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x=0 y=0 z=0

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a)\(a^4+a^3+a^3b+a^2b=\left(a^4+a^3b\right)+\left(a^3+a^2b\right)\)

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

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

\(=a^2\left(a+1\right)\left(a+b\right)\)

b)\(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)

\(=\left[\left(x-y+4\right)-\left(2x+3y-1\right)\right]\left[\left(x-y+4\right)+\left(2x+3y-1\right)\right]\)

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

\(=\left(-x-4y+5\right)\left(4x+2y+3\right)\)

c)\(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)

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

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

\(=\left(y-z\right)\left(x^2-y^2\right)-\left(x-y\right)\left(y^2-z^2\right)\)

\(=\left(y-z\right)\left(x-y\right)\left(x+y\right)-\left(x-y\right)\left(y-z\right)\left(y+z\right)\)

\(=\left(y-z\right)\left(x-y\right)\left(x+y-y-z\right)\)

\(=\left(y-z\right)\left(x-y\right)\left(x-z\right)\)

Lời giải:
Theo bài ra ta có:

$3x=2y; 4y=5z$
$\Rightarrow \frac{x}{2}=\frac{y}{3}; \frac{y}{5}=\frac{z}{4}$

$\Rightarrow \frac{x}{10}=\frac{y}{15}=\frac{z}{12}$

Đặt $\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=k$

$\Rightarrow x=10k; y=15k; z=12k$
Khi đó:

$3x^2-y^2+z^2=876$

$\Rightarrow 3(10k)^2-(15k)^2+(12k)^2=876$

$\Rightarrow 219k^2=876$

$\Rightarrow k^2=4$
$\Rightarrow k=\pm 2$

Nếu $k=2$ thì $x=10k=20; y=15k=30; z=12k=24$

Nếu $k=-2$ thì $x=10k=-20; y=15k=-30; z=12k=-24$