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b) Ta có: \(5^{x+4}-3\cdot5^{x+3}=2\cdot5^{11}\)
\(\Leftrightarrow2\cdot5^{x+3}=2\cdot5^{11}\)
\(\Leftrightarrow x+3=11\)
hay x=8
c) Ta có: \(2\cdot3^{x+2}+4\cdot3^{x+1}=10\cdot3^6\)
\(\Leftrightarrow18\cdot3^x+12\cdot3^x=10\cdot3^6\)
\(\Leftrightarrow30\cdot3^x=30\cdot3^5\)
Suy ra: x=5
d) Ta có: \(6\cdot8^{x-1}+8^{x+1}=6\cdot8^{19}+8^{21}\)
\(\Leftrightarrow6\cdot\dfrac{8^x}{8}+8^x\cdot8=6\cdot8^{19}+64\cdot8^{19}\)
\(\Leftrightarrow8^x\cdot\dfrac{35}{4}=70\cdot8^{19}\)
\(\Leftrightarrow8^x=8^{20}\)
Suy ra: x=20
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\(x^2=1\Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
\(x^2=3\Rightarrow\left[{}\begin{matrix}x=-\sqrt{3}\\x=\sqrt{3}\end{matrix}\right.\)
\(x^2=5\Rightarrow\left[{}\begin{matrix}x=-\sqrt{5}\\x=\sqrt{5}\end{matrix}\right.\Rightarrow x=-\sqrt{5}\left(vì.x< 0\right)\)
\(x^2=7\Rightarrow\left[{}\begin{matrix}x=-\sqrt{7}\\x=\sqrt{7}\end{matrix}\right.\Rightarrow x=-\sqrt{7}\left(vì.x< 0\right)\)
\(x^2=9\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
\(\left(x-2\right)^2=2\Rightarrow\left[{}\begin{matrix}x-2=-\sqrt{2}\\x-2=\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2-\sqrt{2}\\x=2+\sqrt{2}\end{matrix}\right.\)
\(\left(x-4\right)^2=4\Rightarrow\left[{}\begin{matrix}x-2=-2\\x-2=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(\left(x-6\right)^2=6\Rightarrow\left[{}\begin{matrix}x-6=-\sqrt{6}\\x-6=\sqrt{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6-\sqrt{6}\\x=6+\sqrt{6}\end{matrix}\right.\)
\(\left(x-8\right)^2=8\Rightarrow\left[{}\begin{matrix}x-8=-2\sqrt{2}\\x-8=2\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8-2\sqrt{2}\\x=2+2\sqrt{2}\end{matrix}\right.\)
\(\left(x-10\right)^2=10\Rightarrow\left[{}\begin{matrix}x-10=-\sqrt{10}\\x-10=\sqrt{10}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10-\sqrt{10}\\x=10+\sqrt{10}\end{matrix}\right.\)
\(\left(x-\sqrt{3}\right)^2=3\Rightarrow\left[{}\begin{matrix}x-\sqrt{3}=-\sqrt{3}\\x-\sqrt{3}=\sqrt{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\sqrt{3}\end{matrix}\right.\)
\(\left(x-\sqrt{5}\right)^2=5\Rightarrow\left[{}\begin{matrix}x-\sqrt{5}=-\sqrt{5}\\x-\sqrt{5}=\sqrt{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\sqrt{5}\end{matrix}\right.\)
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Đề câu a sai rồi.
`a,(x-3)^{x+8}=4(x-3)^{x+6}`
`=>(x-3)^{x+6}[(x-3)^2-4]=0`
`=>` $\left[ \begin{array}{l}x-3=0\\(x-3)^2=4\end{array} \right.$
`=>` $\left[ \begin{array}{l}x=3\\x-3=2\\x-3=-2\end{array} \right.$
`=>` $\left[ \begin{array}{l}x=3\\x=5\\x=1\end{array} \right.$
Vậy x=1 hoặc x=3 hoặc x=5.
`b,(x-3)^{x+10}=9(x-3)^{x+8}`
`=>(x-3)^{x+8}[(x-3)^2-9]=0`
`=>` $\left[ \begin{array}{l}x-3=0\\(x-3)^2=9\end{array} \right.$
`=>` $\left[ \begin{array}{l}x=3\\x-3=3\\x=-3=-3\end{array} \right.$
`=>` $\left[ \begin{array}{l}x=3\\x=6\\x=0\end{array} \right.$
Vậy x=0 hoặc x=3 hoặc x=6
![](https://rs.olm.vn/images/avt/0.png?1311)
a,
\(5^{x+4}-3.5^{x+3}=2.5^{11}\)
\(\Rightarrow5^{x+3}\left(5-3\right)=2.5^{11}\)
\(\Rightarrow5^{x+3}2=2.5^{11}\)
\(\Rightarrow5^{x+3}=5^{11}\)
\(\Rightarrow x+3=11\)
\(\Rightarrow x=8\)
b, (Check lai xem de sai o dau khong nhe)
\(3.5^{x+2}+4.5^{x+3}=19.5^{10}\)
Dat 5x ra ben ngoai
\(\Rightarrow5^x.5^23+5^x:5^{-3}.4\)
\(\Rightarrow5^x\left(5^2.3+5^{-3}.4\right)\)
\(\Rightarrow5^x\left(5^{-3}.5^5.3+5^{-3}.4\right)\)
\(\Rightarrow5^x[5^{-3}\left(5^53+4\right)\)
\(\Rightarrow5^x[5^{-3}\left(3125.3+4\right)\)
\(\Rightarrow5^x\left(5^{-3}\right).9379\)
=> Khong tim duoc gia tri cua x \(\Rightarrow x\in\varnothing\)
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a, 2^x=8^4/16^3
<=> 2^x = (2^3)^4 / (2^4)^3
<=> 2^x = 2^12 / 2^12
<=> 2^x = 1
<=> 2^x = 2^0
<=> x = 0
Vậy x = 0
b,2^x=2^6/4^3
<=> 2^x = 2^6 / (2^2)^3
<=> 2^x = 2^6 / 2^6
<=> 2^x = 1
<=> 2^x = 2^0
<=> x = 0
Vậy x = 0
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a)
\(\left(x+5\right)^3=-64=\left(-4\right)^3\)
=> x + 5 = -4
=> x = -9
b)
\(\left(2x-3\right)^2=9=\left(\pm3\right)^2\)
+) 2x - 3 = 3
2x = 6
x = 3
+) 2x - 3 = -3
2x = 0
x = 0
c)
\(M=\frac{8^{10}+4^{10}}{8^4+4^{11}}\)
\(M=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}\)
\(M=\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(1+2^{10}\right)}\)
\(M=\frac{2^{12}\cdot2^8}{2^{12}}\)
\(M=2^8=256\)
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ai k mình k lại [ chỉ 3 người đầu tiên mà trên 10 điểm hỏi đáp ]
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Ta có : \(\frac{3}{2}.4^x+\frac{5}{4}.4^{x+2}=\frac{3}{2}.4^8+\frac{5}{3}.4^{10}\)
\(\Rightarrow4^x\left(\frac{3}{2}+\frac{5}{3}.4^2\right)\)=\(4^8\left(\frac{3}{2}+\frac{5}{3}.4^2\right)\)
\(\Rightarrow4^x=4^8\)
\(\Rightarrow x=8\)
\(Can\) \(you\) \(k\)\(for\)\(me?\)