Tìm X: \(\dfrac{30-x}{30}=\dfrac{8}{15};\dfrac{x+30}{72}=\dfrac{5}{8}\)
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T
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
2 tháng 4 2023
\(\dfrac{30-x}{30}\) = \(\dfrac{5}{15}\)
\(\dfrac{30-x}{30}\) = \(\dfrac{1}{3}\)
30 - \(x\) = \(\dfrac{1}{3}\) \(\times\) 30
30 - \(x\) = 10
\(x\) =30 - 10
\(x\) = 20
\(\dfrac{x+30}{72}\) = \(\dfrac{5}{8}\)
\(x+30\) = \(\dfrac{5}{8}\) \(\times\) 72
\(x+30\) = 45
\(x\) = 45 - 30
\(x\) = 15
NT
1
16 tháng 5 2022
\(\Leftrightarrow3\left(5x-1\right)+5\left(2x+3\right)>2\left(x-8\right)-x+1\)
=>15x-3+10x+15>2x-16-x+1
=>25x+12>x-15
=>24x>-27
hay x>-9/8
3 tháng 1 2022
\(\Leftrightarrow2x-\dfrac{1}{3}=\left(\dfrac{12}{30}-\dfrac{4}{15}\right):\dfrac{3}{5}=\dfrac{2}{9}\)
=>2x=5/9
hay x=5/18
13 tháng 7 2023
=>\(1\cdot\dfrac{2}{4}\cdot\dfrac{3}{6}\cdot...\cdot\dfrac{31}{62}\cdot\dfrac{1}{64}=2^x\)
=>\(2^x=\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot...\cdot\dfrac{1}{2}\cdot\dfrac{1}{64}=\left(\dfrac{1}{2}\right)^{30}\cdot\left(\dfrac{1}{2}\right)^6=\dfrac{1}{2^{36}}\)
=>x=-36
HT
1
31 tháng 1 2018
Khỏi ghi đề nha :
\(\Leftrightarrow3\left(5x-1\right)-5\left(2x+3\right)=2\left(x-8\right)-x\)
\(\Leftrightarrow15x-3-10x-15-2x+16+x=0\)
\(\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}.\)
4 tháng 1 2018
\(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot...\cdot\dfrac{14}{30}.\dfrac{15}{32}=\dfrac{1}{2^x}\)
\(\Rightarrow\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot14\cdot15}{4\cdot6\cdot8\cdot10\cdot...\cdot30\cdot32}=\dfrac{1}{2^x}\)
\(\Rightarrow\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot14\cdot15}{2\cdot4\cdot6\cdot8\cdot10\cdot...\cdot30\cdot32}=\dfrac{1}{2^{x+1}}\)
\(\Rightarrow\dfrac{1}{2^{15}\cdot32}=\dfrac{1}{2^{x+1}}\)
\(\Rightarrow2^{15}.2^5=2^{x+1}\)
\(\Rightarrow2^{20}=2^{x+1}\)
\(\Rightarrow x+1=20\Rightarrow x=19\)
Vậy x = 19.
ND
25 tháng 2 2018
a.
\(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)
\(\Leftrightarrow x\left(x+1\right).\left(x-1\right)\left(x+2\right)-24=0\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)
Đặt \(a=x^2+x-1\) , ta có pt:
\(\left(a+1\right)\left(a-1\right)-24=0\)
\(\Leftrightarrow a^2-1-24=0\)
\(\Leftrightarrow a^2-25=0\)
\(\Leftrightarrow\left(a-5\right)\left(a+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=5\\a=-5\end{matrix}\right.\)
*Với a = 5 ta được:
\(x^2+x-1=5\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow x^2+3x-2x-6=0\)
\(\Leftrightarrow\left(x^2+3x\right)-\left(2x+6\right)=0\)
\(\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
*Với a = -5 ta được:
\(x^2+x-1=-5\)
\(\Leftrightarrow x^2+x+4=0\)
\(\Leftrightarrow x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{15}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\) ( loại)
Vậy pt có tập nghiệm là: \(s=\left\{-3;2\right\}\)
25 tháng 2 2018
c)(ĐKXĐ: x khác 30;29)
\(\Leftrightarrow\dfrac{x-29}{30}-1+\dfrac{x-30}{29}-1=\dfrac{29}{x-30}-1+\dfrac{30}{x-29}-1\)
\(\Leftrightarrow\dfrac{x-59}{30}+\dfrac{x-59}{29}=\dfrac{x-59}{30-x}+\dfrac{x-59}{29-x}\)
\(\Leftrightarrow x=59\)(tm) or \(\dfrac{1}{30}+\dfrac{1}{29}-\dfrac{1}{30-x}-\dfrac{1}{29-x}=0\)
\(\Leftrightarrow\dfrac{-x}{30\left(30-x\right)}+\dfrac{-x}{29\left(29-x\right)}=0\)
\(\Leftrightarrow x=0\)(tm) or \(\dfrac{1}{30\left(30-x\right)}+\dfrac{1}{29\left(29-x\right)}=0\)
\(\Leftrightarrow1741-59x=0\)
\(\Leftrightarrow x=\dfrac{1741}{59}\left(tm\right)\)
Vậy S={0;\(\dfrac{1741}{59}\);59}
\(\dfrac{30-x}{30}=\dfrac{8}{15}\)
\(\Rightarrow\left(30-x\right)15=30\times8\)
\(\Rightarrow\left(30-x\right)15=240\)
\(\Rightarrow30-x=240:15\)
\(\Rightarrow30-x=16\)
\(\Rightarrow x=30-16\)
\(\Rightarrow x=14\)
__
\(\dfrac{x+30}{72}=\dfrac{5}{8}\)
\(\Rightarrow\left(x+30\right)8=5\times72\)
\(\Rightarrow\left(x+30\right)8=360\)
\(\Rightarrow x+30=360:8\)
\(\Rightarrow x+30=45\)
\(\Rightarrow x=45-30\)
\(\Rightarrow x=15\)