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2
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GV Nguyễn Trần Thành Đạt
Giáo viên
3 tháng 8 2023
\(x^3-2x^2+x-2=0\\ \Leftrightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x^2+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ Vậy:x=2\\ ---\\ 2x\left(3x-5\right)=10-6x\\ \Leftrightarrow6x^2-10x-10+6x=0\\ \Leftrightarrow6x^2-4x-10=0\\ \Leftrightarrow6x^2+6x-10x-10=0\\ \Leftrightarrow6x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(6x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}6x-10=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
3 tháng 8 2023
\(4-x=2\left(x-4\right)^2\\ \Leftrightarrow4-x=2\left(x^2-8x+16\right)\\ \Leftrightarrow2x^2-16x+32+x-4=0\\ \Leftrightarrow2x^2-15x+28=0\\ \Leftrightarrow2x^2-8x-7x+28=0\\ \Leftrightarrow2x\left(x-4\right)-7\left(x-4\right)=0\\ \Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\\ ---\\ 4-6x+x\left(3x-2\right)=0\\ \Leftrightarrow4-6x+3x^2-2x=0\\ \Leftrightarrow3x^2-8x+4=0\\ \Leftrightarrow3x^2-6x-2x+4=0\\ \Leftrightarrow3x\left(x-2\right)-2\left(x-2\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)
1 tháng 9 2021
\(x^2\left(x-5\right)+x-5=0\)
\(\Rightarrow x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x^2-1\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x-1=0\\x+1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=1\\x=-1\end{matrix}\right.\)
1 tháng 9 2021
\(x^2\left(x-5\right)+x-5=0\)
\(\Leftrightarrow x-5=0\)
hay x=5
2 tháng 8 2017
a) (2x+12x−1−2x−12x+1):4x10x−5=(2x+1)2−(2x−1)2(2x−1)(2x+1).10x+54x(2x+12x−1−2x−12x+1):4x10x−5=(2x+1)2−(2x−1)2(2x−1)(2x+1).10x+54x
=4x2+4x+1−4x2+4x−1(2x−1)(2x+1).5(2x+1)4x4x2+4x+1−4x2+4x−1(2x−1)(2x+1).5(2x+1)4x
=8x.5(2x+1)(2x−1)(2x+1).4x=102x−18x.5(2x+1)(2x−1)(2x+1).4x=102x−1
b) (1x2+x−2−xx+1):(1x+x−2)(1x2+x−2−xx+1):(1x+x−2)
=(1x(x+1)+x−2x+1):1+x2−2xx(1x(x+1)+x−2x+1):1+x2−2xx
=1+x(x−2)x(x+1).xx2−2x+11+x(x−2)x(x+1).xx2−2x+1
=(x2−2x+1)xx(x+1)(x2−2x+1)=1x+1(x2−2x+1)xx(x+1)(x2−2x+1)=1x+1
c) 1x−1−x3−xx2+1.(1x2−2x+1+11−x2)1x−1−x3−xx2+1.(1x2−2x+1+11−x2)
=1x−1−x3−xx2+1.[1(x−1)2−1(x−1)(x+1)]
TN
1 tháng 8 2017
a) (2x+12x−1−2x−12x+1):4x10x−5(2x+12x−1−2x−12x+1):4x10x−5
= 0 - 0
= 0
b) (1x2+x−2−xx+1):(1x+x−2);(1x2+x−2−xx+1):(1x+x−2)
= (x-xx+1) : (2x-2) : (x-xx+1) : (2x-2)
c) 1x−1−x3−xx2+1.(1x2−2x+1+11−x2)
= -2x-1-xx2+1. (14 - 4x)
= -x2-1-xx2+14-4x
= -6x-xx2+13
27 tháng 2 2021
`a,x(x-1)-(x+2)^2=1`
`<=>x^2-x-x^2-4x-4=1`
`<=>-5x=5`
`<=>x=-1`
`b,(x+5)(x-3)-(x-2)^2=-1`
`<=>x^2+2x-15-x^2+4x-4+1=0`
`<=>6x-18=0`
`<=>x-3=0`
`<=>x=3`
`c,x(2x-4)-(x-2)(2x+3)=0`
`<=>2x(x-2)-(x-2)(2x+3)=0`
`<=>(x-2)(2x-2x-3)=0`
`<=>-3(x-2)=0`
`<=>x-2=0`
`<=>x=2`
`d,x(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12`
`<=>3x^2+2x+x^2+2x+1-4x^2+25=-12`
`<=>4x+26=-12`
`<=>4x=-38`
`<=>x=-19/2`
TN
1
30 tháng 7 2023
a: =>(x-2)(2x+5)=0
=>x=2 hoặc x=-5/2
c: \(\dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\)
=>\(\dfrac{2x^2+2x-x^2+x}{x^2-1}=1\)
=>x^2+3x=x^2-1
=>3x=-1
=>x=-1/3
10 tháng 12 2021
Câu 9:
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=-1\end{matrix}\right.\)
10 tháng 12 2021
\(9,\Leftrightarrow x^2\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow x^2+5x-x-5=0\\ \Leftrightarrow\left(x+5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\\ 12,\Leftrightarrow\left(x+1\right)^2-36=0\\ \Leftrightarrow\left(x+7\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\\ 13,\Leftrightarrow x^3-25x-x^3-8=17\\ \Leftrightarrow-25x=25\Leftrightarrow x=-1\\ 14,\Leftrightarrow x\left(2x^2+8x-3x-12\right)=0\\ \Leftrightarrow x\left(x+4\right)\left(2x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=\dfrac{3}{2}\end{matrix}\right.\)
31 tháng 8 2021
a: ta có: \(\left(2x-5\right)\left(x+2\right)-2x\left(x-1\right)=15\)
\(\Leftrightarrow2x^2+4x-5x-10-2x^2+2x=15\)
\(\Leftrightarrow x=25\)
b: Ta có: \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)
\(\Leftrightarrow4x^2-25+\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5+2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-3\end{matrix}\right.\)
c: Ta có: \(x\left(4x-5\right)-\left(2x+1\right)^2=0\)
\(\Leftrightarrow4x^2-5x-4x^2-4x-1=0\)
\(\Leftrightarrow-9x=1\)
hay \(x=-\dfrac{1}{9}\)
a: x(x+5)^2=x+5
=>(x+5)(x^2+5x-1)=0
=>x+5=0 hoặc x^2+5x-1=0
=>\(x\in\left\{-5;\dfrac{-5+\sqrt{29}}{2};\dfrac{-5-\sqrt{29}}{2}\right\}\)
b: x(x-2)=(x-2)
=>(x-2)(x-1)=0
=>x=2 hoặc x=1