Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài làm
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{x.\left(x+2\right)}=\frac{2015}{2016}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2015}{2016}\)
\(1-\frac{1}{x+2}=\frac{2015}{2016}\)
\(\frac{1}{x+2}=\frac{1}{2016}\)
\(\Rightarrow x+2=2016\)
\(x=2014\)
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{93}\)
\(2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\right)=2.\frac{15}{93}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{\left(2x+1\right).\left(2x+3\right)}=\frac{10}{31}\)
\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{93}\)
\(\Rightarrow2x+3=93\)
\(\Rightarrow2x=90\)
\(\Rightarrow x=45\)
a) (7x - 11)3 = 25 x 52 + 200
(7x - 11)3 = 800 + 200
(7x - 11)3 = 1000
(7x - 11)3 = 103
=> 7x - 11 = 10
=> 7x = 10 + 11
=> 7x = 21
=> x = 3
b) \(3\frac{1}{3}x+16\frac{3}{4}=-13,25\)
\(3\frac{1}{3}x=-13,25-16\frac{3}{4}\)
\(\frac{10}{3}x=-30\)
\(x=-9\)
= 1/2 - 1/3 + 1/3 - 1/4 + ...+ 1/x - 1/x + 1
= 1/2 - 1 /x + 1 =199/200
=1/x+1 = 1/2 - 199/200
1/ x + 1 = ....
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x.\left(x+1\right)}=\frac{199}{200}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{199}{200}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{199}{200}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{199}{200}\)
\(\Rightarrow\frac{1}{x+1}=-\frac{99}{200}\)
\(\Rightarrow\frac{-99}{-\left(x+1\right).99}=\frac{-99}{200}\)
\(\Rightarrow-99x+-99=200\)
PT \(\Rightarrow2x^2+2x-3x-6=2x^2-x+4x-8-2\)
\(\Rightarrow-4x=-4\) \(\Leftrightarrow x=1\)
Vậy \(x=1\)
Ta có: \(2x\left(x+1\right)-3\left(x+2\right)=x\left(2x-1\right)+4\left(x-2\right)-2\)
\(\Leftrightarrow2x^2+2x-3x-6=2x^2-x+4x-8-2\)
\(\Leftrightarrow2x^2-x-6=2x^2+3x-10\)
\(\Leftrightarrow2x^2-x-6-2x^2-3x+10=0\)
\(\Leftrightarrow-4x+4=0\)
\(\Leftrightarrow-4x=-4\)
hay x=1
Vậy: x=1
Ta có : \(\left(2.x-1\right)^2=3^2.5^2\)
\(\Leftrightarrow\left(2.x-1\right)^2=\left(3.5\right)^2\)
\(\Leftrightarrow\left(2.x-1\right)^2=15^2\)
\(\Leftrightarrow2.x-1=15\)
\(\Leftrightarrow2.x=15+1\)
\(\Leftrightarrow2.x=16\)
\(\Leftrightarrow x=16:2\)
\(\Leftrightarrow x=8\)
Vậy \(x=8\)
\(\left(2x-1\right)^2=3^2.5^2\)
\(\left(2x-1\right)^2=225\)
\(\left(2x-1\right)^2=\left(\pm15\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2x-1=15\\2x-1=-15\end{cases}\Rightarrow\orbr{\begin{cases}2x=16\\2x=-14\end{cases}\Rightarrow}\orbr{\begin{cases}x=8\\x=-7\end{cases}}}\)
\(\Rightarrow x\in\left\{-7;8\right\}\)