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a: (x+2)(x+5)<0
=>x+5>0 và x+2<0
=>-5<x<2
mà x là số nguyên
nên \(x\in\left\{-4;-3;-2;-1;0;1\right\}\)
b: \(\left(x^2-8\right)\left(x^2+10\right)< 0\)
\(\Leftrightarrow x^2-8< 0\)
\(\Leftrightarrow x^2< 8\)
mà x là số nguyên
nên \(x\in\left\{0;1;-1;2;-2\right\}\)
c: (x-2)(x+1)<0
=>x+1>0 và x-2<0
=>-1<x<2
mà x là số nguyên
nên \(x\in\left\{0;1\right\}\)
a) ( x + 3 )3 : 3 - 1 = -10
( x + 3 )3 : 3 = -10 + 1
( x + 3 )3 = -9 * 3
x + 3 = \(\sqrt[3]{-27}\)
x = -3 - 3
x = -6
b) 3 | x - 1 | + 5 = 17
3 | x - 1 | = 17 - 5
| x - 1 | = 12 : 3
| x - 1 | = 4
( 1 ) x - 1 > 0 => x - 1 = 4 => x = 5
( 2 ) x - 1 < 0 => x - 1 = -4 => x = -3
Vậy S = { -3 ; 5 }
a) x + 15 = 36 - 2x
x + 15 = 36 - (x + x )
15 =36 - ( x + x) - x
15 = 36 - x - x - x
15 = 36 - 3x
3x = 36 - 15
3x = 21
x = 21 : 3
=> x = 7
b) (x - 7) - (2x +5) = -14
x - 7 -( 2x + 5) = -14
x - (2x + 5) = -14 + 7 = -7
x - 2x - 5 = -7
x - 2x = -7 + 5 = -2
x - x + x = 2
x = 2 (-x + x cũng bằng chính nó)
=> x = 2
c) (x - 12) - 15 = (-7 + 20) - (18+x)
(x - 12) - 15 = 13 - (18 + x)
(x - 12) - 15 = 13 - 18 - x
(x - 12) - 15 = -5 - x
15 = (x - 12 ) - (-5 - x)
15 = x - 12 + 5 + x
15 = x + (-12) + 5 + x
15 = 2x + [(-12) + 5]
15 = 2x + -7
2x = -7 + 15
2x = 8
x = 8 : 2
=> x = 4
..................
a ) \(5\left(x^2\right)+7x+2\)
\(\Leftrightarrow5x^2+7x+2=0\)
\(\Leftrightarrow5x^2+5x+2x+2=0\)
\(\Leftrightarrow\left(5x+2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=-1\end{matrix}\right.\)
Vậy .............
b ) \(\dfrac{x+1}{17}+\dfrac{x+2}{16}=\dfrac{x+3}{15}+\dfrac{x+4}{14}\)
\(\Leftrightarrow\dfrac{x+1}{17}+1+\dfrac{x+2}{16}+1=\dfrac{x+3}{15}+1+\dfrac{x+4}{14}+1\)
\(\Leftrightarrow\dfrac{x+18}{17}+\dfrac{x+18}{16}=\dfrac{x+18}{15}+\dfrac{x+18}{14}\)
\(\Leftrightarrow\dfrac{x+18}{17}+\dfrac{x+18}{16}-\dfrac{x+18}{15}-\dfrac{x+18}{14}=0\)
\(\Leftrightarrow\left(x+18\right)\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)=0\)
Vì \(\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)\ne0\)
Ta có : \(x+18=0\Leftrightarrow x=-18\)
Vậy ......
c ) \(\dfrac{x-1}{x-3}=\dfrac{x-4}{x-7}\)
\(\Leftrightarrow\left(x-1\right)\left(x-7\right)=\left(x-3\right)\left(x-4\right)\)
\(\Leftrightarrow x^2-7x-x+7=x^2-4x-3x+12\)
\(\Leftrightarrow-x=5\)
\(\Leftrightarrow x=-5\)
Vậy ..
\(a)\dfrac{1}{3}x+\dfrac{2}{5}\left(x+1\right)=0\)
\(\Leftrightarrow\dfrac{1}{3}x+\dfrac{2}{5}x+\dfrac{2}{5}=0\)
\(\Leftrightarrow x\left(\dfrac{5}{15}+\dfrac{6}{15}\right)=\dfrac{-2}{5}\)
\(\Leftrightarrow x.\dfrac{11}{15}=\dfrac{-2}{5}\)
\(\Leftrightarrow x=\dfrac{-2}{5}.\dfrac{15}{11}\)
\(\Leftrightarrow x=\dfrac{-6}{11}\)
\(a,\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-5< 0\\x+2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-5>0\\x+2< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 5\\x>-2\end{matrix}\right.\\\left\{{}\begin{matrix}x>5\\x< -2\end{matrix}\right.\end{matrix}\right.\Rightarrow-2< x< 5\\ \Rightarrow x\in\left\{-1;0;1;2;3;4\right\}\\ b,\Rightarrow5< x^2< 14\\ \Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)