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29 tháng 9 2018
\(2x^3-50x=0\)
<=> \(2x\left(x^2-25\right)=0\)
<=> \(2x\left(x-5\right)\left(x+5\right)=0\)
đến đây
bạn tự giải nhé
hk tốt
LH
3
KN
30 tháng 7 2019
\(\left(y-2\right)\left(y-3\right)+\left(y-2\right)-1=0\)
\(\Leftrightarrow\left(y-2\right)\left(y-3\right)+\left(y-3\right)=0\)
\(\Leftrightarrow\left(y-3\right)^2=0\)
\(\Leftrightarrow y=3\)
\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow\left(x+3\right)x\left(x-2\right)=0\)
\(\Leftrightarrow x\in\left\{0;-3;2\right\}\)
PN
21 tháng 8 2020
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
PN
21 tháng 8 2020
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 : \(\Delta'=\frac{2116}{4}-477=529-477=52\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-23+\sqrt{52}}{9}\\x=\frac{-23-\sqrt{52}}{9}\end{cases}}\)
N
2 tháng 11 2018
\(x^2-3x+2.\left(x-3\right)=0\)
\(x.\left(x-3\right)+2.\left(x-3\right)=0\)
\(\left(x-3\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
\(x.\left(x-3\right)-3x+9=0\)
\(x.\left(x-3\right)-3.\left(x-3\right)=0\)
\(\left(x-3\right)^2=0=>x=3\)
2 tháng 11 2018
a,\(x^2-3x+2\left(x-3\right)=0.\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
Y
11 tháng 1 2023
\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
2
11 tháng 1 2023
`8)1-(x-5)=4(2-2x)`
`<=>1-x+5=8-6x`
`<=>5x=2<=>x=2/5`
`9)(3x-2)(x+5)=0`
`<=>[(x=2/3),(x=-5):}`
`10)(x+3)(x^2+2)=0`
Mà `x^2+2 > 0 AA x`
`=>x+3=0`
`<=>x=-3`
`11)(5x-1)(x^2-9)=0`
`<=>(5x-1)(x-3)(x+3)=0`
`<=>[(x=1/5),(x=3),(x=-3):}`
`12)x(x-3)+3(x-3)=0`
`<=>(x-3)(x+3)=0`
`<=>[(x=3),(x=-3):}`
`13)x(x-5)-4x+20=0`
`<=>x(x-5)-4(x-5)=0`
`<=>(x-5)(x-4)=0`
`<=>[(x=5),(x=4):}`
`14)x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>(x+5)(x-1)=0`
`<=>[(x=-5),(x=1):}`
SP
3
17 tháng 9 2018
\(a)\)\(x^3-x^2-x+1=0\)
\(\Leftrightarrow\)\(x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\)\(\left(x-1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\)\(\left(x-1\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\)\(\left(x-1\right)^2\left(x+1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\left(x-1\right)^2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}}\)
Vậy \(x=1\) hoặc \(x=-1\)
Chúc bạn học tốt ~
VL
0
\(\left(3x-1\right)\left(x-3\right)-9+x^2=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)+\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x-1+x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(4x+2\right)=0\Leftrightarrow x=-\frac{1}{2};x=3\)
Vậy tập nghiệm của phương trình là S = [ -1/2 ; 3 }
ta có: (3xX-1)x(X-3)-9 + X2=0
<=> (3X-1)(X-3)+(X2-9)=0
<=> (3X-1)(X-3)+(X-3)(X+3)=0
<=> (3X-1+X+3)(X-3)=0
<=>(4X+2)(X-3)=0
<=>\(\hept{\begin{cases}X=\frac{-1}{2}\\X=3\end{cases}}\)
Vậy....