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Bài 1:
Ta có: \(4-2\left(x+1\right)=2\)
\(\Leftrightarrow2\left(x+1\right)=2\)
\(\Leftrightarrow x+1=1\)
hay x=0
Bài 2:
Ta có: \(\left|2x-3\right|-1=2\)
\(\Leftrightarrow\left|2x-3\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
\(a,2^{x+1}=32\\ 2^{x+1}=2^5\\ x+1=5\\ x=4\\ b,2^{2x}+2^{2x+1}=48\\ 2^{2x}+2\cdot2^{2x}=48\\ 3\cdot2^{2x}=48\\ 2^{2x}=16\\ 2^{2x}=2^4\\ 2x=4\\ x=2\)
\(c,3^x+5\cdot3^{x+1}=144\\ 3^x+15\cdot3^x=144\\ 16\cdot3^x=144\\ 3^x=9\\ 3^x=3^2\\ x=2\\ d,3^{x+5}=9^{x+1}\\ 3^{x+5}=3^{2x+2}\\ x+5=2x+2\\ x=3\)
1.
| x + 2 | = | 2 - 3x |
xét 2 trường hợp :
+) TH1 :
2 - 3x = x + 2
-3x + x = 2 + 2
2x = 4
x = 4 : 2 = 2
+) TH2 :
2 - 3x = - ( x + 2 )
2 - 3x = -x - 2
-3x - x = 2 - 2
-4x = 0
x = 0 : ( -4 )
x = 0
bài còn lại tương tự
a)\(\frac{5}{3}-\frac{2}{3}\times x=1\)
=>\(\frac{2}{3}\times x=\frac{5}{3}-1\)
=>\(\frac{2}{3}\times x=\frac{2}{3}\)
=>\(x=\frac{2}{3}:\frac{2}{3}\)
=>\(x=1\)
b)\(\frac{1}{2}+\frac{5}{7}:x=\frac{1}{6}\)
=>\(\frac{5}{7}:x=\frac{1}{6}-\frac{1}{2}\)
=>\(\frac{5}{7}:x=-\frac{1}{3}\)
=>\(x=-\frac{1}{3}\times\frac{5}{7}\)
=>\(x=-\frac{5}{21}\)
\(\frac{5}{n+1}=\frac{n+1}{5}\)
\(\Leftrightarrow\left(n+1\right)^2=5^2\)
\(\Leftrightarrow\sqrt{\left(n+1\right)^2}=\sqrt{5^2}\)
\(\Leftrightarrow n+1=5\)
\(\Leftrightarrow n=5-1\)
\(\Leftrightarrow n=4\)
\(\frac{2x+1}{5}=\frac{x+2}{8}\)
\(\Leftrightarrow8\left(2x+1\right)=5\left(x+2\right)\)
\(\Leftrightarrow16x+8=5x+10\)
\(\Leftrightarrow11x=2\)
\(\Leftrightarrow x=\frac{2}{11}\)