(x+3\right)^8\left[\left(x+3\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+3\right)^8=0\\\left(x+3\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\\left(x+3\right)^2=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x+3=1\\x+3=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\\x=-4\end{matrix}\right.\)

\(1.6x\left(x-10\right)-2x+20=0\)

⇔\(6x\left(x-10\right)-2\left(x-10\right)=0\)

⇔ \(2\left(x-10\right)\left(3x-1\right)=0\)

⇔ x = 10 hoặc x = \(\dfrac{1}{3}\)

KL....

\(2.3x^2\left(x-3\right)+3\left(3-x\right)=0\)

⇔ \(3\left(x-3\right)\left(x^2-1\right)=0\)

⇔ \(x=+-1\) hoặc \(x=3\)

KL....

\(3.x^2-8x+16=2\left(x-4\right)\)

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

⇔ \(\left(x-4\right)\left(x-6\right)=0\)

⇔ \(x=4\) hoặc \(x=6\)

KL.....

\(4.x^2-16+7x\left(x+4\right)=0\)

\(\text{⇔}4\left(x+4\right)\left(2x-1\right)=0\)

⇔ \(x=-4hoacx=\dfrac{1}{2}\)

KL.....

\(5.x^2-13x-14=0\)

⇔ \(x^2+x-14x-14=0\)

\(\text{⇔}\left(x+1\right)\left(x-14\right)=0\)

\(\text{⇔}x=14hoacx=-1\)

KL......

Còn lại tương tự ( dài quá ~ )

Ta có:\(x^2+4y+4=0;y^2+4z+4=0;z^2+4x+4=0\)

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

\(\Leftrightarrow x^2+4x+4+y^2+4y+4+z^2+4z+4=0\)

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

Mà\(\left(x+2\right)^2\ge0;\left(y+2\right)^2\ge0;\left(z+2\right)^2\ge0\)

\(\Leftrightarrow\left(x+2\right)^2+\left(y+2\right)^2+\left(z+2\right)^2\ge0\)

Dấu "=" xảy ra\(\Leftrightarrow\hept{\begin{cases}x+2=0\\y+2=0\\z+2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-2\\y=-2\\z=-2\end{cases}\Leftrightarrow}x=y=z=-2}\)

Vậy\(x^{10}+y^{10}+z^{10}=x^{10}+x^{10}+x^{10}\)                         

                    \(=3\cdot x^{10}=3\cdot\left(-2\right)^{10}=3\cdot1024=3072\)

a) \(\left(x+1\right).10=0\)

                \(x+1=0\)

                \(x=0-1\)

                \(x=-1\)

b) \(15\left(x+2\right)=0\)

            \(x+2=0\)

                    \(x=0-2\)

                    \(x=-2\)

c) \(\left(x-1\right).10=10\)

             \(x-1=10:10\)

              \(x-1=1\)

                 \(x=1+1\)

                  x   =   2

d) \(\left(x-2\right):5=0\)

             \(x-2=0\)

                  \(x=2\)

ai làm có thưởng 2điem

\(x^3-x=0\Leftrightarrow x\left(x^2-1\right)=0\Leftrightarrow\left(x-1\right)x\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\\x=-1\end{matrix}\right..Vậy:x\in\left\{-1;0;-1\right\}\)

\(x^3+4x=0\Leftrightarrow x\left(x^2+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+4=0\end{matrix}\right.mà:x^2+4\ge0+4=4\Rightarrow x=0\)

\(\left(x+2\right)^2=x+2\Leftrightarrow\left(x+2\right)\left(x+2-1\right)=0\Leftrightarrow\left(x+1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)

a) Ta có: \(x^2-2x+1=0\)

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

\(\Leftrightarrow x-1=0\)hay x=1

Vậy: S={1}

c) Ta có: \(x+x^4=0\)

\(\Leftrightarrow x\left(x^3+1\right)=0\)

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

mà \(x^2-x+1>0\forall x\)

nên x(x+1)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Vậy: S={0;-1}

\(\left|3x+2\right|=\left|x-8\right|\)

\(\Leftrightarrow\orbr{\begin{cases}3x+2=x-8\\3x+2=8-x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x-x=-8-2\\3x+x=8-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=-10\\4x=6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{3}{2}\end{cases}}\)

\(x\left(3x+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\3x=-6\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)

\(\left(2x-4\right)\left(x^3-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-4=0\\x^3-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=4\\x^3=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=1\end{cases}}\)

\(\left(4x-8\right)\left(5x+10\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x-8=0\\5x+10=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4x=8\\5x=-10\end{cases}}\)

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

\(\left|3x-10\right|=5\)

\(\Leftrightarrow\orbr{\begin{cases}3x-10=5\\3x-10=-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=15\\3x=5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{5}{3}\end{cases}}\)