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YN
12 tháng 12 2021
Answer:
\(3x^2-4x=0\)
\(\Rightarrow x\left(3x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)
\(\left(x^2-5x\right)+x-5=0\)
\(\Rightarrow x\left(x-5\right)+\left(x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)
\(x^2-5x+6=0\)
\(\Rightarrow x^2-2x-3x+6=0\)
\(\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)
\(\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
\(5x\left(x-3\right)-x+3=0\)
\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Rightarrow\left(5x-1\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)
\(x^2-2x+5=0\)
\(\Rightarrow\left(x^2-2x+1\right)+4=0\)
\(\Rightarrow\left(x-1\right)^2=-4\) (Vô lý)
Vậy không có giá trị \(x\) thoả mãn
\(x^2+x-6=0\)
\(\Rightarrow x^2+3x-2x-6=0\)
\(\Rightarrow x.\left(x+3\right)-2\left(x+3\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)
NH
1
M
4 tháng 7 2019
a, \(4x\left(x+1\right)-5\left(x+1\right)=0\)
\(\left(x+1\right)\left(4x-5\right)\)=0
\(\left\{{}\begin{matrix}x+1=0\\4x-5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\left(-1\right)\\4x=5\Rightarrow x=\frac{5}{4}\end{matrix}\right.\)
b, \(5x\left(x-20\right)+5x-100=0\)
\(5x\left(x-20\right)+\left(5x-100\right)=0\)
\(5x\left(x-20\right)+5\left(x-20\right)=0\)
\(\left(x-20\right)\left(5x+5\right)\)= 0
\(\left\{{}\begin{matrix}x-20=0\\5x+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=20\\5x=-5\Rightarrow x=-1\end{matrix}\right.\)
c, \(2\left(x-2\right)+\left(x-2\right)^2=0\)
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tập xác định của chương trình
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Rút gọn thừa số chung
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Giải phương trình
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Giải phương trình
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Biệt thức
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Biệt thức
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Nghiệm
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Lời giải thu được
Vậy x= 0 và x = 2
d, \(\left(x-3\right)^2-5x-x^2=12\)
\(\left(x^2-2.x.3+3^2\right)-5x-x^2=12\)
\(x^2-6x+9-5x-x^2=12\)
\(-11x+9=12\)
\(-11x=3\)
=> \(x=-\frac{3}{11}\)
HT
2
10 tháng 10 2018
a) (5x+1)2 - (5x-3).(5x+3) = 0
25x2 + 10x + 1 - 25x2 + 9 = 0
10x + 10 = 0
10.(x+1) = 0
=> x + 1 = 0 => x = - 1
b) (x+3).(x2 - 3x + 9) - x.(x-2).(x+2) = 0
x3 + 27 - x.(x2 - 4) = 0
x3 + 27 - x3 + 4x = 0
27 + 4x = 0
4x = - 27
x = -27/4
10 tháng 10 2018
c) 3x.(x-2) - x + 2= 0
3x.(x-2) - (x-2) = 0
(x-2).(3x-1) = 0
=> x - 2 =0 => x = 2
3x-1 = 0 => 3x = 1 => x = 1/3
d) x.(2x-3) - 2.(3-2x) = 0
x.(2x-3) + 2.(2x-3) = 0
(2x-3).(x+2) = 0
=> 2x - 3 = 0 => 2x = 3 => x = 3/2
x+ 2 = 0 => x = -2
KL:...\
NT
1
AT
28 tháng 6 2018
a/ \(\left(5x-1\right)^2-5x\left(5x-1\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(5x-1-5x\right)=0\Leftrightarrow1-5x=0\Leftrightarrow x=\dfrac{1}{5}\)
Vaayj........
b/ \(x\left(x+1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\Leftrightarrow x=-1\\x+2=0\Leftrightarrow x=-2\end{matrix}\right.\)
Vay......
c/ \(\left(3x+2\right)x-3\left(3x+2\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=3\end{matrix}\right.\)
Vậy.....
NT
2
28 tháng 6 2018
\(a)\) \(\left(5x-1\right)^2-5x\left(5x-1\right)=0\)
\(\Leftrightarrow\)\(\left(5x-1\right)\left(5x-1-5x\right)=0\)
\(\Leftrightarrow\)\(\left(5x-1\right).\left(-1\right)=0\)
\(\Leftrightarrow\)\(5x-1=0\)
\(\Leftrightarrow\)\(5x=1\)
\(\Leftrightarrow\)\(x=\frac{1}{5}\)
Vậy \(x=\frac{1}{5}\)
\(b)\) \(x\left(x+1\right)\left(x+2\right)=0\)
Suy ra \(x=0\) hoặc \(x+1=0\) hoặc \(x+2=0\)
\(\Leftrightarrow\)\(x=0\) hoặc \(x=-1\) hoặc \(x=-2\)
Vậy \(x=0\) hoặc \(x=-1\) hoặc \(x=-2\)
\(c)\) \(\left(3x+2\right)x-3\left(3x+2\right)=0\)
\(\Leftrightarrow\)\(\left(3x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+2=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=0-2\\x=0+3\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=-2\\x=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-2}{3}\\x=3\end{cases}}}\)
Vậy \(x=\frac{-2}{3}\) hoặc \(x=3\)
Chúc bạn học tốt ~
HH
28 tháng 6 2018
a/ \(\left(5x-1\right)^2-5x\left(5x-1\right)=0\)
<=> \(\left(5x-1\right)\left(5x-1-5x\right)=0\)
<=> \(-1\left(5x-1\right)=0\)
<=> \(5x-1=0\)
<=> \(5x=1\)
<=> \(x=\frac{1}{5}\)
b/ \(x\left(x+1\right)\left(x+2\right)=0\)
<=> \(x=0\) hoặc \(\orbr{\begin{cases}x+1=0\\x+2=0\end{cases}}\)
<=> \(x=0\)hoặc \(\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)
c/ \(\left(3x+2\right)x-3\left(3x+2\right)=0\)
<=> \(\left(3x+2\right)\left(x-3\right)=0\)
<=> \(\orbr{\begin{cases}3x+2=0\\x-3=0\end{cases}}\)
<=> \(\orbr{\begin{cases}3x=-2\\x=3\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-\frac{2}{3}\\x=3\end{cases}}\)
5 tháng 12 2017
1)⇔x2+1x-3x+3=0
⇔x(x+1)-3(x+1)=0
⇔(x+1)(x-3)=0
⇔x+1=0 hoặc x-3=0
⇔x=-1 hoặc x=3
5 tháng 12 2017
4)⇔x(1+5x)=0
⇔x=0 hoặc 1+5x=0
⇔x=0 hoặc 5x=-1
⇔x=0 hoặc x=-0.2
LN
15 tháng 4 2020
1) (x+6)(3x-1)+x+6=0
⇔(x+6)(3x-1)+(x+6)=0
⇔(x+6)(3x-1+1)=0
⇔3x(x+6)=0
2) (x+4)(5x+9)-x-4=0
⇔(x+4)(5x+9)-(x+4)=0
⇔(x+4)(5x+9-1)=0
⇔(x+4)(5x+8)=0
3)(1-x)(5x+3)÷(3x-7)(x-1)
=\(\frac{\left(1-x\right)\left(5x+3\right)}{\left(3x-7\right)\left(x-1\right)}=\frac{\left(1-x\right)\left(5x+3\right)}{\left(7-3x\right)\left(1-x\right)}=\frac{\left(5x+3\right)}{\left(7-3x\right)}\)
TK
6
26 tháng 8 2020
a) 5x( x - 1 ) = x - 1
<=> 5x2 - 5x = x - 1
<=> 5x2 - 5x - x + 1 = 0
<=> 5x2 - 6x + 1 = 0
<=> 5x2 - 5x - x + 1 = 0
<=> 5x( x - 1 ) - 1( x - 1 ) = 0
<=> ( x - 1 )( 5x - 1 ) = 0
<=> \(\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
b) 2( x + 5 ) - x2 - 5x = 0
<=> 2x + 10 - x2 - 5x = 0
<=> -x2 - 3x + 10 = 0
<=> -x2 - 5x + 2x + 10 = 0
<=> -x( x + 5 ) + 2( x + 5 ) = 0
<=> ( x + 5 )( 2 - x ) = 0
<=> \(\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)
c) x2 - 2x - 3 = 0
<=> x2 + x - 3x - 3 = 0
<=> x( x + 1 ) - 3( x + 1 ) = 0
<=> ( x + 1 )( x - 3 ) = 0
<=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
d) 2x2 + 5x - 3 = 0
<=> 2x2 - x + 6x - 3 = 0
,<=> x( 2x - 1 ) + 3( 2x - 1 ) = 0
<=> ( 2x - 1 )( x + 3 ) = 0
<=> \(\orbr{\begin{cases}2x-1=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}\)
PN
26 tháng 8 2020
a) 5x ( x - 1 ) = x - 1 <=> 5x2 - 5x - x + 1 = 0
<=> 5x2 - 6x + 1 = 0 <=> 5x2 - x - ( 5x - 1 ) = 0
<=> x ( 5x - 1 ) - ( 5x - 1 ) = 0 <=> ( x - 1 )( 5x - 1 ) = 0
<=> x = 1 hoặc x = 1/5
b) 2 ( x + 5 ) - x2 - 5x = 0 <=> 2 ( x + 5 ) - x ( x + 5 ) = 0
<=> ( 2 - x ) ( x + 5 ) = 0 <=> x = 2 hoặc x = -5
c) x2 - 2x - 3 = 0 <=> x2 + x - 3x - 3 = 0
<=> x ( x + 1 ) - 3 ( x + 1 ) = 0 <=> ( x - 3 ) ( x + 1 ) = 0
<=> x = 3 hoặc x = -1
d) 2x2 + 5x - 3 = 0
Ta có : delta = 52 - 4.2.3 = 25 - 24 = 1
Khi đó : x = -1 hoặc x = 3/2
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 : \(\Delta'=\frac{2116}{4}-477=529-477=52\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-23+\sqrt{52}}{9}\\x=\frac{-23-\sqrt{52}}{9}\end{cases}}\)
\(x\left(x+2\right)+x^2-5x=0\)
\(\Leftrightarrow x^2+2x+x^2-5x=0\)
\(\Leftrightarrow2x^2-3x=0\Leftrightarrow x\left(2x-3\right)=0\)
\(\Leftrightarrow x=0;\frac{3}{2}\)
X(x+2)+ x2 -5x= 0
x2 +2x +x2 - 5x= 0
2x2- 3x= 0
x( 2x-3 )= 0
x= 0 hoặc 2x-3= 0
x=0 hoặc x= 1,5