A=B=0(vì A mũ chẵn, B mũ chẵn)                                                                                                                                                                       => x=0

Ta có :

A + B = 81x²⁰y¹² + 32x¹⁰z²⁰ 

vì 81x²⁰y¹² \(\ge\)0 ; 32x¹⁰z²⁰ \(\ge\)0

nên A + b = 0 \(\Leftrightarrow\)\(\hept{\begin{cases}x^{20}y^{12}=0\\x^{10}z^{20}=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\y=z=0\end{cases}}\)

\(A+B=\left(-3x^5y^3\right)^4+\left(2x^2z^4\right)^5=81x^{20}y^{12}+32x^{10}z^{20}\)

Ta thấy \(81x^{20}y^{12}\ge0;32x^{10}z^{20}\ge0\) => \(81x^{20}y^{12}+32x^{10}z^{20}\ge0\)

Mà A + B = 0 \(\Rightarrow\hept{\begin{cases}x^{20}y^{12}=0\\x^{10}z^{20}=0\end{cases}}\)=> x = 0 ; y và z bất kỳ hoặc y = z = 0 ; x bất kỳ

Ta có :

A=\(\left(-3x^5y^3\right)^4\ge0\forall x,y\)

B=\(\left(2x^2z^4\right)^5=\left(2xz^2\right)^{10}\ge0\forall x,z\)

Mà A+B = 0

\(\Rightarrow\left\{{}\begin{matrix}A=0\\B=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x^5y^3\\2xz^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=0\\z=0\end{matrix}\right.\end{matrix}\right.\)

Vậy x =0 ; y = 0 ; z = 0 là các giá trị cần tìm

a)\(\left|2x-3y\right|+\left|2y-4z\right|=0\)

\(\left\{{}\begin{matrix}\left|2x-3y\right|\ge0\forall x;y\\\left|2y-4z\right|\ge0\forall y;z\end{matrix}\right.\) \(\Rightarrow\left|2x-3y\right|+\left|2y-4z\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|2x-3y\right|=0\\\left|2y-4z\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=3y\\2y=4z\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{2}\\\dfrac{y}{4}=\dfrac{z}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{6}=\dfrac{y}{4}\\\dfrac{y}{4}=\dfrac{z}{2}\end{matrix}\right.\)

\(\Rightarrow\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{2}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{2}=\dfrac{x+y+z}{6+4+2}=\dfrac{7}{12}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{12}.6=\dfrac{7}{2}\\y=\dfrac{7}{12}.4=\dfrac{7}{3}\\z=\dfrac{7}{12}.2=\dfrac{7}{6}\end{matrix}\right.\)

b)\(\left|x-2\right|+\left|x-3\right|+\left|x-4\right|=0\)

\(\left\{{}\begin{matrix}\left|x-2\right|\ge0\\\left|x-3\right|\ge0\\\left|x-4\right|\ge0\end{matrix}\right.\) \(\Leftrightarrow\left|x-2\right|+\left|x-3\right|+\left|x-4\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|x-2\right|=0\\\left|x-3\right|=0\\\left|x-4\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=3\\x=4\end{matrix}\right.\)

Vì \(2\ne3\ne4\) nên \(x\in\varnothing\)

c)

\(\left|x+1\right|+\left|x+2\right|+...+\left|x+8\right|+\left|x+9\right|\)

Với mọi \(x\ge0\) ta có:

\(\left\{{}\begin{matrix}\left|x+1\right|=x+1\\\left|x+2\right|=x+2\\\left|x+8\right|=x+8\\\left|x+9\right|=x+9\end{matrix}\right.\)\(\Leftrightarrow x+1+x+2+...+x+8+x+9=x-1\)

\(\Leftrightarrow9x+90=x-1\)

\(\Leftrightarrow9x=x-89\)

\(\Leftrightarrow-8x=89\)

\(\Leftrightarrow x=\dfrac{89}{-8}\left(KTM\right)\)

Với mọi \(x< 0\) ta có:

\(\left\{{}\begin{matrix}x+1=-x-1\\x+2=-x-2\\x+8=-x-8\\x+9=-x-9\end{matrix}\right.\) \(\Leftrightarrow\left(-x-1\right)+\left(-x-2\right)+...+\left(-x-8\right)+\left(-x-9\right)=x-1\)

\(\Leftrightarrow-9x-90=x-1\)

\(\Leftrightarrow-9x=x+89\)

\(\Leftrightarrow-10x=89\)

\(\Leftrightarrow x=\dfrac{89}{-10}\left(TM\right)\)

d)\(\left|2x-3y\right|+\left|5y-2z\right|+\left|2z-6\right|=0\)

\(\left\{{}\begin{matrix}\left|2x-3y\right|\ge0\\ \left|5y-2z\right|\ge0\\ \left|2z-6\right|\ge0\end{matrix}\right.\) \(\Leftrightarrow\left|2x-3y\right|+\left|5y-2z\right|+\left|2z-6\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|2x-3y\right|=0\\\left|5y-2z\right|=0\\\left|2z-6\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}z=3\\y=\dfrac{6}{5}\\x=\dfrac{9}{5}\end{matrix}\right.\)

cái đây mà Toán 6 ak ????