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\(a,\left(4\frac{1}{2}-\frac{2}{5}x\right):1\frac{3}{4}=\frac{11}{14}\)
\(\Rightarrow\left(\frac{9}{2}-\frac{2}{5}x\right):\frac{7}{4}=\frac{11}{4}\)
\(\Rightarrow\left(\frac{9}{2}-\frac{2}{5}x\right)=\frac{11}{4}\cdot\frac{7}{4}\)
\(\Rightarrow\left(\frac{9}{2}-\frac{2}{5}x\right)=\frac{77}{16}\)
\(\Rightarrow\frac{9}{2}-\frac{2}{5}x=\frac{77}{16}\)
\(\Rightarrow-\frac{2}{5}x=\frac{77}{16}-\frac{9}{2}\)
\(\Rightarrow-\frac{2}{5}x=\frac{5}{16}\)
\(\Rightarrow x=\frac{5}{16}:\left(-\frac{2}{5}\right)\)
\(\Rightarrow x=-\frac{25}{32}\)
\(b,\frac{2}{3}\cdot x-\frac{2}{5}x=\frac{9}{3}\)
\(\Rightarrow x\left(\frac{2}{3}-\frac{2}{5}\right)=\frac{8}{3}\)
\(\Rightarrow x\cdot\frac{4}{15}=\frac{8}{3}\)
\(\Rightarrow x=\frac{8}{3}:\frac{4}{15}\)
\(\Rightarrow x=10\)
\(c,\frac{-2}{3}|x|+1\frac{1}{2}=\frac{2}{5}\)
\(\Rightarrow\frac{-2}{3}|x|+\frac{3}{2}=\frac{2}{5}\)
\(\Rightarrow\frac{-2}{3}|x|=\frac{2}{5}-\frac{3}{2}\)
\(\Rightarrow\frac{-2}{3}|x|=-\frac{11}{10}\)
\(\Rightarrow|x|=\frac{-11}{10}:\frac{-2}{3}\)
\(\Rightarrow|x|=\frac{33}{20}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{33}{20}\\x=-\frac{33}{20}\end{cases}}\)
\(d,|2x-\frac{1}{3}|+\frac{1}{6}=\frac{3}{4}\)
\(\Rightarrow|2x-\frac{1}{3}|=\frac{3}{4}-\frac{1}{6}\)
\(\Rightarrow|2x-\frac{1}{3}|=\frac{7}{12}\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{3}=\frac{7}{12}\\2x-\frac{1}{3}=-\frac{7}{12}\end{cases}\Rightarrow\orbr{\begin{cases}2x=\frac{11}{12}\\2x=-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{11}{24}\\x=-\frac{1}{8}\end{cases}}}\)
a) 3x.812x+1 = 81
3x.(34)2x+1 = 81
3x.38x+4 = 81
3x+8x+4 = 81 = 34
=> x + 8x + 4 = 4
9x + 4 = 4
9x = 0
x = 0
b) 2.3x+3x = 27
3x.(2+1) = 27
3x.3 = 27
3x = 9 = 32
=> x = 2
c) 4x -25 = 89
4x = 114
=>...
( câu c bn thử tìm xem có 4 mũ bao nhiêu = 114 ko nha! câu này mk ko tìm giúp bn đk)
3x.812x + 1 = 81
3x.38x + 4 = 81
39x + 4 = 81
9x + 4 = 4
9x = 0
x = 0
a, \(5^x=625\Rightarrow5^x=5^4\Rightarrow x=4\)
b, \(4^{2x-6}=1\Rightarrow4^{2x-6}=4^0\)
\(\Rightarrow2x-6=0\Rightarrow x=3\)
c, \(\left(3x-1\right)^3=8\Rightarrow3x-1=2\)
\(\Rightarrow x=1\)
d, \(49.7^n=2401\Rightarrow7^{n+2}=7^4\)
\(\Rightarrow n+2=4\Rightarrow n=2\)
e, \(x^4.x-27.x=0\)
\(\Rightarrow x\left(x^4-27\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^4-27=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt[4]{27}\end{cases}}\)
f, \(\left(x-6\right)^3=\left(x-6\right)^2\)
\(\Rightarrow\left(x-6\right)^3-\left(x-6\right)^2=0\)
\(\Rightarrow\left(x-6\right)^2\left(x-6-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-6=0\\x-7=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=7\end{cases}}\)
Chúc bạn học tốt!!!
\(a,5^x=625\) \(b,4^{2x-6}=1\)
\(\Rightarrow5^x=5^4\) \(\Rightarrow2x-6=0\) (Vì mọi số mũ không bằng 1)
\(\Rightarrow x=4\) \(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
\(c,\left(3x-1\right)^3=8\) \(d,49.7^n=2401\)
\(\Rightarrow\left(3x-1\right)^3=2^3\) \(\Rightarrow7^n=2401:49\)
\(\Rightarrow3x-1=2\) \(\Rightarrow7^n=49\)
\(\Rightarrow3x=3\) \(\Rightarrow7^n=7^2\)
\(\Rightarrow x=1\) \(\Rightarrow n=2\)
\(e,x^4.x-27.x=0\)
\(\Rightarrow x.\left(x^4-27\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x^4-27=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^4=27\end{cases}}}\)
\(f,\left(x-6\right)^3=\left(x-6\right)^2\)
\(\Rightarrow x.\left(x^4-27\right)=0\) \(\Rightarrow\hept{\begin{cases}\left(x-6\right)^3=\left(x-6\right)^2\\\left(x-6\right)^3=\left(x-6\right)^2\end{cases}\Rightarrow\hept{\begin{cases}\left(x-6\right)^3:\left(x-6\right)^2=1\\\left(x-6\right)^3-\left(x-6\right)^2=0\end{cases}\Rightarrow}\hept{\begin{cases}x-6=1\\x-6=0\end{cases}}\Rightarrow\hept{\begin{cases}x=7\\x=6\end{cases}}}\)
\(a,2x+15=27\)
\(\Rightarrow2x=27-15\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=12\div2\)
\(\Rightarrow x=6\)
a) \(\left(\frac{5}{6}-\frac{2}{3}\right)+\frac{1}{4}:x=-4\)
\(\Rightarrow\frac{1}{6}+\frac{1}{4}:x=-4\)
\(\Rightarrow\frac{1}{4}:x=-4-\frac{1}{6}\)
\(\Rightarrow\frac{1}{4}:x=-\frac{25}{6}\)
\(\Rightarrow\)\(x=\frac{1}{4}:-\frac{25}{6}\)
\(\Rightarrow x=-\frac{3}{50}\)
b) \(\left|2x-\frac{1}{3}\right|+1=\frac{5}{6}\)
\(\Rightarrow\left|2x-\frac{1}{3}\right|=\frac{5}{6}-1\)
\(\Rightarrow\left|2x-\frac{1}{3}\right|=-\frac{1}{6}\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{3}=-\frac{1}{6}\\2x-\frac{1}{3}=\frac{1}{6}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=-\frac{1}{6}+\frac{1}{3}\\2x=\frac{1}{6}+\frac{1}{3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=\frac{1}{6}\\2x=\frac{1}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{6}:2\\x=\frac{1}{2}:2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{1}{4}\end{cases}}\)
\(\left(3x-1\right)⋮\left(x+1\right)\)
\(\Rightarrow\left(3x+3-4\right)⋮\left(x+1\right)\)
\(\Rightarrow\left(-4\right)⋮\left(x+1\right)\)
\(\Rightarrow x+1\inƯ\left(-4\right)=\left\{-4;-1;1;4\right\}\)
\(\Rightarrow x\in\left\{-5;-2;0;3\right\}\)
a, \(x^2\) = \(x^3\)
\(x^3\) - \(x^2\) = 0
\(x^2\)( \(x\) -1) = 0
\(\left[{}\begin{matrix}x^2=0\\x-1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy \(x\) \(\in\) { 0; 1}
e, 32\(x+1\) = 27
\(3^{2x}\)+1 = 33
2\(x\) + 1 = 3
2\(x\) = 2
\(x\) = 1
g, 62 = 6\(x-3\)
2 = \(x-3\)
\(x\) = 3 + 2
\(x\) = 5
\(a,x^2=x^3\\ \Rightarrow x^2-x^3=0\\ \Rightarrow x^2\left(1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2=0\\1-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(b,3^{2x+1}=27\\ \Rightarrow3^{2x+1}=3^3\\ \Rightarrow2x+1=3\\ \Rightarrow2x=3-1\\ \Rightarrow2x=2\\ \Rightarrow x=2:2\\ \Rightarrow x=1\)
\(c,6^2=6^{x-3}\\ \Rightarrow6^{x-3}=6^2\\ \Rightarrow x-3=2\\ \Rightarrow x=2+3\\ \Rightarrow x=5\)