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a) x - 1/2 = 3/5
x = 3/5 + 1/2
x = 11/10
b) x - 1/2 = -2/3
x = -2/3 + 1/2
x = -1/6
c) 2/5 - x = 0,25
x = 2/5 - 0,25
x = 2/5 - 1/4
x = 3/20
1/2x + 1/5 = 2/3x - 1/4
=> 1/2x - 2/3x = -1/4 - 1/5
=> -1/6x = -9/20
=> x = -9/20 : (-1/6)
=> x = 27/10
\(\frac{1}{2}x+\frac{1}{5}=\frac{2}{3}x-\frac{1}{4}\)
\(\frac{1}{2}x=\frac{2}{3}x-\frac{1}{4}-\frac{1}{5}\)
\(\frac{1}{2}x=-\frac{9}{20}+\frac{2x}{3}\)
\(\frac{1}{2}x=-\frac{9}{20}+\frac{2x}{3}-\frac{2x}{3}\)
\(-\frac{x}{6}=-\frac{9}{20}\)
\(6\left(-\frac{x}{6}\right)=6\left(-\frac{9}{20}\right)\)
\(-x=-\frac{27}{10}\)
\(\Rightarrow x=-\frac{27}{10}\)
\(\dfrac{20142014}{20152015}\times x+7986=1+3+5+...+199\)
Vì các số ở vế 2 đều cách nhau 2 đơn vị
=> Số số hạng của vế 2 là \(\left(199-1\right)\div2+1=100\) ( số hạng )
=> Tổng của vế 2 là \(\left(199+1\right)\times100\div2=10000\)
Thay vào biểu thức, ta có:
\(\dfrac{20142014}{20152015}\times x+7986=10000\)
\(\dfrac{2014}{2015}\times x=10000-7986=2014\)
\(x=2014\div\dfrac{2014}{2015}\)
\(x=2015\)
\(\dfrac{20142014}{20152015}\)\(x\)+ 7986 = 1 + 3 + 5 + ...+ 197 + 199
\(\dfrac{2014}{2015}\)\(x\) + 7986 = (199 + 3){ (199 -1): 2 + 1}: 2
\(\dfrac{2014}{2015}\)\(x\) + 7986 = 202. 100: 2
\(\dfrac{2014}{2015}x\) = 10000
\(\dfrac{2014}{2015}\)\(x\) = 10000 - 7986
\(\dfrac{2014}{2015}\)\(x\) = 2014
\(x\) = 2014 : \(\dfrac{2014}{2015}\)
\(x\) = 2015
Bài 7:
a, \(x\) = \(\dfrac{1}{5}\) + \(\dfrac{2}{11}\)
\(x\) = \(\dfrac{11}{55}\) + \(\dfrac{10}{55}\)
\(x=\dfrac{21}{55}\)
b, \(\dfrac{x}{15}\) = \(\dfrac{3}{5}\) - \(\dfrac{2}{3}\)
\(\dfrac{x}{15}\) = \(\dfrac{9}{15}\) - \(\dfrac{10}{15}\)
\(\dfrac{x}{15}\) = \(\dfrac{1}{15}\)
\(x\) = 1
c, \(\dfrac{11}{8}\) + \(\dfrac{13}{6}\)= \(\dfrac{85}{x}\)
\(\dfrac{33}{24}\) + \(\dfrac{52}{24}\) = \(\dfrac{85}{x}\)
\(\dfrac{85}{24}\) = \(\dfrac{85}{x}\)
24 = \(x\)
5.(x + 4) - 3.(x - 2) = x
5x + 20 - (3x - 6) =x
5x + 20 - 3x + 6 = x
2x + 26 = x
2x - x = (-26)
x = (-26)
Vậy x = (-26)
\(3\left(x+2\right)^2-5^2=2.5^2\)
\(\Rightarrow3\left(x+2\right)^2=2.5^2+5^2\)
\(\Rightarrow3\left(x+2\right)^2=5^2\left(2+1\right)\)
\(\Rightarrow3\left(x+2\right)^2=5^2.3\)
\(\Rightarrow\left(x+2\right)^2=5^2\)
\(\Rightarrow\left[{}\begin{matrix}x+2=5\\x+2=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\) \(\Rightarrow x=3\left(x\inℕ\right)\)
3(x + 2)² - 5² = 2.5²
3(x + 2)² - 25 = 50
3(x + 2)² = 50 + 25
3(x + 2)² = 75
(x + 2)² = 75 : 3
(x + 2)² = 25
x + 2 = 5 hoặc x + 2 = -5
*) x + 2 = 5
x = 5 - 2
x = 3 (nhận)
*) x + 2 = -5
x = -5 - 2
x = -7 (loại)
Vậy x = 3
a: \(\Leftrightarrow x\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;9;-9;12;-12;18;-18;36;-36\right\}\)
mà -3<x<30
nên \(x\in\left\{-2;-1;1;2;3;4;6;9;12;18\right\}\)
b: \(\Leftrightarrow x\in\left\{0;4;-4;8;-8;12;-12;...\right\}\)
mà -16<=x<20
nên \(x\in\left\{-16;-12;-8;-4;0;4;8;12;16\right\}\)
c: \(\Leftrightarrow x-1+4⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{2;0;3;-1;5;-3\right\}\)
d: \(\Leftrightarrow2x+4-5⋮x+2\)
\(\Leftrightarrow x+2\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{-1;-3;3;-7\right\}\)
\(\dfrac{1}{2}\) \(\times\) ( \(x\) - \(\dfrac{2}{3}\)) - \(\dfrac{1}{3}\) \(\times\) ( 2\(x\) - 3) = \(x\)
\(\dfrac{1}{2}\) \(\times\) \(\dfrac{3x-2}{3}\) - \(\dfrac{2x-3}{3}\) = \(x\)
\(\dfrac{3x-2}{6}\) - \(\dfrac{4x-6}{6}\) = \(\dfrac{6x}{6}\)
3\(x-2-4x\) + 6 = 6\(x\)
-\(x\) + 4 - 6\(x\) = 0
7\(x\) = 4
\(x\) = \(\dfrac{4}{7}\)