\(a,\frac{x-1}{9}=\frac{8}{3}\)
\(\Leftrightarrow x-1=24\)
\(\Rightarrow x=25\)
\(b,-\frac{x}{4}=-\frac{9}{x}\)
\(\Leftrightarrow x^2=36\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
\(c,\frac{x}{4}=\frac{18}{x+1}\)
\(\Leftrightarrow x^2+x=72\)
\(\Leftrightarrow x\left(x+1\right)=72..\)
ấn nhầm: lm tiếp nhé!
\(x\left(x+1\right)=72\)
\(\text{Mà x thuộc Z nên }x\left(x+1\right)=8\left(8+1\right)\)
\(\Leftrightarrow x=8\)
a) \(\frac{x-1}{9}=\frac{8}{3}\)
\(\Rightarrow3x-3=72\)
\(\Rightarrow3x=75\)
\(\Rightarrow x=25\)
b) \(\frac{-x}{4}=\frac{-9}{x}\)
\(\Rightarrow\left(-x\right).\left(-x\right)=4.9\)
\(\Rightarrow x^2=36\)
\(\Rightarrow x=\pm6\)
c) \(\frac{x}{4}=\frac{18}{x+1}\)
\(\frac{x}{4}-\frac{18}{x+1}=0\)
\(\frac{x\left(x+1\right)}{4\left(x+1\right)}-\frac{72}{4\left(x+1\right)}=0\)
\(\frac{x^2+x-72}{4\left(x+1\right)}=0\)
đến đây chịu :))
a) \(\frac{x-1}{9}=\frac{8}{3}\Rightarrow\left(x-1\right)\cdot3=8\cdot9\)
\(\Rightarrow3x-3=72\)
\(\Rightarrow x=\left(72+3\right):3=25\)
b) \(\frac{-x}{4}=-\frac{9}{x}\Rightarrow-x\cdot x=-9\cdot4\)
\(\Rightarrow-x^2=-36\Rightarrow x=6\)
c) \(\frac{x}{4}=\frac{18}{x+1}\Rightarrow\left(x+1\right)\cdot x=18\cdot4\)
\(\Rightarrow x^2+x=72\)
\(\Rightarrow x=8\)
a)\(\frac{x-1}{9}=\frac{8}{3}\)
\(\Rightarrow\frac{x-1}{9}=\frac{24}{9}\)
\(\Rightarrow x-1=24\)
\(\Rightarrow x=25\)
b)\(-\frac{x}{4}=-\frac{9}{x}\)
\(\Rightarrow\left(-x\right).x=-9.4\)
\(\Rightarrow-x^2=-36\)
\(\Rightarrow x^2=36\)
\(\Rightarrow x^2=6^2\)
\(\Rightarrow x=6\)
c)\(\frac{x}{4}=\frac{18}{x+1}\)
\(\Rightarrow x.\left(x+1\right)=4.18\)
\(\Rightarrow x.\left(x+1\right)=72\)
\(\Rightarrow x.\left(x+1\right)=8.9\)
\(\Rightarrow x.\left(x+1\right)=8.\left(8+1\right)\)
\(\Rightarrow x=8\)
\(\frac{x-1}{9}=\frac{8}{3}\)
\(\Rightarrow\left(x-1\right)\cdot3=9\cdot8\)
\(\Rightarrow x-1=\frac{9\cdot8}{3}\)
\(\Rightarrow x-1=24\)
\(\Rightarrow x=24+1\)
\(\Rightarrow x=25\)
