Theo đầu bài ta có:
\(A=\frac{-5x}{21}+\frac{-5y}{21}+\frac{-5z}{21}\)
\(=\frac{-5x+-5y+-5z}{21}\)
\(=\frac{-5\left(x+y+z\right)}{21}\)
\(=\frac{-5\left(-z+z\right)}{21}\)
\(=\frac{-5\cdot0}{21}\)
\(=\frac{0}{21}=0\)

\(A=\frac{-5x}{21}+\frac{-5y}{21}+\frac{-5z}{21}\)

=>\(A=\frac{\left(-5x\right)+\left(-5y\right)+\left(-5z\right)}{21}\)

=>\(A=\frac{\left(-5\right)\left(x+y+z\right)}{21}\)

=>\(A=\frac{\left(-5\right)\left(-z+z\right)}{21}\)

=>\(A=\frac{\left(-5\right).0}{21}\)

=>\(A=\frac{0}{21}\)

=>A=0

\(A=\frac{-5x}{21}+\frac{-5y}{21}+\frac{-5z}{21}\)

\(=\frac{-5x-5y-5z}{21}\)

\(=\frac{-5\left(x+y+z\right)}{21}\)

Do  \(x+y=-z\) =>    \(x+y+z=0\)

Như vậy  \(A=0\)

\(A=\dfrac{-5x}{21}+\dfrac{-5y}{21}+\dfrac{-5z}{21}\)

\(=\dfrac{-5x-5y-5z}{21}\\ =\dfrac{-5\left(x+y\right)-5z}{21}\\ =\dfrac{-5\cdot\left(-z\right)-5z}{21}\\ =\dfrac{5z-5z}{21}\\ =\dfrac{0}{21}\\ =0\)

`Answer:`

\(A=\frac{-5x}{21}+\frac{-5y}{21}+\frac{-5z}{21}\)

\(=\frac{-5x-5y-5z}{21}\)

\(=\frac{-5\left(x+y\right)-5z}{21}\)

\(=\frac{-5\left(-z\right)-5z}{21}\)

\(=\frac{5z-5z}{21}\)

\(=\frac{0}{21}\)

\(=0\)

Cảm ơn bạn nha

x + y = -z => x+y +z = -z + z =0

\(\frac{-5x}{21}+\frac{-5y}{21}+\frac{-5z}{21}=\frac{\left(-5x\right)+\left(-5y\right)+\left(-5z\right)}{21}=\frac{-5.\left(x+y+z\right)}{21}.\frac{-5.0}{21}=\frac{0}{21}=0\)

-5x/21+-5y/21+-5z/21=-5/21

(x+y+z)=-5/21.(-z+z)=0

bài 1:rất dễ,nhân chéo sẽ giải đc

bài 2: x+y=-x

=>x+y+z=0

Ta có: \(A=\frac{-5x}{21}+\frac{-5y}{21}+\frac{-5z}{21}=\frac{\left(-5x\right)+\left(-5y\right)+\left(-5z\right)}{21}=\frac{-5.\left(x+y+z\right)}{21}=\frac{0}{21}=0\)

bài 1:

\(\frac{1}{2a^2+1}:x=2\)

\(\Leftrightarrow\frac{1}{2a^2+1}.\frac{1}{x}=2\)

\(\Leftrightarrow\frac{1}{\left(2a^2+1\right).x}=2\)

\(\Leftrightarrow x=\frac{1}{\frac{\left(2a^2+1\right)}{2}}=\frac{1}{2a^2+1}.\frac{1}{2}=\frac{1}{\left(2a^2+1\right).2}=\frac{1}{4a^2+2}\)

\(A=\dfrac{-5x}{21}+\dfrac{-5y}{21}+\dfrac{-5z}{21}\)

\(A=\dfrac{-5x+\left(-5y\right)}{21}+\dfrac{-5z}{21}\)

\(A=\dfrac{-5\cdot\left(x+y\right)}{21}+\dfrac{-5z}{21}\)

\(A=\dfrac{-5\cdot\left(-z\right)}{21}+\dfrac{-5z}{21}\)

\(A=\dfrac{5z}{21}+\dfrac{-5z}{21}\)

\(A=\dfrac{5z+\left(-5z\right)}{21}=\dfrac{0}{21}=0\)

Vậy \(A=0\)

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