Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left[{}\begin{matrix}\dfrac{x}{3}-\dfrac{12}{x}=0\\5:x-1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{x^2-36}{3x}=0\\\dfrac{5}{x}=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}=>x=6và\left(-6\right)\\=>x=5\end{matrix}\right.\)
a) \(\left|x+\frac{1}{2}\right|=\left|2x+3\right|\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=2x+3\\x+\frac{1}{2}=-\left(2x+3\right)\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}2x-x=\frac{1}{2}-3\\x+\frac{1}{2}=-2x-3\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\x+2x=-3-\frac{1}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\3x=\frac{-7}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\x=\frac{-7}{6}\end{array}\right.\)
Vậy \(x\in\left\{\frac{-5}{2};\frac{-7}{6}\right\}\)
\(\left|x+\frac{1}{2}\right|=\left|2x+3\right|\)
\(Ta\) \(có\): \(x+\frac{1}{2}=2x+3\)
\(x+\frac{1}{2}=x+x+3\\\)
\(x+\frac{1}{2}=x+\left(x+3\right)\)
\(\Rightarrow\frac{1}{2}=x+3\)
\(\Rightarrow x=\frac{1}{2}-3\)
\(\Rightarrow x=-\frac{5}{2}\)
Vậy \(x=-\frac{5}{2}\)
b, \(\left|x+\frac{1}{5}\right|+\left|x+\frac{2}{5}\right|+\left|x+1\frac{2}{5}\right|=4x\)
\(Ta\) \(có\)
\(x+\frac{1}{5}+x+\frac{2}{5}+x+1\frac{2}{5}\)\(=4x\)
\(3x+\left(\frac{1}{5}+\frac{2}{5}+1\frac{2}{5}\right)=4x\)
\(3x+2=4x\)
\(3x+2=3x+x\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
a, \(P=\left(x-4\right)^{\left(x-5\right)^{\left(x-6\right)^{\cdot\left(x+6\right)^{\left(x+5\right)}}}}\)
Thay x = 7 ta được:
\(P=\left(7-4\right)^{\left(7-5\right)^{\left(7-6\right)^{\left(7+6\right)^{\left(7+5\right)}}}}\)
\(P=3^{2^{1^{13^{12}}}}=3^2.1^{13^{12}}=9.1=9\)
b, Vì \(x-1=x-1\) nên lũy thừa của nó phải giống nhau
mà \(x+2\ne x+4\)
\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\) có nghiệm \(\Leftrightarrow\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-2\end{matrix}\right.\)
bị sai đề
sai đề ????????????????????????????????