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22 tháng 7 2021
b) 5x(x-2000)-x+2000=0
\(\Rightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\\ \Rightarrow\left(x-2000\right)\left(5x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-2000=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+2000\\5x=0+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\5x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\)
14 tháng 2 2022
a: =>x-3=2 hoặc x-3=-2
=>x=5 hoặc x=1
b: =>x2=0
hay x=0
c: =>(3x-5-x+1)(3x-5+x-1)=0
=>(2x-4)(4x-6)=0
=>x=2 hoặc x=3/2
d: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(2x-1-x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-4\right)=0\)
hay \(x\in\left\{1;-1;4\right\}\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
14 tháng 2 2022
\(a,\left(x-3\right)^2=4\\\Leftrightarrow\left(x-3\right)^2-2^2=0\\ \Leftrightarrow \left(x-3-2\right).\left(x-3+2\right)=0\\ \Leftrightarrow\left(x-5\right).\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\\\Rightarrow S=\left\{1;5\right\}\\ b,x^2.\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=0\\x^2+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=-1\left(vô.lí\right)\end{matrix}\right.\\ \Rightarrow S=\left\{0\right\}\\ c,\left(3x-5\right)^2-\left(x-1\right)^2=0\\ \Leftrightarrow\left(3x-5-x+1\right).\left(3x-5+x-1\right)=0\\ \Leftrightarrow\left(2x-4\right).\left(4x-6\right)=0\\ \Leftrightarrow2.\left(x-2\right).2.\left(2x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{2}\end{matrix}\right.\\ \Rightarrow S=\left\{\dfrac{3}{2};2\right\}\)
\(d,\left(x^2-1\right).\left(2x-1\right)=\left(x^2-1\right).\left(x+3\right)\\ \Leftrightarrow\left(x^2-1\right).\left(2x-1-x-3\right)=0\\ \Leftrightarrow\left(x^2-1\right).\left(x-4\right)=0\\ \Leftrightarrow\left(x-1\right).\left(x+1\right).\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=4\end{matrix}\right.\\ \Rightarrow S=\left\{-1;1;4\right\}\)
16 tháng 2 2022
\(a,\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x-6=0\Leftrightarrow x=6\\2x-5=0\Leftrightarrow x=\dfrac{5}{2}\\3x+9=0\Leftrightarrow x=-3\end{matrix}\right.\)
\(b,2x\left(x-3\right)+5\left(x-3\right)=0\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\2x+5=0\Leftrightarrow x=-\dfrac{5}{2}\end{matrix}\right.\)
\(c,x^2-4-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(x=-7\left(2m-5\right)x-2m^2+8\Leftrightarrow x+7\left(2m-5\right)=8-2m^2\Leftrightarrow x\left(14m-34\right)=8-2m^2\)
\(ycđb\Leftrightarrow14m-34\ne0\Leftrightarrow m\ne\dfrac{34}{14}\)\(\Rightarrow x=\dfrac{8-2m^2}{14m-34}\)
\(3.17\Leftrightarrow4x^2-4x+1-2x-1=0\Leftrightarrow4x^2-6x=0\Leftrightarrow x\left(4x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
DT
16 tháng 2 2022
3.15:
a, \(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\2x-5=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=\dfrac{5}{2}\\x=-\dfrac{9}{3}=-3\end{matrix}\right.\)
b, \(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
c, \(\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
3.16
\(\Leftrightarrow\left(2m-5\right).-7-2m^2+8=0\)
\(\Leftrightarrow-14m+35-2m^2+8=0\)
\(\Leftrightarrow-14m-2m^2+43=0\)
\(\Leftrightarrow-2\left(7m+m^2\right)=-43\)
\(\Leftrightarrow m\left(7-m\right)=\dfrac{43}{2}\)
\(\Leftrightarrow\dfrac{m\left(7-m\right)}{1}-\dfrac{43}{2}=0\)
\(\Leftrightarrow\dfrac{14m-2m^2}{2}-\dfrac{43}{2}=0\)
pt vô nghiệm
25 tháng 2 2022
1: \(\Leftrightarrow\left(x-4\right)^2+14=-9\left(x-4\right)\)
\(\Leftrightarrow x^2-8x+16+14+9x-36=0\)
\(\Leftrightarrow x^2+x-6=0\)
=>(x+3)(x-2)=0
=>x=-3(nhận) hoặc x=2(nhận)
2: \(\Leftrightarrow\left(8x+1\right)\left(2x-1\right)-2x\left(2x+1\right)-12x^2+9=0\)
\(\Leftrightarrow16x^2-8x+2x-1-4x^2-2x-12x^2+9=0\)
=>-8x+8=0
hay x=1(nhận)
c: \(\dfrac{1}{2\left(x-3\right)}-\dfrac{3x-5}{\left(x-3\right)\left(x-1\right)}=\dfrac{1}{2}\)
\(\Leftrightarrow x-1-2\left(3x-5\right)=\left(x-3\right)\left(x-1\right)\)
\(\Leftrightarrow x^2-4x+3=x-1-6x+10=-5x+9\)
\(\Leftrightarrow x^2+x-6=0\)
=>(x+3)(x-2)=0
=>x=-3(nhận) hoặc x=2(nhận)
20 tháng 3 2023
a: =>x(x+3)=0
=>x=0 hoặc x=-3
b: =>x(1-2x)=0
=>x=0 hoặc x=1/2
c: =>(x-7)(2x+3-x)=0
=>(x-7)(x+3)=0
=>x=7 hoặc x=-3
d: =>(x-2)(3x-1-x-3)=0
=>(x-2)(2x-4)=0
=>x=2
NH
20 tháng 3 2023
a)
`x^2 +3x=0`
`<=>x(x+3)=0`
\(< =>\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
b)
`x-2x^2 =0`
`<=>x(1-2x)=0`
\(< =>\left[{}\begin{matrix}x=0\\1-2x=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)
c)
`(x-7)(2x+3)=x(x-7)`
`<=>(x-7)(2x+3)-x(x-7)=0`
`<=>(x-7)(2x+3-x)=0`
`<=>(x-7)(x+3)=0`
\(< =>\left[{}\begin{matrix}x-7=0\\x+3=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
d)
`(x-2)(x+3)=(x-2)(3x-1)`
`<=>(x-2)(x+3)-(x-2)(3x-1)=0`
`<=>(x-2)(x+3-3x+1)=0`
`<=>(x-2)(-2x+4)=0`
\(< =>\left[{}\begin{matrix}x-2=0\\-2x+4=0\end{matrix}\right.\\ < =>x=2\)
5 tháng 3 2022
a: (2x+1)(3-x)(4-2x)=0
=>(2x+1)(x-3)(x-2)=0
hay \(x\in\left\{-\dfrac{1}{2};3;2\right\}\)
b: 2x(x-3)+5(x-3)=0
=>(x-3)(2x+5)=0
=>x=3 hoặc x=-5/2
c: =>(x-2)(x+2)+(x-2)(2x-3)=0
=>(x-2)(x+2+2x-3)=0
=>(x-2)(3x-1)=0
=>x=2 hoặc x=1/3
d: =>(x-2)(x-3)=0
=>x=2 hoặc x=3
e: =>(2x+5+x+2)(2x+5-x-2)=0
=>(3x+7)(x+3)=0
=>x=-7/3 hoặc x=-3
f: \(\Leftrightarrow2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
hay \(x\in\left\{0;-3;\dfrac{1}{2}\right\}\)
5 tháng 2 2022
Bài 3:
b: \(\Leftrightarrow x^2\left(x+1\right)^2=0\)
hay \(x\in\left\{0;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=0\)
=>x-1=0
hay x=1
d: \(\Leftrightarrow6x^2-3x-4x+2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0\)
hay \(x\in\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)
CM
31 tháng 1 2019
a) Đa thức thương x 2 – 6x + 9.
b) Đa thức thương 2 x 2 – 5.
c) Đa thức thương x 2 + 4x + 3 và đa thức dư -12.
d) Đa thức x + 5 và đa thức dư x – 4.
a) \(x-\left(5x+3\right)=2x-4\)
\(\Leftrightarrow x-5x-3=2x-4\)
\(\Leftrightarrow x-5x-2x=-4+3\)
\(\Leftrightarrow-6x=-1\)
\(\Leftrightarrow x=\frac{1}{6}\)
b) \(2x^3-18x=0\)
\(\Leftrightarrow2x\left(x^2-9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x^2-9=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}\)
c)\(\left(x-3\right)^2=\left(2x+7\right)^2\)
\(\Leftrightarrow x-3=2x+7\)
\(\Leftrightarrow x-2x=7+3\)
\(\Leftrightarrow-x=10\)
\(\Leftrightarrow x=-10\)