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1 tháng 9 2020
( 2x - 3 )( x + 1 ) - 2x2 + 6x = 0
<=> 2x2 - x - 3 - 2x2 + 6x = 0
<=> 5x - 3 = 0
<=> 5x = 3
<=> x = 3/5
( x2 - x + 1 )( x - 3 ) - x3 + 4x2 = 0
<=> x3 - 4x2 + 4x - 3 - x3 + 4x2 = 0
<=> 4x - 3 = 0
<=> 4x = 3
<=> x = 3/4
( x2 - 2 )( x2 + 2 ) - x4 - 2x + 5 = 0
<=> ( x2 )2 - 4 - x4 - 2x + 5 = 0
<=> x4 + 1 - x4 - 2x = 0
<=> 1 - 2x = 0
<=> 2x = 1
<=> x = 1/2
( x - 3 )( x2 - 3x + 2 ) - ( x2 - 2x - 7 )( x - 2 ) + 2x2 - 2x = 0
<=> x3 - 6x2 + 11x - 6 - ( x3 - 4x2 - 3x + 14 ) + 2x2 - 2x = 0
<=> x3 - 6x2 + 11x - 6 - x3 + 4x2 + 3x - 14 + 2x2 - 2x = 0
<=> 12x - 20 = 0
<=> 12x = 20
<=> x = 20/12 = 5/3
A
1 tháng 9 2020
a, \(\left(2x-3\right)\left(x+1\right)-2x^2+6x=0\)
\(\Leftrightarrow2x^2+2x-3x-3-2x^2+6x=0\Leftrightarrow5x-3=0\Leftrightarrow x=\frac{3}{5}\)
b, \(\left(x^2-x+1\right)\left(x-3\right)-x^3+4x^2=0\)
\(\Leftrightarrow x^3-3x^2-x^2+3x+x-3-x^3+4x^2=0\Leftrightarrow4x-3=0\Leftrightarrow x=\frac{3}{4}\)
c ; d tương tự nhé !
KN
28 tháng 7 2015
1) (2x-1)(x+3)(2-x)=0
=>2x-1 =0 hoặc x+3=0 hoặc 2-x=0
=>x=1/2 hoặc x=-3 hoặc x=2
2)x^3 + x^2 + x + 1 = 0
=>.x^2(x+1)+(x+1)=0
=>(x^2+1)(x+1)=0
=>x^2+1=0 hoặc x+1=0
=> x =-1
3) 2x(x-3)+5(x-3) =0
=>(2x+5)(x-3)=0
=>2x+5=0 hoặc x-3=0
=>x=-5/2 hoặc x=3
4)x(2x-7)-(4x-14)=0
=> (x-2)(2x-7)=0
=> x-2 =0 hoặc 2x-7=0
=>x=2 hoặc x=7/2
5)2x^3+3x^2+2x+3=0
=>x^2(2x+3)+2x+3=0
=>(x^2+1)(2x+3)=0
=>x^2+1=0 hoặc 2x+3=0
=> x =-3/2
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 9 2020
a/
\(\Leftrightarrow x-2x^2+2x^2-3x-4x+6=0\)
\(\Leftrightarrow-6x+6=0\)
\(\Leftrightarrow x=1\)
b/
\(\Leftrightarrow2x^2-4x-2x^2-6x=0\)
\(\Leftrightarrow-10x=0\)
\(\Leftrightarrow x=0\)
c/
\(\Leftrightarrow\left(2x+3\right)\left(2x+3+x-3\right)=0\)
\(\Leftrightarrow3x\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\frac{3}{2}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 9 2020
c/
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(9y^2+30y+25\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(3y+5\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\3x+5=0\end{matrix}\right.\)
\(\Leftrightarrow x=y=-\frac{5}{3}\)
d/
\(\Leftrightarrow4x^2-4x+1+4x^2+4x+1-2\left(4x^2-2x-2\right)+x=12\)
\(\Leftrightarrow8x^2+x+2-8x^2+4x+4=12\)
\(\Leftrightarrow5x=6\)
\(\Leftrightarrow x=\frac{6}{5}\)
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 9 2019
a) 3x(4x - 3) - 2x(5 - 6x) = 0
=> 6x2 - 9x - 10x + 12x2 = 0
=> 18x2 - 19x = 0
=> x(18x - 19) = 0
=> \(\orbr{\begin{cases}x=0\\18x-19=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{19}{18}\end{cases}}\)
b) 5(2x - 3) + 4x(x - 2) + 2x(3 - 2x) = 0
=> 10x - 15 + 4x2 - 8x + 6x - 4x2 = 0
=> 8x - 15 = 0
=> 8x = 15
=> x = 15 : 8 = 15/8
c) 3x(2 - x) + 2x(x - 1) = 5x(x + 3)
=> 6x - 3x2 + 2x2 - 2x = 5x2 + 15x
=> 4x - x2 - 5x2 - 15x = 0
=> -6x2 - 11x = 0
=> -x(6x - 11) = 0
=> \(\orbr{\begin{cases}-x=0\\6x-11=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{11}{6}\end{cases}}\)
2T
5 tháng 9 2019
a) \(3x\left(4x-3\right)-2x\left(5-6x\right)=0\)
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow-19x=0\Leftrightarrow x=0\)
b) \(5\left(2x-3\right)+4x\left(x-2\right)+2x\left(3-2x\right)=0\)
\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)
\(\Leftrightarrow8x-15=0\Leftrightarrow x=\frac{15}{8}\)
14 tháng 8 2020
a) Ta có: 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{19}{24}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{19}{24}\right\}\)
b) Ta có: \(5\left(2x-3\right)+4x\left(x-2\right)+2x\left(3-2x\right)=0\)
\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)
\(\Leftrightarrow8x-15=0\)
\(\Leftrightarrow8x=15\)
hay \(x=\frac{15}{8}\)
Vậy: \(x=\frac{15}{8}\)
c) Ta có: \(3x\left(2-x\right)+2x\left(x-1\right)=5x\left(x+3\right)\)
\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\)
\(\Leftrightarrow-x^2+4x-5x^2-15x=0\)
\(\Leftrightarrow-6x^2-11x=0\)
\(\Leftrightarrow6x^2+11x=0\)
\(\Leftrightarrow x\left(6x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-11}{6}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{-11}{6}\right\}\)
d) Ta có: \(3x\left(x+1\right)-5x\left(3-x\right)+6\left(x^2+2x+3\right)=0\)
\(\Leftrightarrow3x^2+3x-15x+5x^2+6x^2+12x+18=0\)
\(\Leftrightarrow14x^2+18=0\)
\(\Leftrightarrow14x^2=-18\)
mà \(14x^2\ge0\forall x\)
nên \(x\in\varnothing\)
Vậy: \(x\in\varnothing\)
6 tháng 2 2017
c(x-1)^2=4
x^2-2x+1=4
x^2-2x+1-4=0
x^2-2x-3=0
x^2-3x+x-3=0
x(x-3)+(x-3)=0
(x-3)(x+1)=0
\(\Rightarrow\hept{\begin{cases}x-3=0\\x+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\x=-1\end{cases}}}\)
6 tháng 2 2017
d, x^3+2x^2-x-2=0
x^2(x+2)-(x+2)=0
(x+2)(x^2-1)=0
\(\Rightarrow\hept{\begin{cases}x=-2\\x=+-1\end{cases}}\)
L
4
18 tháng 11 2019
Suy ra (2x-4)-(3x-3×5)=1 Suy ra(2x-4)-3x+15=1 Suy ra 2x-4-3x+15=1 Suy ra (2x-3x)+(15-4)=1 -1x+11=1 1-11=-1x -1x=-10 X=10
3 tháng 6 2022
a: \(\Leftrightarrow x^2+6x+9+x^2-4-2x-2=7\)
\(\Leftrightarrow2x^2+4x-4=0\)
\(\Leftrightarrow x^2+2x-2=0\)
\(\Leftrightarrow x^2+2x+1-3=0\)
\(\Leftrightarrow\left(x+1\right)^2=3\)
hay \(x\in\left\{-\sqrt{3}-1;\sqrt{3}-1\right\}\)
b: \(\Leftrightarrow2x^2-x-\left(2x^2+3x-4x-6\right)=0\)
\(\Leftrightarrow2x^2-x-2x^2+x+6=0\)
=>6=0(vô lý)
c: \(\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=0\)
=>x=-2 hoặc x=2
đ: \(\Rightarrow2x^2-2x-5x+5=0\)
=>(x-1)(2x-5)=0
=>x=1 hoặc x=5/2
DT
4
KN
5 tháng 9 2020
a. \(x\left(x-2\right)-x\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow x^2-2x-x^3+4x^2-3x=0\)
\(\Leftrightarrow-x^3+5x^2-5x=0\)
\(\Leftrightarrow-x\left(x^2-5x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-x=0\\x^2-5x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x-\frac{5}{2}\right)^2-\frac{5}{4}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x-\frac{5}{2}\right)^2=\frac{5}{4}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=0\\x-\frac{5}{2}=\frac{\sqrt{5}}{2}\\x-\frac{5}{2}=-\frac{\sqrt{5}}{2}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{5+\sqrt{5}}{2}\\x=\frac{5-\sqrt{5}}{2}\end{cases}}\)
HT
5 tháng 9 2020
a) \(x\left(x-2\right)-x\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow x\left(x-2-x^2+4x-3\right)=0\)
\(\Leftrightarrow x\left(-x^2+5x-5\right)=0\)
\(\Leftrightarrow x\left(x-\frac{5+\sqrt{5}}{2}\right)\left(x-\frac{5-\sqrt{5}}{2}\right)=0\)
=> \(x\in\left\{0;\frac{5+\sqrt{5}}{2};\frac{5-\sqrt{5}}{2}\right\}\)
b) \(\left(2x-5\right)\left(x+3\right)-\left(x-1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow2x^2+x-15-2x^2-x+3=0\)
\(\Leftrightarrow-12=0\left(vn\right)\)
c) \(\left(x-2\right)\left(x^2+2x+8\right)-x^3-2x+1=0\)
\(\Leftrightarrow x^3+4x-16-x^3-2x+1=0\)
\(\Leftrightarrow2x=15\)
\(\Rightarrow x=\frac{15}{2}\)
Đề như thế này phải không ??
\(\left(2x-3\right)^2-\left(2x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-2\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=5\end{cases}}\)
( 2x - 3 )^2 - ( 2x - 3 )( x + 2 ) = 0