Tìm x thuộc Z, biết:
a) (x\(^2\) - 5) (x\(^2\)-25) <0
b) ( x+5) (9+x\(^2\) )<0
c) ( x+3) ( x\(^2\) +1) =0
d) (x+ 5) (x\(^2\) - 4) =0
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a)
\(x+\left(x+2\right)+\left(x+4\right)+...+\left(x+98\right)=0\)
\(x+x+2+x+4+...+x+98=0\)
\(50x+\left(98+2\right).\left[\left(98-2\right):2+1\right]:2=0\)
\(50x+100.49:2=0\)
\(50x+49.50=0\)
\(50x=0-49.50\)
\(50x=-2450\)
\(x=-2450:50\)
\(x=-49\)
b)
\(\left(x-5\right)+\left(x-4\right)+\left(x-3\right)+...+\left(x+11\right)+\left(x+12\right)=99\)
\(x+x+x+...+x-5-4-3-...+11+12=99\)
\(18x+6+7\text{+ 8 + 9 + 10 + 11 + 12 = 99}\)
\(18x+63=99\)
\(18x=99-63\)
\(18x=36\)
\(x=36:18\)
\(x=2\)
-29-9(2x-1)\(^2\)= -110
(=) 9(2x-1)2 = (-29) +110
(=) 9(2x-1)2 = 81
(=) (2x-1)2 =81: 9
(=) (2x-1)2 =9
(=) (2x-1)2 = 32 =(-3)2
\(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)
\(\orbr{\begin{cases}2x=4\\2x=-2\end{cases}}\)
\(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
vậy : ........
a,\(-29-9\left(2x-1\right)^2=-110\)
\(=>-29+110=9.\left(2x-1\right)^2\)
\(=>81=9.\left(2x-1\right)^2\)
\(=>\left(2x-1\right)^2=9\)
\(=>\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}=>\orbr{\begin{cases}x=\frac{4}{2}=2\\x=\frac{-2}{2}=-1\end{cases}}}\)
\(a,\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-5< 0\\x+2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-5>0\\x+2< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 5\\x>-2\end{matrix}\right.\\\left\{{}\begin{matrix}x>5\\x< -2\end{matrix}\right.\end{matrix}\right.\Rightarrow-2< x< 5\\ \Rightarrow x\in\left\{-1;0;1;2;3;4\right\}\\ b,\Rightarrow5< x^2< 14\\ \Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
a: =>\(\dfrac{x}{-5}=\dfrac{y}{-7}=\dfrac{z}{2}=\dfrac{x-y+z}{-5+7+2}=\dfrac{-28}{4}=-7\)
=>x=35; y=49; z=-14
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
`x/-5=y/-7=z/2=(x-y+z)/((-5)-(-7)+2)=-28/4=-7`
`-> x/-5=y/-7=z/2=-7`
`-> x=-7*-5=35, y=-7*-7=49, z=-7*2=-14`
\(\dfrac{x}{2}=\dfrac{y}{3}\text{⇒}\dfrac{x}{10}=\dfrac{y}{15}\)
\(\dfrac{y}{5}=\dfrac{z}{4}\text{⇒}\dfrac{y}{15}=\dfrac{z}{12}\)
⇒\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-21}{-3}=7\)
⇒x=70;y=105;z=84
a: =>3x+3=5x-25
=>-2x=-28
hay x=14
b: =>3x+6=-4x+20
=>7x=14
hay x=2
\(a,\Leftrightarrow2^x\left(1+2^4\right)=544\\ \Leftrightarrow2^x=\dfrac{544}{17}=32=2^5\\ \Leftrightarrow x=5\\ b,\Leftrightarrow\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\3x-\dfrac{2}{5}=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-\dfrac{1}{5}\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)