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b) 5x(x-2000)-x+2000=0
\(\Rightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\\ \Rightarrow\left(x-2000\right)\left(5x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-2000=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+2000\\5x=0+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\5x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\)
a) \(5x\left(x-2000\right)-x+2000=0\)
\(\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-2000\right)=0\)
\(\Leftrightarrow x\in\left\{\frac{1}{5};2000\right\}\)
a ) \(5x\left(x-2000\right)-x+2000=0\)
\(\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)
\(\Leftrightarrow\left(x-2000\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2000=0\\5x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy \(x=2000\) và \(x=\dfrac{1}{5}\)
b ) \(x^3-13x=0\)
\(\Leftrightarrow x\left(x^2-13\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=\sqrt{13}\end{matrix}\right.\)
Vậy \(x=0\) và \(x=\sqrt{13}\)
c ) \(x+5x^2=0\)
\(\Leftrightarrow x\left(1+5x\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\1+5x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=-\dfrac{1}{5}\end{matrix}\right.\)
Vậy \(x=0\) và \(x=-\dfrac{1}{5}\)
d ) \(\left(x+1\right)=\left(x+1\right)^2\)
\(\Leftrightarrow\left(x+1\right)-\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)\left[1-\left(x+1\right)\right]=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy \(x=0\) và \(x=-1\)
e ) \(x^3+x=0\)
\(\Leftrightarrow x\left(x^2+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x^2+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\\left(loại\right)\end{matrix}\right.\)
Vậy \(x=0\)
a, \(5x\left(x-2000\right)-x+2000=0\)
\(\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-2000\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\x-2000=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2000\end{matrix}\right.\)
b,\(x^3-13x=0\)
\(\Leftrightarrow x\left(x ^2-13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{13}\end{matrix}\right.\)
c,\(x+5x^2=0\)
\(\Leftrightarrow x\left(5x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\5x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{5}\end{matrix}\right.\)
d,\(x+1=\left(x+1\right)^2\)
\(\Leftrightarrow\left(x+1\right)-\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)\left(1-x-1\right)=0\)
\(\Leftrightarrow-x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
e,\(x^3+x=0\)
\(\Leftrightarrow x\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
CHÚC BẠN HỌC TỐT........
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
Trả lời:
Bài 2:
a, \(x^3-13x=0\)
\(\Leftrightarrow x\left(x^2-13\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-13=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=13\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{13}\end{cases}}\)
Vậy ...
b, \(5x\left(x-2000\right)-x+2000=0\)
\(\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-2000\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\x-2000=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=2000\end{cases}}\)
Vậy ...
c, \(2x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+3=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=2\end{cases}}\)
Vậy ...
d, \(\left(x+1\right)=\left(x+1\right)^2\)
\(\Leftrightarrow\left(x+1\right)-\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)\left(1-x-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\-x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}\)
Vậy ...
Trả lời:
Bài 1:
\(C=x-x^2=-\left(x^2-x\right)=-\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\right)=-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]\)
\(=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x\)
Dấu "=" xảy ra khi x - 1/2 = 0 <=> x = 1/2
Vậy GTLN của C = 1/4 khi x = 1/2
\(E=4x^2+8x+y^2-4y+32=\left(2x\right)^2+8x+y^2-4y+4+4+24\)
\(=\left[\left(2x\right)^2+8x+4\right]+\left(y^2-4y+4\right)+24=\left(2x+2\right)^2+\left(y-2\right)^2+24\ge24\forall x\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}2x+2=0\\y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\y=2\end{cases}}}\)
Vậy GTNN của E = 24 khi x = - 1; y = 2
a) \(5x\left(x-2000\right)-x+2000=0\)
\(\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-2000\right)=0\)
\(\Leftrightarrow x\in\left\{\frac{1}{5};2000\right\}\)
a) 5x(x - 2000) - x + 2000 = 0
=> 5x(x - 2000) - (x - 2000) = 0
=> (x - 2000).(5x - 1) = 0
=> x - 2000 = 0 hoặc 5x - 1 = 0
=> x = 2000 hoặc 5x = 1
=> x = 2000 hoặc x = 1/5
b) x3 - 13x = 0
=> x.(x2 - 13) = 0
=> x = 0 hoặc x2 - 13 = 0
=> x = 0 hoặc x2 = 13, vô lí
=> x = 0
a) 5x(x-2000)-(x-2000)=(5x-1)(x-2000)=0 nên x=1/5 hoặc x=2000
b)\(x^3-13x=x\left(x^2-13\right)=0\)\(\Rightarrow\)x=0 hoặc x^2=13 hay x=\(\sqrt{13}\)
1,\(5x^2=13x\Leftrightarrow5x^2-13x=0\Leftrightarrow x\left(5x-13\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{13}{5}\end{cases}}\)
2,\(\left(5x^2+3x-2\right)^2=\left(4x^2-3x-2\right)^2\Leftrightarrow\orbr{\begin{cases}5x^2+3x-2=4x^2-3x-2\\5x^2+3x-2=-4x+3x+2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+6x=0\\9x^2-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x\left(x+6\right)=0\\\left(3x\right)^2=2^2\end{cases}\Leftrightarrow}}\orbr{\begin{cases}x=0or-6\\x=-\frac{2}{3}or\frac{2}{3}\end{cases}}\)
3,\(x^3+27+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2+3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+3x+9+x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2+4x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x^2+4x=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x\left(x+4\right)=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=0or-4\end{cases}}\)
4,\(5x\left(x-2000\right)-x+2000=0\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)
\(\Leftrightarrow\left(x-2000\right)\left(5x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=2000\\x=\frac{1}{5}\end{cases}}\)
5,\(5x\left(x-2\right)-x+2=0\Leftrightarrow5x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(5x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x-2=0\\5x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=\frac{1}{5}\end{cases}}\)
6,\(4x\left(x+1\right)=8\left(x+1\right)\Leftrightarrow4x\left(x+1\right)-8\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(4x-8\right)=0\Leftrightarrow\orbr{\begin{cases}x+1=0\\4x-8=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
7,\(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(2x-4\right)=0\Leftrightarrow\orbr{\begin{cases}x-4=0\\2x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=2\end{cases}}\)
tí làm nửa kia
8,\(x^2-6x+8=0\Leftrightarrow x^2-6x+9-1=0\Leftrightarrow\left(x-3\right)^2-1^2=0\)
\(\Leftrightarrow\left(x-3-1\right)\left(x-3+1\right)=0\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=2\end{cases}}\)
9,\(9x^2+6x-8=0\Leftrightarrow9x^2+6x+1-9=0\Leftrightarrow\left(3x+1\right)^2-3^2=0\)
\(\Leftrightarrow\left(3x+1-3\right)\left(3x+1+3\right)=0\Leftrightarrow\left(3x-2\right)\left(3x+4\right)=0\Leftrightarrow\orbr{\begin{cases}3x-2=0\\3x+4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{4}{3}\end{cases}}\)
10,\(x^3+x^2+x+1=0\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2+1=0\end{cases}\Leftrightarrow}x=-1\)
11,\(x^3-x^2-x+1=0\Leftrightarrow\left(x-1\right)\left(x^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}x-1=0\\x^2-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
12,\(\left(5-2x\right)\left(2x+7\right)=4x^2-25\Leftrightarrow\left(5-2x\right)\left(2x+7\right)-4x^2+25=0\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)-\left(5-2x\right)\left(5+2x\right)=0\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7-5-2x\right)=0\Leftrightarrow\left(5-2x\right).2=0\Leftrightarrow5-2x=0\Leftrightarrow x=\frac{5}{2}\)
13,\(x\left(2x-1\right)+\frac{1}{3}.\frac{2}{3}x=0\Leftrightarrow x\left(2x-1\right)+\frac{2}{9}x=0\)
\(\Leftrightarrow x\left(2x-1+\frac{2}{9}\right)=0\Leftrightarrow x\left(2x-\frac{7}{9}\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\2x=\frac{7}{9}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{7}{18}\end{cases}}\)
14,\(4\left(2x+7\right)-9\left(x+3\right)^2=0\Leftrightarrow8x+28-9x^2-54x-81=0\)
\(\Leftrightarrow-9x^2+\left(8x-54x\right)+\left(28-81\right)=0\Leftrightarrow-9x^2-46x-53=0\)
\(\Leftrightarrow9x^2+46x+53=0\)Ta có : \(\Delta'=\frac{2116}{4}-477=529-477=52\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-23+\sqrt{52}}{9}\\x=\frac{-23-\sqrt{52}}{9}\end{cases}}\)
\(a,\Rightarrow\left(x-2000\right)\left(5x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\\ b,\Rightarrow x\left(x^2-13\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{13}\\x=-\sqrt{13}\end{matrix}\right.\\ c,\Rightarrow3x\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ d,\Rightarrow\left(x-5\right)\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\\ e,\Rightarrow\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)