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`a)25/(x+1)-1 1/6=-1/3-0,5`
`=>25/(x+1)=-1/3-1/2+1+1/6`
`=>25/(x+1)=1/3`
`=>75=x+1`
`=>x=74`
Vậy `x=74`
`b)(2x+25 3/5)^2-9/25=0`
`=>(2x+128/5)=9/25`
`**2x+128/5=3/5`
`=>2x=-125/5=-25`
`=>x=-25/2`
`**2x+128/5=-3/5`
`=>2x=-131/5`
`=>x=-131/10`
Giải:
a) \(\dfrac{25}{x+1}-1\dfrac{1}{6}=\dfrac{-1}{3}-0,5\)
\(\dfrac{25}{x+1}=\dfrac{-5}{6}+\dfrac{7}{6}\)
\(\dfrac{25}{x+1}=\dfrac{1}{3}\)
\(\Rightarrow1.\left(x+1\right)=25.3\)
\(\Rightarrow x+1=75\)
\(\Rightarrow x=75-1\)
\(\Rightarrow x=74\)
b) \(\left(2x+25\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)
\(\left(2x+\dfrac{128}{5}\right)^2=0+\dfrac{9}{25}\)
\(\left(2x+\dfrac{128}{5}\right)^2=\dfrac{9}{25}\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x+\dfrac{128}{5}\right)^2=\left(\dfrac{3}{5}\right)^2\\\left(2x+\dfrac{128}{5}\right)^2=\left(\dfrac{-3}{5}\right)^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{128}{5}=\dfrac{3}{5}\\2x+\dfrac{128}{5}=\dfrac{-3}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-25}{2}\\x=\dfrac{-131}{10}\end{matrix}\right.\)
Chúc bạn học tốt!
\(3^x+4\cdot3^{x-2}=333\)
\(\Rightarrow3^{x-2+2}+4\cdot3^{x-2}=333\)
\(\Rightarrow3^{x-2}\cdot\left(3^2+4\right)=333\)
\(\Rightarrow3^{x+2}\cdot\left(9+4\right)=333\)
\(\Rightarrow3^{x+2}\cdot13=333\)
\(\Rightarrow3^{x+2}=333:13\)
\(\Rightarrow3^{x+2}=\dfrac{333}{13}\)
Không có x nào thỏa mãn
⇒ x ∈ ∅
a, 5x - 19 = 106
5x = 106 + 19
5x = 125
Mà 125 = 53
=> x = 3
Vậy x = 3
S= 1+ 3+ 3^2+ 3^3+...+ 3^48+ 3^49
3S= 3.(1+3+ 3^2+ 3^3+...+ 3^49)
3S= 3+ 3^2+ 3^3+ 3^4+....+3^50
3S-S= (3+ 3^2+ 3^3+ 3^4+...+3^50)-(1+3+ 3^2+ 3^3+...+3^49)
2S= 3+3^2+ 3^3+ 3^4+...+3^50- 1-3- 3^2- 3^3-...-3^49
2S=(3-3)+ (3^2- 3^2)+ ...+(3^49-3^49)+ 3^50-1
2S= 3^50-1
S= (3^50-1):2
S=1+3+32+33+...+348+349
=> 3S=3+32+33+34+....+349+350
=> 2S=350-1
=> \(S=\frac{3^{50}-1}{2}\)