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a) Ta có: \(x^2-25=0\)
\(\Leftrightarrow x^2=25\)
hay \(x\in\left\{5;-5\right\}\)
Vậy: \(x\in\left\{5;-5\right\}\)
b) Ta có: \(\left(x-4\right)^2-36=0\)
\(\Leftrightarrow\left(x-4\right)^2=36\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=6\\x-4=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{10;-2\right\}\)
c) Ta có: \(\left(6-x\right)^2=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
Vậy: x=6
d) Ta có: \(\left(x+\frac{1}{4}\right)^2-\frac{1}{9}=0\)
\(\Leftrightarrow\left(x+\frac{1}{4}\right)^2=\frac{1}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{1}{4}=\frac{1}{3}\\x+\frac{1}{4}=\frac{-1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{12}\\x=\frac{-7}{12}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{12};\frac{-7}{12}\right\}\)
a) \(\left(x^2-4\right)\left(x^2-9\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\x^2-9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x^2=2^2\\x^2=3^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm2\\x=\pm3\end{matrix}\right.\)
b) \(\left(x^2-4\right)\left(x^2-9\right)\le0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-4\ge0;x^2-9\le0\\x^2-4\le0;x^2-9\ge0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2\ge4;x^2\le9\\x^2\le4;x^2\ge9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4\le x^2\le9\left(tm\right)\\9\le x^2\le4\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow x^2\in\left\{4;5;...;9\right\}\)
\(\Rightarrow x\in\left\{\pm2;\pm\sqrt{5};...;\pm3\right\}\).
a) ( x2 - 4 ) . ( x2 - 9 ) = 0
=> \(\left[{}\begin{matrix}x^2-4=0\\x^2-9=0\end{matrix}\right.=>\left[{}\begin{matrix}x^2=4\\x^2=9\end{matrix}\right.\)
= > \(\left[{}\begin{matrix}x=-2\\x=2\\x=3\\x=-3\end{matrix}\right.\)
e)
A = \(\frac{x+5}{x-2}\) = \(\frac{\left(x-2\right)+7}{x-2}=1+\frac{7}{x-2}\)
Muốn A nguyên thì:
=> \(\frac{7}{x-2}\) ∈ Z
=> 7 ⋮ x - 2
=> x - 2 ∈ Ư (7)
=> x - 2 ∈ { 1; 7; -1; -7 }
=> x ∈ { 3; 9; -5; 1 }
a) (5x - 1)(2x - 1/3) = 0
\(\Rightarrow5x-1=0\) hoặc \(2x-\frac{1}{3}=0\)
\(\Rightarrow\left[{}\begin{matrix}5x=0+1\\2x=0+\frac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1:5=\frac{1}{5}\\x=\frac{1}{3}:2=\frac{1}{3}.\frac{1}{2}=\frac{1}{6}\end{matrix}\right.\)
Vậy x = 1/5 hoặc x = 1/6
1.a) x=0
x+2 = 0=> x= -2
vậy x=0 hoặc -2
b) x-1 =0=>x=1
x-2= 0 => x=2
\(a,\left(x-2\right)\left(3x-9\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\3x-9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\3x=9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
\(b,\left(3-x\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3-x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
\(c,\left(x+1\right)\left(4-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\4-2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\2x=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
a)(x-2 ).(3x-9)=0
x-2 =0
x=0+2
x=2
Hoặc :3x-9 =0
3x =0+9
3x =9
x =9\(\div\)3
x =3
Vậy : x bằng 2 hoặc bằng 3
b)(3-x).(x+5)=0
3-x =0
x =3-0
x =3
Hoặc : x+5 =0
x =0-5
x =-5
Vậy x bằng 3 hoặc bằng -5
c)(x+1).(4-2x)=0
x+1 =0-1
=-1
Hoặc : 4-2x =0
2x =4-0
2x =4
x =4\(\div\)2
x =2
Vậy x bằng -1 hoặc 2
Bài 2:
a: (x-3)(x+2)>0
=>x-3>0 hoặc x+2<0
=>x>3 hoặc x<-2
b: (2x-4)(x+4)<0
=>x+4>0 và x-2<0
=>-4<x<2
a)\( 3x . ( x- 4 ) = 0\)
\(\Rightarrow\) \(\left[{}\begin{matrix}3x=0 \\x-4=0\end{matrix}\right.\)
\(\Rightarrow\) \(\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=4\)
b)\(( x- 2 ) . ( 2x - 4) =0\)
\(\Rightarrow\) \(\left[{}\begin{matrix}x-2=0\\2x-4=0\end{matrix}\right.\)
\(\Rightarrow\) \(x=2\)
Vậy \(x=2\)
:v
a: =>x+3>0
hay x>-3
b: =>4-x<0
hay x>4
c: =>x2-1=0 hoặc x+5=0
hay \(x\in\left\{1;-1;-5\right\}\)