Tìm x thuộc Z biết:
a/ x^2 - x = 0
b/ x^2 - 5x + 4 = 0
tìm x biết:
a.(x+3)^2-(x+3)(x-3)=0
b.5x(x^2+4)=0
\(a,\Leftrightarrow\left(x+3\right)\left(x+3-x+3\right)=0\Leftrightarrow x=-3\\ b,\Leftrightarrow x=0\left(x^2+4>0\right)\)
\(a,x^2+2.x.3+3^2-\left(x^2-3^2\right)=0\)
\(x^2+6x+9-x^2+9=0\)
\(6x+18=0\)
\(6x=-18\)
\(x=-3\)
Vậy x=-3
\(b,5x^3+20x=0\)
\(5x\left(x^2+4\right)=0\)
\(Th1:5x=0=>x=0\)
\(Th2:x^2+4=0\)
\(x^2=-4\)(vô lý)
Vậy x=0
Tìm x biết:
a)x^2-100x=0
b)x^2+5x+6=0
Lời giải:
a. $x^2-100x=0$
$\Leftrightarrow x(x-100)=0$
$\Rightarrow x=0$ hoặc $x-100=0$
$\Leftrightarrow x=0$ hoặc $x=100$
b.
$x^2+5x+6=0$
$\Leftrightarrow (x^2+2x)+(3x+6)=0$
$\Leftrightarrow x(x+2)+3(x+2)=0$
$\Leftrightarrow (x+2)(x+3)=0$
$\Leftrightarrow x+2=0$ hoặc $x+3=0$
$\Leftrightarrow x=-2$ hoặc $x=-3$
Tìm x biết:
a)x^2-100x=0
b)x^2+5x+6=0
Tìm x,biết:
a)5x.(x+1)-5.(x+1).(x-2)=0
b)(4x+1).(x-2)-(2x-3)2=4
a)5(x+1)(x-x-2)=0
=>5(x+1).-2=0
=>5(x+1)=0
=>x+1=0
=>x=-1
a)5x.(x+1)-5.(x+1).(x-2)=0
⇒5x(x+1)-(5x-10)(x+1)=0
⇒(x+1)(5x-5x+10)=0
⇒10(x+1)=0
⇒x+1=0⇒x=-1
a) \(5x\left(x+1\right)-5\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow5\left(x+1\right)\left(x-x+2\right)=0\)
\(\Leftrightarrow10\left(x+1\right)=0\)
\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
b) \(\left(4x+1\right)\left(x-2\right)-\left(2x-3\right)^2=4\)
\(\Leftrightarrow4x^2-7x-2-4x^2+12x-9=4\)
\(\Leftrightarrow5x=15\Leftrightarrow x=3\)
Tìm x, biết:
a) 5x(x – 200) – x + 200 = 0
b) x3 – 11x = 0
a) \(\Rightarrow5x\left(x-200\right)-\left(x-200\right)=0\)
\(\Rightarrow\left(x-200\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=200\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\Rightarrow x\left(x^2-11\right)=0\)
\(\Rightarrow x\left(x-\sqrt{11}\right)\left(x+\sqrt{11}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{11}\\x=-\sqrt{11}\end{matrix}\right.\)
a) 5x(x-200)-(x-200)=0
(x-200)(5x-1)=0
Th1 : x-200=0
X=200
Th2 : 5x-1=0
5x=1
X=1/5
Vậy S={200;1/5}
\(a,5x.\left(x-200\right)-x+200=0\)
\(\Rightarrow5x.\left(x-200\right)-\left(x-200\right)=0\)
\(\Rightarrow\left(5x-x\right).\left(x-200\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\x-200=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=1\\x=200\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=200\end{matrix}\right.\)
\(b.x^3-11x=0\)
\(\Rightarrow x.\left(x^2-11\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-11=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=11\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x=\sqrt{11}\\x=-\sqrt{11}\end{matrix}\right.\end{matrix}\right.\)
tìm x biết:
a) 4x2-(x-3)2=0
b)x2-4+(x+2)2=0
a ,\(4x^2-\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(2x-x+3\right)\left(2x+x-3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(3x-3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+3=0\\3x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\3x=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)
Vậy
b,\(x^2-4+\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x+2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy ...
Bài 7. Tìm x,biết:
a) x-3x2=0 e) 5x(3x-1)+x(3x-1)-2(3x-1)=0
b) (x+3)2-x(x-2)=13 c) (x-4)2-36=0
d) x2-7x+12=0 g) x2-2018x-2019=0
Bài 8. Tìm x, biết
a) (2x-1)2=(x+5)2 b) x2-x+1/4
c) 4x4-101x2+25=0 d) x3-3x2+9x-91=0
tìm x biết:
a)2(x+3)+x(3+x)=0
b)(2x-3)^2-(4x-6)(x+2)+x^2+4x+4=0
\(\Rightarrow\left(x+3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x+3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
\(2\left(x+3\right)+x\left(3+x\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
<=> (x+3)(x+2)=0
TH1 x+3=0 <=> x=-3
TH2 x+2=0 <=> x=-2
Vậy....
Tìm x biết:
a) 2x2 - 4 = 0
b) (x + 1)2 = 4
c) (2x - 1)2 - 9 = 0
d) x2 - x = 0
a: =>2x^2=4
=>x^2=2
=>\(x=\pm\sqrt{2}\)
b: =>(x+1)^2-4=0
=>(x+1+2)(x+1-2)=0
=>(x+3)(x-1)=0
=>x=1 hoặc x=-3
c: =>(2x-1)^2-3^2=0
=>(2x-1-3)(2x-1+3)=0
=>(2x-4)(2x+2)=0
=>x=2 hoặc x=-1
d: x^2-x=0
=>x(x-1)=0
=>x=0 hoặc x=1
Tìm x∈Z, biết:
a)x.(x-6)=0
b)(-7-x).(-x+5)=0
c)(x+3).(x-7)=0
d)(x-3).(x2+12)=0
e)(x+1).(2-x) ≥0
f)(x-3).(x-5) ≤0
a) \(x\left(x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b) \(\left(-7-x\right)\left(-x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)
c) \(\left(x+3\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)
d) \(\left(x-3\right)\left(x^2+12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)
\(\Rightarrow x=3\)
e) \(\left(x+1\right)\left(2-x\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)
\(\Rightarrow-1\le x\le2\)
f) \(\left(x-3\right)\left(x-5\right)\le0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow3\le x\le5\)
a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)
d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3
c) => \(\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)