=  \(\frac{-10z}{21} + \frac{-5z}{21} \)  = \(\frac{-15z}{21} \)

Ta có : 

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

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

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

   = \(\frac{-5}{21}.0\)

    = 0

Vậy A = 0

A=\(\frac{-5x}{21}+\frac{-5y}{21}+\frac{-5z}{21}\)=\(\frac{\left(-5x\right)+\left(-5y\right)+\left(-5z\right)}{21}=\frac{-5.\left(x+y+z\right)}{21}\)vì x+y=z \(\Rightarrow\)x+y là số đối của z

\(\Rightarrow\)x+y+z=0

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

\(\Rightarrow\)A=0

Ngu từng mà ko biết

`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

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:

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

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

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

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

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

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

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

Do  \(x+y=-z\) =>    \(x+y+z=0\)

Như vậy  \(A=0\)

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