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19 tháng 7 2016
a) \(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2x+10-x^2-5x=0\)
\(\Leftrightarrow-x^2-3x+10=0\)
\(\Leftrightarrow x^2+3x-10=0\)
\(\Leftrightarrow x^2-2x+5x-10=0\)
\(\Leftrightarrow x\left(x-2\right)+5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}}\)
b) \(x^3-6x^2+12x-8=0\)
\(\Leftrightarrow\left(x^3-8\right)-\left(6x^2-12x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-6x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4-6x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
c)\(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x\right)^2-\left[3\left(x+1\right)\right]^2=0\)
\(\Leftrightarrow\left(4x-3x-1\right)\left(4x+3x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\7x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{7}\end{cases}}}\)
d) \(x^3+x=0\)
\(\Leftrightarrow x^2\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
e)\(x^2-2x-3=0\)
\(\Leftrightarrow x^2+x-3x-3=0\)
\(\Leftrightarrow x\left(x+1\right)-3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)
2
10 tháng 1 2023
`a)2x^2+3(x-1)(x+1)=5x(x+1)`
`<=>2x^2+3x^2-3=5x^2+5x`
`<=>5x=-3`
`<=>x=-3/5`
__________________________________________
`b)(x-3)^3+3-x=0` nhỉ?
`<=>(x-3)^3-(x-3)=0`
`<=>(x-3)(x^2-1)=0`
`<=>[(x=3),(x^2=1<=>x=+-1):}`
__________________________________________
`c)5x(x-2000)-x+2000=0`
`<=>5x(x-2000)-(x-2000)=0`
`<=>(x-2000)(5x-1)=0`
`<=>[(x=2000),(x=1/5):}`
__________________________________________
`d)3(2x-3)+2(2-x)=-3`
`<=>6x-9+4-2x=-3`
`<=>4x=2`
`<=>x=1/2`
__________________________________________
`e)x+6x^2=0`
`<=>x(1+6x)=0`
`<=>[(x=0),(x=-1/6):}`
TN
10 tháng 6 2017
a)\(x\left(x-3\right)-2x+6=0\)
\(\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
b)\(\left(3x-5\right)\left(5x-7\right)+\left(5x+1\right)\left(2-3x\right)=4\)
\(\Leftrightarrow15x^2-46x+35-15x^2+7x+2-4=0\)
\(\Leftrightarrow33-39x=0\Leftrightarrow33=39x\Leftrightarrow x=\frac{33}{39}\)
10 tháng 6 2017
a) \(x\left(x-3\right)-2x+6=0\)
\(x\left(x-3\right)-2\left(x-3\right)=0\)
\(\left(x-3\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}}\)
b) \((3x-5)(5x-7)+(5x+1)(2-3x)=4\)
\(15x^2-46x+35+10x-15x^2+2-3x-4=0\)
\(33-39x=0\)
\(3\left(11-13x\right)=0\)
\(11-13x=0\)
\(13x=11\)
\(x=\frac{11}{13}\)
12 tháng 4 2016
a) |2-x|=|x-5|
\(\Rightarrow2-x=x-5\)hoặc \(2-x=-\left(x-5\right)\)
\(\Rightarrow-x-x=-5-2\)hoặc \(-x+x=5-2\)(vô lý)
\(\Rightarrow x=3,5\)
GT
3
8 tháng 5 2020
Cj lm 2 cách nha,e kham khảo cách nào cx đc.
\(\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)=0\)
TH1 : \(2x+1=0\Leftrightarrow2x=-1\Leftrightarrow x=-\frac{1}{2}\)
TH2 : \(\left(x+1\right)^2=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
TH3 : \(2x+3=0\Leftrightarrow2x=-3\Leftrightarrow x=-\frac{3}{2}\)
\(\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)=0\)
\(\left(2x^3+4x^2+2x+x^2+2x+1\right)\left(2x+3\right)=0\)
\(\left(2x^3+5x^2+4x+1\right)\left(2x+3\right)=0\)
\(4x^4+6x^3+10x^3+15x^2+8x^2+12x+2x+3=0\)
\(4x^4+16x^3+23x^2+14x+3=0\)
\(\left(4x^2+6x+2x+3\right)\left(x+1\right)\left(x+1\right)=0\)
\(\left(2x+3\right)\left(2x-1\right)\left(x+1\right)^2=0\)
Tương tự như trên ....
8 tháng 5 2020
\(\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)=0\)
Th1: \(2x+1=0\Rightarrow2x=-1\Rightarrow x=-\frac{1}{2}\)
Th2: \(\left(x+1\right)^2=0\Rightarrow x+1=0\Rightarrow x=-1\)
Th3: \(2x+3=0\Rightarrow2x=-3\Rightarrow x=-\frac{3}{2}\)
( x - 1 )( x2 + 5x - 2 ) - ( x3 - 1 ) = 0
<=> ( x - 1 )( x2 + 5x - 2 ) - ( x - 1 )( x2 + x + 1 ) = 0
<=> ( x - 1 )( x2 + 5x - 2 - x2 - x - 1 ) = 0
<=> ( x - 1 )( 4x - 3 ) = 0
<=> x - 1 = 0 hoặc 4x - 3 = 0
<=> x = 1 hoặc x = 3/4
Vậy pt có tập nghiệm S = { 1 ; 3/4 }