\(lim\left(50.\dfrac{1-\left(\dfrac{4}{5}\right)^n}{1-\dfrac{4}{5}}+\dfrac{4}{5}.50.\dfrac{1-\left(\dfrac{4}{5}\right)^{n-1}}{1-\dfrac{4}{5}}\right)\) \(=50.\dfrac{1}{\dfrac{1}{5}}+\dfrac{4}{5}.50.\dfrac{1}{\dfrac{1}{5}}=450\)

\(x+\left(13-15\right)=5+\left(10-7\right)\)

\(< =>x-2=8=>x=10\)

Ta có: \(x+\left(13-15\right)=5+10-7\)

\(\Leftrightarrow x-2=15-7=8\)

hay x=10

x + (13 - 15) = 5 + (10 - 7)

x + (13 - 15) = 5 + 3

x + (13 - 15) = 8

x + (-2) = 8

x = 8 - (-2)

x = 10

Chúc bạn học tốt!

\(\dfrac{x}{10}=\dfrac{-3}{5}\\ \Rightarrow x=\dfrac{10.\left(-3\right)}{5}=-6\\ ------------\\ x+\dfrac{5}{12}=\dfrac{-2}{3}\\ x=\dfrac{-2}{3}-\dfrac{5}{12}\\ x=\dfrac{-8}{12}-\dfrac{5}{12}\\ x=\dfrac{-13}{12}\)

`x`=`-6`

`x` = `-13/12`

x/10=-3/5

x/10=-3.2/5.2

x/10=-6/10

x=-6

x+5/12=-2/3

x=-2/3-5/12

x=8/12-5/12

x=3/12

x=1/4

10900

x = 218 x 50

x = 10900

x=10900

A

A

A

a) xx là x^2 hả ??? (tính sau nha)

b)Ta có  \(\left|x-100\right|\ge0;\left|y+200\right|\ge0\)

\(\Rightarrow\left|x-100\right|+\left|y+200\right|\ge0\)

\(\Rightarrow B\ge-1\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left|x-100\right|=0\\\left|y+200\right|=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x-100=0\\y+200=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=100\\y=-200\end{cases}}\)

Vậy \(B_{min}=-1\Leftrightarrow\hept{\begin{cases}x=100\\y=-200\end{cases}}\)

c)pt o có GTLN 

Tham khảo(nếu a ko có xx)

https://olm.vn/hoi-dap/detail/97637814260.html

a, Ta có: x + 2 ≥ 0⇒A = x + 2 + 50 ≥ 50

Dấu "=" xảy ra khi x=-2

Vậy GTNN của A=50 khi x=-2

b, Ta có: x − 100 ≥ 0; y + 200 ≥ 0

⇒ x − 100 + y + 200 ≥ 0⇒B= x − 100 + y + 200 − 1 ≥ −1

Dấu "=" xảy ra khi x=100,y=-200

Vậy GTNN của B=-1 khi x=100,y=-200

c, Đặt C = 2015-|x+5|

Ta có: x + 5 ≥ 0⇒ − x + 5 ≤ 0

⇒C = 2015 − x + 5 ≤ 2015

Dấu "=" xảy ra khi x=-5

Vậy GTLN của C = 2015 khi x = -5 

\(x^3-2x^2+x-2=0\\ \Leftrightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x^2+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ Vậy:x=2\\ ---\\ 2x\left(3x-5\right)=10-6x\\ \Leftrightarrow6x^2-10x-10+6x=0\\ \Leftrightarrow6x^2-4x-10=0\\ \Leftrightarrow6x^2+6x-10x-10=0\\ \Leftrightarrow6x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(6x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}6x-10=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)

\(4-x=2\left(x-4\right)^2\\ \Leftrightarrow4-x=2\left(x^2-8x+16\right)\\ \Leftrightarrow2x^2-16x+32+x-4=0\\ \Leftrightarrow2x^2-15x+28=0\\ \Leftrightarrow2x^2-8x-7x+28=0\\ \Leftrightarrow2x\left(x-4\right)-7\left(x-4\right)=0\\ \Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\\ ---\\ 4-6x+x\left(3x-2\right)=0\\ \Leftrightarrow4-6x+3x^2-2x=0\\ \Leftrightarrow3x^2-8x+4=0\\ \Leftrightarrow3x^2-6x-2x+4=0\\ \Leftrightarrow3x\left(x-2\right)-2\left(x-2\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)