(x-3)8 = (x-3)10
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T
1
H
26 tháng 10 2023
\(a,\left(x+2\right)^{10}+\left(x+2\right)^8=0\\ \Leftrightarrow\left(x+2\right)^8\left[\left(x+2\right)^2+1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)^8=0\\\left(x+2\right)^2+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+2=0\\\left(x+2\right)^2=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\end{matrix}\right.\\ b,\left(x+3\right)^{10}-\left(x+3\right)^8=0\\ \Leftrightarrow\left(x+3\right)^8\left[\left(x+3\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+3\right)^8=0\\\left(x+3\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\\left(x+3\right)^2=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x+3=1\\x+3=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\\x=-4\end{matrix}\right.\)
2
1
27 tháng 2 2021
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
K
15 tháng 5 2022
Đặt A = \(\dfrac{3}{6\cdot8}+\dfrac{3}{8\cdot10}+...+\dfrac{3}{158\cdot160}\)
A = \(\dfrac{3}{2}\cdot\left(\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}+...+\dfrac{2}{158\cdot160}\right)\)
A = \(\dfrac{3}{2}\cdot\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+...+\dfrac{1}{158}-\dfrac{1}{160}\right)\)
A = \(\dfrac{3}{2}\cdot\left(\dfrac{1}{6}-\dfrac{1}{160}\right)\)
A = \(\dfrac{3}{2}\cdot\dfrac{77}{480}\)
A = \(\dfrac{77}{320}\)
Vậy: A = \(\dfrac{77}{320}\)
5 tháng 8 2021
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
MN
28 tháng 5 2021
\(a.\)
\(\left(x-8\right)\left(x^3+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
\(S=\left\{8,-2\right\}\)
\(b.\)
\(\left(4x-3\right)-\left(x+5\right)=3\cdot\left(10-x\right)\)
\(\Leftrightarrow4x-3-x-5-30+3x=0\)
\(\Leftrightarrow6x-38=0\)
\(\Leftrightarrow x=\dfrac{38}{6}\)
\(S=\left\{\dfrac{38}{6}\right\}\)
<=> ( x - 3) = 1 hoặc ( x- 3) = 0
\(\orbr{\begin{cases}x-3=1\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=3\end{cases}}}\)
Vậy.............
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