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`Answer:`
\(A=\frac{-5x}{21}+\frac{-5y}{21}+\frac{-5z}{21}\)
\(=\frac{-5x-5y-5z}{21}\)
\(=\frac{-5\left(x+y\right)-5z}{21}\)
\(=\frac{-5\left(-z\right)-5z}{21}\)
\(=\frac{5z-5z}{21}\)
\(=\frac{0}{21}\)
\(=0\)
\(\dfrac{-5x}{21}+\dfrac{-5y}{21}+\dfrac{-5z}{21}=\dfrac{-5x-5y-5z}{21}\)
= \(\dfrac{-5\left(x+y\right)-5z}{21}=\dfrac{-5\left(-z\right)-5z}{21}=\dfrac{5z-5z}{21}=\dfrac{0}{21}=0\)
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:
\(\frac{1}{2a^2+1}:x=2\)
\(\Leftrightarrow\frac{1}{2a^2+1}.\frac{1}{x}=2\)
\(\Leftrightarrow\frac{1}{\left(2a^2+1\right).x}=2\)
\(\Leftrightarrow x=\frac{1}{\frac{\left(2a^2+1\right)}{2}}=\frac{1}{2a^2+1}.\frac{1}{2}=\frac{1}{\left(2a^2+1\right).2}=\frac{1}{4a^2+2}\)
\(A=\frac{-5x}{21}+\frac{-5y}{21}+\frac{-5o}{21}=-\frac{5}{21}\left(x+y+o\right)\)
\(=-\frac{5}{21}\left(-o+o\right)=0\)
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
a. \(\Rightarrow\left\{\begin{matrix}\dfrac{-10}{15}=\dfrac{x}{-9}\\\dfrac{-10}{15}=\dfrac{-8}{y}\\\dfrac{-10}{15}=\dfrac{z}{-21}\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=6\\y=12\\z=14\end{matrix}\right.\)
b. \(\Rightarrow\left\{\begin{matrix}\dfrac{-7}{6}=\dfrac{x}{18}\\\dfrac{-7}{6}=\dfrac{-98}{y}\\\dfrac{-7}{6}=\dfrac{-14}{z}\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=-21\\y=84\\z=-12\end{matrix}\right.\)
a) Ta có: \(\dfrac{-10}{15}=\dfrac{x}{-9}\)
\(\Rightarrow15x=-10.\left(-9\right)\)
\(\Rightarrow15x=90\)
\(\Rightarrow x=6\)
Khi đó: \(\dfrac{6}{-9}=\dfrac{-8}{y}=\dfrac{z}{-21}\)
\(\Rightarrow y=\dfrac{-8\left(-9\right)}{6}=12\)
và \(z=\dfrac{-8\left(-21\right)}{12}\) \(=14\)
Vậy \(\left[{}\begin{matrix}x=6\\y=12\\z=14\end{matrix}\right.\)
b) Lại có: \(\dfrac{-7}{6}=\dfrac{x}{18}\)
\(\Rightarrow6x=-7.18\)
\(\Rightarrow6x=-126\)
\(\Rightarrow x=-21\)
Khi đó \(\dfrac{-21}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}\)
\(\Rightarrow y=\dfrac{-98.18}{-21}=84\)
và \(z=\dfrac{-14.84}{-98}=12\)
Vậy \(\left[{}\begin{matrix}x=-21\\y=84\\z=12\end{matrix}\right.\)
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
\(A=\dfrac{-5x}{21}+\dfrac{-5y}{21}+\dfrac{-5z}{21}\)
\(A=\dfrac{-5x+\left(-5y\right)}{21}+\dfrac{-5z}{21}\)
\(A=\dfrac{-5\cdot\left(x+y\right)}{21}+\dfrac{-5z}{21}\)
\(A=\dfrac{-5\cdot\left(-z\right)}{21}+\dfrac{-5z}{21}\)
\(A=\dfrac{5z}{21}+\dfrac{-5z}{21}\)
\(A=\dfrac{5z+\left(-5z\right)}{21}=\dfrac{0}{21}=0\)
Vậy \(A=0\)
AI NHANH MÌNH TICK CHO