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\(a,2^{x+1}=32\\ 2^{x+1}=2^5\\ x+1=5\\ x=4\\ b,2^{2x}+2^{2x+1}=48\\ 2^{2x}+2\cdot2^{2x}=48\\ 3\cdot2^{2x}=48\\ 2^{2x}=16\\ 2^{2x}=2^4\\ 2x=4\\ x=2\)
\(c,3^x+5\cdot3^{x+1}=144\\ 3^x+15\cdot3^x=144\\ 16\cdot3^x=144\\ 3^x=9\\ 3^x=3^2\\ x=2\\ d,3^{x+5}=9^{x+1}\\ 3^{x+5}=3^{2x+2}\\ x+5=2x+2\\ x=3\)
Mình giải phần 1 ) thôi
\(1)\)
\(a)\frac{3}{2}x-\frac{1}{3}=1-x\)
\(\Rightarrow\frac{3}{2}x+x=1-\frac{1}{3}\)
\(\Rightarrow\frac{5}{2}x=\frac{2}{3}\)
\(\Rightarrow x=\frac{2}{3}:\frac{5}{2}\)
\(\Rightarrow x=\frac{2}{3}.\frac{2}{5}\)
\(\Rightarrow x=\frac{4}{15}\)
b ) \(\left(\frac{1}{3}+x\right)^3=27\)
\(\Rightarrow\frac{1}{3}+x=3\)
\(\Rightarrow x=3-\frac{1}{3}\)
\(\Rightarrow x=\frac{9}{3}-\frac{1}{3}\)
\(\Rightarrow x=\frac{8}{3}\)
Chúc bạn học tốt !!!
a, 3.(2\(x\) + 4) + 198 = (-3)2.10
3.(2\(x\) + 4) + 198 = 90
3.(2\(x\) + 4) = 90 - 198
3.(2\(x\) + 4) = - 108
2\(x\) + 4 = -108 : 3
2\(x\) + 4 = -36
2\(x\) = - 36 - 4
2\(x\) = - 40
\(x\) = -40 : 2
\(x\) = - 20
b, 2.(\(x\) + 7) - 6 = 18
2.(\(x\) + 7) = 18 + 6
2.(\(x\) + 7) =24
\(x\) + 7 = 24 : 2
\(x\) + 7 = 12
\(x\) = 12 - 7
\(x\) = 5
a, 12 - (2\(x^2\) - 3) = 7
2\(x^2\) - 3 = 12 - 7
2\(x^2\) - 3 = 5
2\(x^2\) = 8
\(x^2\) = 4
\(\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)
c)\(\Leftrightarrow\)(x+1)+2 chia hết x+1
\(\Rightarrow\)2 chia hết x+1
\(\Rightarrow\)x+1 ∈ {1,-1,2,-2}
\(\Rightarrow\)x ∈ {0,-2,1,-3}
c) \(x+3⋮x+1\)
\(\Rightarrow x+1+2⋮x+1\)
\(\Rightarrow2⋮x+1\) ( vì \(x+1⋮x+1\) )
\(\Rightarrow x+1\in\text{Ư}_{\left(2\right)}\)
\(\text{Ư}_{\left(2\right)}=\text{ }\left\{1;-1;2;-2\right\}\)
| \(x+1\) | \(1\) | \(-1\) | \(2\) | \(-2\) |
| \(x\) | \(0\) | \(-2\) | \(1\) | \(-3\) |
vậy................
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: \(\left|2x-3\right|-1=2\)
\(\Leftrightarrow\left|2x-3\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
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