a) \(\left|x-1\right|-1=2\)

\(\Rightarrow\left|x-1\right|=3\Rightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)

Vậy......

b) \(\left|5x+1\right|+\left|6y-3\right|\le0\)

Vì \(\left\{{}\begin{matrix}\left|5x+1\right|\ge0\forall x\\\left|6y-3\right|\ge0\forall y\end{matrix}\right.\) Để biểu thức <= 0

\(\Rightarrow\left\{{}\begin{matrix}\left|5x+1\right|=0\\\left|6y-3\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{5}\\y=\dfrac{1}{2}\end{matrix}\right.\)

Vậy........

c) \(\left|3x-1\right|+\left(2y-1\right)^{20}=0\)

Vì \(\left\{{}\begin{matrix}\left|3x-1\right|\ge0\forall x\\\left(2y-1\right)^{20}\ge0\forall y\end{matrix}\right.\)

Để biểu thức = 0

\(\Rightarrow\left\{{}\begin{matrix}\left|3x-1\right|=0\\\left(2y-1\right)^{20}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=\dfrac{1}{2}\end{matrix}\right.\)

Vậy........

d/ \(\left|x-3\right|+\left|x+10\right|=13\)

1. Đề có thiếu chỗ nào k?

3. \(2x\left(x-3\right)\left(2x+7\right)-4x\left(x^2+2\right)\)

\(=\left(2x^2-6x\right)\left(2x+7\right)-4x^3-8x\)

\(=4x^3+14x^2-12x^2-42x-4x^3-8x\)

\(=2x^2-50x\)

Vậy hệ số của x³ là 0

\(VT:\left(x+y\right)^2-y^2=\left(x+y-y\right)\left(x+y+y\right)=x\left(x+2y\right)=VP\left(đpcm\right)\)