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giúp mik mik đang cần gấp
nhưng phả có lời giải đừng cho mỗi đáp án
a:Ta có: \(\left(x-9\right)^7=\left(x-9\right)^4\)
\(\Leftrightarrow\left(x-9\right)^4\cdot\left[\left(x-9\right)^3-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-9=0\\x-9=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=10\end{matrix}\right.\)
b: ta có: \(\left(3x-15\right)^{15}=\left(3x-15\right)^{10}\)
\(\Leftrightarrow\left(3x-15\right)^{10}\cdot\left[\left(3x-15\right)^5-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-15=0\\3x-15=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{16}{3}\end{matrix}\right.\)
a, `(x-9)^4=(x-9)^7`
`(x-9)^4-(x-9)^7=0`
`(x-9)^4 . [(1-(x-9)^3]=0`
TH1: `(x-9)^4=0`
`x-9=0`
`x=9`
TH2: `1-(x-9)^3=0`
`(x-9)^3=1^3`
`x-9=1`
`x=10`
b, `(3x-15)^10=(3x-15)^15`
`(3x-15)^10 . [1-(3x-15)^5]=0`
TH1: `(3x-15)^10=0`
`3x-15=0`
`x=5`
TH2: `1-(3x-15)^5=0`
`(3x-15)^5=1^5`
`3x-15=1`
`x=16/3` (Loại)
c, `(x-8)^3=(x-8)^6`
`(x-8)^3 .[1-(x-8)^3]=0`
TH1: `(x-8)^3=0`
`x=8`
TH2: `1-(x-8)^3=0`
`x-8=1`
`x=9`
a)<=>\(\dfrac{\left(2x-3\right).2}{6}-\dfrac{3.3}{6}=\dfrac{5-2x}{6}-\dfrac{1.3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}=\dfrac{5-2x}{6}-\dfrac{3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}-\dfrac{5-2x}{6}+\dfrac{3}{6}=0\)
<=>\(\dfrac{4x-6-9-5+2x+3}{6}=\dfrac{4x-17}{6}=0\)
<=>\(4x-17=0\)
<=>\(4x=17\)<=>\(x=\dfrac{17}{4}\)
a) \(\left(x-9\right)^4=\left(x-9\right)^7\)
\(\Rightarrow\left[{}\begin{matrix}x-9=1\\x-9=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=10\\x=9\end{matrix}\right.\)
b) \(\left(3x-15\right)^{10}=\left(3x-15\right)^{15}\)
\(\Rightarrow\left[{}\begin{matrix}3x-15=0\\3x-15=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{3}\\x=\dfrac{16}{3}\end{matrix}\right.\)
c) \(\left(x-8\right)^3=\left(x-8\right)^6\)
\(\Rightarrow\left[{}\begin{matrix}x-8=0\\x-8=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=8\\x=9\end{matrix}\right.\)
\(\text{Ta có: }\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+.....+\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.....+\frac{1}{\left(x+2\right)}-\frac{1}{\left(x+5\right)}=\frac{3}{20}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{\left(x+5\right)}=\frac{3}{20}\)
\(\Rightarrow\frac{1}{\left(x+5\right)}=\frac{1}{2}-\frac{3}{20}\)