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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 !
25 tháng 1 2021
a) Ta có: \(3x^2+2x-1=0\)
\(\Leftrightarrow3x^2+3x-x-1=0\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;\dfrac{1}{3}\right\}\)
b) Ta có: \(x^2-5x+6=0\)
\(\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy: S={2;3}
c) Ta có: \(x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: S={1;2}
d) Ta có: \(2x^2-6x+1=0\)
\(\Leftrightarrow2\left(x^2-3x+\dfrac{1}{3}\right)=0\)
mà \(2\ne0\)
nên \(x^2-3x+\dfrac{1}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{23}{12}=0\)
\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{23}{12}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{69}}{6}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{69}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9+\sqrt{69}}{6}\\x=\dfrac{9-\sqrt{69}}{6}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{9+\sqrt{69}}{6};\dfrac{9-\sqrt{69}}{6}\right\}\)
e) Ta có: \(4x^2-12x+5=0\)
\(\Leftrightarrow4x^2-10x-2x+5=0\)
\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{5}{2};\dfrac{1}{2}\right\}\)
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
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
\(\Rightarrow\hept{\begin{cases}x=-2\\x=+-1\end{cases}}\)
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
PA
30 tháng 11 2016
\(2x^2-7x+5=0\)
\(2x^2-2x-5x+5=0\)
\(2x\left(x-1\right)-5\left(x-1\right)=0\)
\(\left(x-1\right)\left(2x-5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\2x-5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\2x=5\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=\frac{5}{2}\end{array}\right.\)
\(x\left(2x-5\right)-4x+10=0\)
\(x\left(2x-5\right)-2\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(x-2\right)=0\)
\(\left[\begin{array}{nghiempt}x-2=0\\2x-5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=2\\2x=5\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=2\\x=\frac{5}{2}\end{array}\right.\)
\(\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\)
\(x^2-25-x^2+2x=15\)
\(2x=15+25\)
\(2x=40\)
\(x=\frac{40}{2}\)
\(x=20\)
\(x^2\left(2x-3\right)-12+8x=0\)
\(x^2\left(2x-3\right)+4\left(2x-3\right)=0\)
\(\left(2x-3\right)\left(x^2+4\right)=0\)
\(2x-3=0\) (vì \(x^2\ge0\Rightarrow x^2+4\ge4>0\))
\(2x=3\)
\(x=\frac{3}{2}\)
\(x\left(x-1\right)+5x-5=0\)
\(x\left(x-1\right)+5\left(x-1\right)=0\)
\(\left(x-1\right)\left(x+5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\x+5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=-5\end{array}\right.\)
\(\left(2x-3\right)^2-4x\left(x-1\right)=5\)
\(4x^2-12x+9-4x^2+4x=5\)
\(-8x=5-9\)
\(-8x=-4\)
\(x=\frac{4}{8}\)
\(x=\frac{1}{2}\)
\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(5x-2x^2+2x^2-2x=13\)
\(3x=13\)
\(x=\frac{13}{3}\)
\(2\left(x+5\right)\left(2x-5\right)+\left(x-1\right)\left(5-2x\right)=0\)
\(\left(2x+10\right)\left(2x-5\right)-\left(x-1\right)\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(2x+10-x+1\right)=0\)
\(\left(2x-5\right)\left(x+11\right)=0\)
\(\left[\begin{array}{nghiempt}2x-5=0\\x+11=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}2x=5\\x=-11\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-11\end{array}\right.\)
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 \(x\in\left\{0;-3;\dfrac{1}{2}\right\}\)
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 11 2017
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
30 tháng 9 2018
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
`x^2-2x >= 0`
`<=>x(x-2) >= 0`
`<=>` $\left[\begin{matrix} \begin{cases} x \ge 0\\x-2 \ge 0 \end{cases}\\ \begin{cases} x \le 0\\x-2 \le 0 \end{cases}\end{matrix}\right.$
`<=>` $\left[\begin{matrix} \begin{cases} x \ge 0\\x \ge 2 \end{cases}\\ \begin{cases} x \le 0\\x \le 2 \end{cases}\end{matrix}\right.$
`<=>` $\left[\begin{matrix} x \ge 2\\ x \le 0\end{matrix}\right.$
Vậy `S={x|x >= 2;x <= 0}`
\(x^2-2x\ge0\Leftrightarrow x\left(x-2\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x-2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-2< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x\ge2\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x< 2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\ge2\\x< 0\end{matrix}\right.\)