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18 tháng 3 2018
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) \(3\frac{1}{3}x+16\frac{3}{4}=-13,25\)
\(3\frac{1}{3}x=-13,25-16\frac{3}{4}\)
\(\frac{10}{3}x=-30\)
\(x=-9\)
NT
0
30 tháng 4 2019
Bài làm
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{x.\left(x+2\right)}=\frac{2015}{2016}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2015}{2016}\)
\(1-\frac{1}{x+2}=\frac{2015}{2016}\)
\(\frac{1}{x+2}=\frac{1}{2016}\)
\(\Rightarrow x+2=2016\)
\(x=2014\)
24 tháng 6 2016
a, \(\frac{4x}{3}=\frac{14x}{3}+5\)
\(\frac{4x}{3}-\frac{14x}{3}=5\)
\(\frac{-10x}{3}=5\)
x=-1,5
b, 3x-5x=2-4
-2x=-2
x=1
c, \(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\right)\) .x=2
\(\left(\frac{1}{1}-\frac{1}{12}\right).x=2\)
\(\frac{11}{12}.x=2\)
x=\(\frac{24}{11}\)
d, \(x=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{101.103}\right)\)
\(x=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{103}\right)\)
x=\(\frac{1}{2}\left(1-\frac{1}{103}\right)\)
x=\(\frac{1}{2}.\frac{102}{103}\)
x=\(\frac{51}{103}\)
9 tháng 8 2016
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{93}\)
\(2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\right)=2.\frac{15}{93}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{\left(2x+1\right).\left(2x+3\right)}=\frac{10}{31}\)
\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{93}\)
\(\Rightarrow2x+3=93\)
\(\Rightarrow2x=90\)
\(\Rightarrow x=45\)
AK
8 tháng 5 2018
\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{3}-\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{33}{99}-\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{99}\)
\(\Rightarrow x+2=99\)
\(\Rightarrow x=99-2\)
\(\Rightarrow x=97\)
Vậy \(x=97\)
WH
8 tháng 5 2018
\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{x\cdot\left(x+2\right)}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{x}-\frac{1}{x+2}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{3}-\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{99}\)
\(\Rightarrow x+2=99\)
\(\Rightarrow x=99-2\)
\(\Rightarrow x=97\)
Vậy x=97
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 \(x=8\)
\(\left(2x-1\right)^2=3^2.5^2\)
\(\left(2x-1\right)^2=225\)
\(\left(2x-1\right)^2=\left(\pm15\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2x-1=15\\2x-1=-15\end{cases}\Rightarrow\orbr{\begin{cases}2x=16\\2x=-14\end{cases}\Rightarrow}\orbr{\begin{cases}x=8\\x=-7\end{cases}}}\)
\(\Rightarrow x\in\left\{-7;8\right\}\)