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\(\frac{2x+3}{6}=\frac{7x-3}{15}\)
\(15\left(2x+3\right)=6\left(7x-3\right)\)
\(30x+45=42x-18\)
\(30x-42x=-18-45\)
\(-12x=-63\)
\(x=\frac{21}{4}\)
\(\frac{2x+3}{6}=\frac{7x-3}{15}\)
\(\Rightarrow\frac{5.\left(2x+3\right)}{30}=\frac{2.\left(7x-3\right)}{30}\)
\(\Rightarrow5.\left(2x+3\right)=2.\left(7x-3\right)\)
\(\Rightarrow10x+15=14x-6\)
\(\Rightarrow10x-14x=-6-15\)
\(\Rightarrow-4x=-21\)
\(\Rightarrow x=-21:\left(-4\right)\)
\(\Rightarrow x=\frac{21}{4}\)
c , d giải nốt:
c) 2x + 3/6 = 7x - 13/15
=> (2x + 3) . 15 = (7x - 13) x 6
=> 30x +45 = 42x - 78
=> 30x - 42x = -78 - 45
=> -12x = -123
=> x = 41/4
d) 2-x/4 = 3x - 1/ - 3
=> (2-x) . -3 = 4. (3x - 1)
=> -6 - (-3x) = 12x - 4
(-6) + 4 = 12x - - (-3x)
=> -2 = 9x
=> x = -2/9
\(3x+\frac{3}{6}=7x+\frac{3}{15}\)
\(\Rightarrow4x=\frac{3}{10}\Rightarrow x=\frac{3}{40}\)
Ta co:
2x+3/6 = 7x + 3/15
=> 3/6 = 5x + 3/15
=> x = (3/6 - 3/15):5 = 3/50
\(1,\frac{x+1}{x-2}=\frac{3}{4}\)
\(\Rightarrow3x-6=4x+4\)
\(\Rightarrow3x-4x=4+6\)
\(\Rightarrow-x=10\Leftrightarrow x=-10\)
\(2,\frac{x-1}{3}=\frac{x+3}{5}\)
\(\Rightarrow5x-5=3x+9\)
\(\Rightarrow5x-3x=9+5\)
\(\Rightarrow2x=14\Leftrightarrow x=7\)
\(3,\frac{2x+3}{24}=\frac{3x-1}{32}\)
\(\Rightarrow64x+96=72x-24\)
\(\Rightarrow72x-64x=24+96\)
\(\Rightarrow8x=120\)
\(\Rightarrow x=15\)
1.
b) \(3^x+3^{x+2}=2430\)
\(\Rightarrow3^x.1+3^x.3^2=2430\)
\(\Rightarrow3^x.\left(1+3^2\right)=2430\)
\(\Rightarrow3^x.10=2430\)
\(\Rightarrow3^x=2430:10\)
\(\Rightarrow3^x=243\)
\(\Rightarrow3^x=3^5\)
\(\Rightarrow x=5\)
Vậy \(x=5.\)
c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=15\\2x-15=\pm1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=15:2\\2x-15=1\\2x-15=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\2x=16\\2x=14\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{15}{2};8;7\right\}.\)
Chúc bạn học tốt!
\(2x+\frac{3}{6}=7x-\frac{3}{15}\)
\(2x+\frac{1}{2}=7x-\frac{1}{5}\)
\(2x-7x=-\frac{1}{5}-\frac{1}{2}\)
\(-5x=-\frac{7}{10}\)
\(x=\frac{7}{50}\)