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Ta có : \(156:\left(7x+135\right):x=\frac{156}{\frac{7x+135}{x}}=156.\frac{x}{7x+135}\)
\(\Rightarrow156.\frac{x}{7x+135}=13\)
\(\Rightarrow\frac{x}{7x+135}=\frac{13}{156}\)
\(\Rightarrow\frac{x}{7x+135}=\frac{1}{12}\)
\(\Rightarrow12x=7x+135\)
\(\Rightarrow5x=135\)
\(\Rightarrow x=27\)
\(156:\left(7x+135\right):x=13\)
\(\left(7x+135\right):x=12\)
\(7x+135=12x\)
\(7x-12x=-135\)
\(-5x=-135\)
\(x=\frac{-135}{-5}\)
\(x=27\)
hok tốt!!

| x - \(\frac{1}{2}\)| = 1
TH1:x - \(\frac{1}{2}\) = 1 TH2:x - \(\frac{1}{2}\) = -1
x = 1 + \(\frac{1}{2}\) x = -1 + \(\frac{1}{2}\)
x = \(\frac{3}{2}\) x = \(\frac{-1}{2}\)
Vậy x thuộc \(\frac{3}{2}\)và \(\frac{-1}{2}\)

a: \(\left(x+10\right)\left(x-5\right)=0\)
=>\(\left[{}\begin{matrix}x+10=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=5\end{matrix}\right.\)
b: \(\left(2x+10\right)\left(4+x\right)=0\)
=>\(\left[{}\begin{matrix}2x+10=0\\4+x=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-4\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-5\end{matrix}\right.\)
c: \(\left(4x+20\right)\left(12x-24\right)=0\)
=>\(\left[{}\begin{matrix}4x+20=0\\12x-24=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-20\\12x=24\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
d: \(\left(x-2024\right)\left(4x+4\right)=0\)
=>\(\left[{}\begin{matrix}x-2024=0\\4x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2024\\4x=-4\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-1\\x=2024\end{matrix}\right.\)
e: \(\left(2x-6\right)\left(7+x\right)=0\)
=>\(\left[{}\begin{matrix}2x-6=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\x=-7\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\)
g: (4x+8)(6-x)=0
=>\(\left[{}\begin{matrix}4x+8=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x=6\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-2\\x=6\end{matrix}\right.\)
h: (2x+2)(4x-8)=0
=>2(x+1)*4*(x-2)=0
=>(x+1)(x-2)=0
=>\(\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
i: (2x-2024)(8x-16)=0
=>\(2\left(x-1012\right)\cdot8\cdot\left(x-2\right)=0\)
=>\(\left(x-1012\right)\left(x-2\right)=0\)
=>\(\left[{}\begin{matrix}x-1012=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1012\\x=2\end{matrix}\right.\)

a, Ta có :
xy=6
yz=-14
xz=-21
=>(xyz)2=1764=>xzy=42 hoặc -42
+)xyz=42
=>z=42:6=7
=>x=-3
=>y=-2
+)xyz=-42
=>z=-7
=>y=2
=>x=3

\(3^{x+2}+3^x=270\\ =>3^x.3^2+3^x=270\)
\(=>2.3^x=270:9=30\)
\(=>3^x=30:2=15\)
\(=>3^x=15\)
Có sai đề ko bn ???
x= -1;1;0
Cảm ơn nhé 🧡💛💚💙💜🤎🖤🤍💕