\(\overline{X=}\dfrac{1\cdot1+2\cdot9+3\cdot8+4\cdot3+5\cdot4}{25}\)

\(\overline{X=3}\)

=>3x+20=50

hay x=10

=>3x+20=50

hay x=10

x+x+x+20      =25x2

x+x+x+20      =50

x+x+x            = 50-20

x+x+x            =30

Vậy x=10 vì 10+10+10=30

\(\Leftrightarrow3x+10=25\\ \Leftrightarrow3x=25-10=15\\ \Leftrightarrow x=\dfrac{15}{3}=5\)

=>3x=15

hay x=5

\(\Leftrightarrow 3x + 10 = 25\)

\(\Leftrightarrow 3x = 25 - 10 = 15\)

\(\Leftrightarrow x = \dfrac{15}{3^0} = 5\)

số 0 ko liên quan nhé!

121-\(12\left(x-1\right)^3=25\)

\(12\left(x-1\right)^3=121-25=96\)

\(\left(x-1\right)^3=8\)

\(x-1=2\)

\(x=3\)

\(121-12(x-1)^3=25\\\Rightarrow 12(x-1)^3=121-25\\\Rightarrow12(x-1)^3=96\\\Rightarrow(x-1)^3=96:12\\\Rightarrow(x-1)^3=8\\\Rightarrow(x-1)^3=2^3\\\Rightarrow x-1=2\\\Rightarrow x=2+1\\\Rightarrow x=3\\Vậy:x=3\)

help

Vì 4x=5y$\Rightarrow \frac{x}{5}=\frac{y}{4}$

      Áp dụng tính chất dãy tỉ số bằng nhau ta có:

$\Rightarrow \frac{x}{5}=\frac{y}{4}=\frac{x+y}{5+4}=\frac{90}{9}=10$

$\Rightarrow \{\begin{array}{c}\frac{x}{5}=10\\ \frac{y}{4}=10\end{array}$$\Rightarrow \{\begin{array}{c}x=50\\ y=40\end{array}$

Vậy x=50;y=40

Ta có: 4x=5y

nên \(\dfrac{x}{5}=\dfrac{y}{4}\)

hay \(\dfrac{x}{35}=\dfrac{y}{28}\left(1\right)\)

Ta có: 3y=7z

nên \(\dfrac{y}{7}=\dfrac{z}{3}\)

hay \(\dfrac{y}{28}=\dfrac{z}{12}\left(2\right)\)

Từ \(\left(1\right),\left(2\right)\) suy ra \(\dfrac{x}{35}=\dfrac{y}{28}=\dfrac{z}{12}\)

hay \(\dfrac{x}{35}=\dfrac{3y}{84}=\dfrac{2z}{24}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được: 

\(\dfrac{x}{35}=\dfrac{3y}{84}=\dfrac{2z}{24}=\dfrac{x-3y+2z}{35-84+24}=\dfrac{-25}{-25}=1\)

Do đó: x=35; y=28; z=12

a, để A<1\(\Rightarrow\dfrac{x-2}{x+3}< 1\Rightarrow x-2< x+3\Rightarrow-5< 0\left(luôn.đúng\right)\)

Vậy x∈R