4x - 12 = 54
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a: Ta có: \(\sqrt{9x-54}-\sqrt{4x-24}=2\)
\(\Leftrightarrow3\sqrt{x-6}-2\sqrt{x-6}=2\)
\(\Leftrightarrow x-6=4\)
hay x=10
b: Ta có: \(\sqrt{4x^2+4x+1}=7\)
\(\Leftrightarrow\left|2x+1\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=7\\2x+1=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)
\(\hept{\begin{cases}\left(2x-3\right)\left(2y+4\right)=4x\left(y-3\right)+54\\\left(x+1\right)\left(3y-3\right)=3y\left(x+1\right)-12\end{cases}}\)
\(\hept{\begin{cases}4xy+8x-6y-12=4xy-12x+54\\3xy-3x+3y-3=3xy+3y-12\end{cases}}\)
\(\hept{\begin{cases}4xy-4xy+8x+12x-6y-12-54=0\\3xy-3xy-3x+3y-3y-3+12=0\end{cases}}\)
\(\hept{\begin{cases}20x-6y-66=0\\-3x+9=0\end{cases}}\)
\(\hept{\begin{cases}2\left(10x-3y\right)=66\\-3\left(x-3\right)=0\end{cases}}\)
\(\hept{\begin{cases}10x-3y=33\\x-3=0\end{cases}}\)
\(\hept{\begin{cases}10x-3y=33\\x=3\end{cases}}\)
\(1,2x+3x-4x=\left(-2\right)^3\)
<=>\(x=-8\)
\(2,x-2x=4^2+4^0\)
<=>\(-x=16+1\)
<=>\(-x=17\)
<=>\(x=-17\)
\(3,2^3x-3^2x=|12-21|\)
<=>\(-x=9\)
<=>\(x=-9\)
\(4,x-45=2x+54\)
<=>\(x-2x=54+45\)
<=>\(-x=99\)
<=>\(x=-99\)
\(5,5x-12+23=6^7:6^5\)
<=>\(5x+11=6^2\)
<=>\(5x+11=36\)
<=>\(5x=25\)
<=>\(x=5\)
1) (-10)-|5-x|=(-12)
|5-x|=(-10)-(-12)
|5-x|=2
\(\Rightarrow\left[{}\begin{matrix}5-x=2\\5-x=\left(-2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=7\end{matrix}\right.\)
Vậy:............
a: \(\Leftrightarrow2x+1\in\left\{1;-1\right\}\)
hay \(x\in\left\{0;-1\right\}\)
4x=66
=>x ko có giá trị
sai j hả mấy chế