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PP
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 7 2023
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
H
19 tháng 7 2023
\(a,x^2=x^3\\ \Rightarrow x^2-x^3=0\\ \Rightarrow x^2\left(1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2=0\\1-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(b,3^{2x+1}=27\\ \Rightarrow3^{2x+1}=3^3\\ \Rightarrow2x+1=3\\ \Rightarrow2x=3-1\\ \Rightarrow2x=2\\ \Rightarrow x=2:2\\ \Rightarrow x=1\)
\(c,6^2=6^{x-3}\\ \Rightarrow6^{x-3}=6^2\\ \Rightarrow x-3=2\\ \Rightarrow x=2+3\\ \Rightarrow x=5\)
14 tháng 8 2017
\(x+\frac{2}{15}=\frac{1}{3}\)
\(x=\frac{1}{3}-\frac{2}{15}\)
\(x=\frac{1}{5}\)
h, \(h,\frac{1}{3}-\frac{2}{3}:x=\frac{1}{4}\)
\(\frac{2}{3}:x\)= \(\frac{1}{3}-\frac{1}{4}\)
\(\frac{2}{3}:x=\frac{1}{12}\)
\(x=\frac{2}{3}:\frac{1}{12}\)
\(x=8\)
UN
7
NN
0
NK
1
12 tháng 7 2019
đề bài là gì vậy bạn uiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
Thui mình giải đại nhennnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn
=(x+x+x)+(1/2+2/3+3/6)
=3x+(3/6+4/6+3/6)
=3x+15
Hết ùiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
Bye nhennnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn
TT
2
19 tháng 8 2018
\(\frac{2}{3}\left(x-1\right)+\frac{4}{7}\left(2x-3\right)=\frac{1}{21}\left(x-6\right)\)
\(\frac{2}{3}x-\frac{2}{3}+\frac{8}{7}x-\frac{12}{7}=\frac{1}{21}x-\frac{6}{21}\)
\(\frac{2}{3}x+\frac{8}{7}x-\frac{1}{21}x=\frac{2}{3}+\frac{12}{7}-\frac{6}{21}\)
\(\frac{14}{21}x+\frac{24}{21}x-\frac{1}{21}x=\frac{14}{21}+\frac{36}{21}-\frac{6}{21}\)
\(x\left(\frac{14}{21}+\frac{24}{21}-\frac{1}{21}\right)=\frac{14+36-6}{21}\)
\(\frac{37}{21}x=\frac{44}{21}\)
\(x=\frac{44}{21}:\frac{37}{21}=\frac{44}{37}\)
=2*6=12