\(32\left(\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+...+\frac{1}{197.200}\right)-x=\frac{1}{2}\)

\(\frac{32}{3}\left(\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+....+\frac{3}{197.200}\right)-x=\frac{1}{2}\)

\(\frac{32}{3}\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\right)-x=\frac{1}{2}\)

\(\frac{32}{3}\left(\frac{1}{8}-\frac{1}{200}\right)-x=\frac{1}{2}\)

x=0.78

khó quá làm sao mà trả lời đc

Vắt óc đi

tự đầu mình vắt óc mà suy nghĩ

\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}=\frac{32}{99}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{32}{99}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{32}{99}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{3}-\frac{32}{99}\)

\(\Rightarrow\frac{1}{x+2}=\frac{33}{99}-\frac{32}{99}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{99}\)

\(\Rightarrow x+2=99\)

\(\Rightarrow x=99-2\)

\(\Rightarrow x=97\)

Vậy \(x=97\)

\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{x\cdot\left(x+2\right)}=\frac{32}{99}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{x}-\frac{1}{x+2}=\frac{32}{99}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{32}{99}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{3}-\frac{32}{99}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{99}\)

\(\Rightarrow x+2=99\)

\(\Rightarrow x=99-2\)

\(\Rightarrow x=97\)

Vậy x=97

chọn câu nào thì giải thích giúp mik luôn nha

\(\frac{4}{x-4}-\frac{x}{x+4}+\frac{32}{16-x^2}.\)

\(=\frac{4\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}-\frac{x\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}-\frac{32}{\left(x-4\right)\left(x+4\right)}\)

\(=\frac{4x+16-x^2+4x-32}{\left(x-4\right)\left(x+4\right)}\)

\(=\frac{-x^2+8x-16}{\left(x-4\right)\left(x+4\right)}=\frac{-\left(x-4\right)^2}{\left(x-4\right)\left(x+4\right)}=\frac{-\left(x-4\right)}{x+4}\)

x - x/2 = 32/15 : -16/5

=> x(1 - 1/2) = -2/3

=> x.1/2 = -2/3

=> x = -4/3

vậy_ 

\(x-\frac{x}{2}=\frac{32}{15}:-\frac{16}{5}\)

\(x\cdot\left(1-\frac{1}{2}\right)=-\frac{2}{3}\)

\(x\cdot\frac{1}{2}=-\frac{2}{3}\)

\(x=-\frac{2}{3}:\frac{1}{2}\)

\(x=-\frac{4}{3}\)

\(x-\frac{x}{2}=\frac{32}{15}:\frac{-16}{5}\)

\(\Leftrightarrow\)\(x-\frac{x}{2}=\frac{-2}{3}\)

\(\Leftrightarrow\)\(x=\frac{-2}{3}+\frac{x}{2}\)

\(\Leftrightarrow\)\(x=\frac{-4}{6}+\frac{3x}{6}\)

\(\Leftrightarrow\)\(x=\frac{-4+3x}{6}\)

\(\Leftrightarrow\)\(6x=-4+3x\)

\(\Leftrightarrow\)\(2x=-4\)

\(\Leftrightarrow\)\(x=-2\)

Vậy.....

\(\frac{3x+2}{x+4}+\frac{2x+1}{x-2}=5-\frac{x-32}{x^2+2x-8}\)

\(\Leftrightarrow\) \(\frac{\left(3x+2\right)\left(x-2\right)}{\left(x+4\right)\left(x-2\right)}+\frac{\left(2x+1\right)\left(x+4\right)}{\left(x+4\right)\left(x-2\right)}=\frac{5\left(x+4\right)\left(x-2\right)}{\left(x+4\right)\left(x-2\right)}-\frac{x-32}{\left(x+4\right)\left(x-2\right)}\)

\(\Rightarrow\) (3x + 2)(x - 2) + (2x + 1)(x + 4) = 5(x + 4)(x - 2) - x + 32

\(\Leftrightarrow\) 3x² - 6x + 2x - 4 + 2x² + 8x + x + 4 = 5x² - 10x + 20x - 40 - x + 32

\(\Leftrightarrow\) 5x² + 5x = 5x² + 9x - 8

\(\Leftrightarrow\) 5x² + 5x - 5x² - 9x + 8 = 0

\(\Leftrightarrow\) -4x + 8 = 0

\(\Leftrightarrow\) x - 2 = 0

\(\Leftrightarrow\) x = 2

Vậy S = {2}

\(\frac{x+2m}{x+3}+\frac{x-m}{x-3}=\frac{mx\left(x+1\right)}{x^2-9}\) (đkxđ: x \(\ne\) \(\pm\) 3)

\(\Leftrightarrow\) \(\frac{\left(x+2m\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x-m\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{mx\left(x+1\right)}{\left(x+3\right)\left(x-3\right)}\)

\(\Rightarrow\) (x + 2m)(x - 3) + (x - m)(x + 3) = mx(x + 1)

\(\Leftrightarrow\) x² - 3x + 2mx - 6m + x² + 3x - mx - 3m - mx² - mx = 0

\(\Leftrightarrow\) (2 - m)x² - 9m = 0

Thay m = 1 ta được:

(2 - 1)x² - 9 . 1 = 0

\(\Leftrightarrow\) x² - 9 = 0

\(\Leftrightarrow\) (x - 3)(x + 3) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(KTM\right)\\x=-3\left(KTM\right)\end{matrix}\right.\)

Vậy S = \(\varnothing\)

Thay m = 2 ta được:

(2 - 2)x² - 9 . 2 = 0

\(\Leftrightarrow\) -18 = 0

\(\Rightarrow\) Pt vô nghiệm

Vậy S = \(\varnothing\)

Chúc bn học tốt!!

\(\frac{x+2}{2}=\frac{32}{x+2}\)

\(\Rightarrow\left(x+2\right)^2=32.2=64=8^2\)

\(\Rightarrow\orbr{\begin{cases}x+2=8\\x+2=-8\end{cases}\Rightarrow}\orbr{\begin{cases}x=6\\x=-10\end{cases}}\)

P/s: Hoq chắc :<

\(\Rightarrow x=6\)

​\(\frac{x+2}{2}\)= \(\frac{32}{x+2}\)

=> (x+2).(x+2)=2.32

      \(\left(x+2^{ }\right)^2\)=64

=> \(\left(x+2\right)^2\)=\(8^2\)

=> x+2=8

     x = 8-2

Vậy x=6

Sory còn 1 trường hợp nữa

=> x+2=-8

     x =-8-2

Vậy x = -10

=^_^=