Tìm x
( x - 4)( 5x - 2 ) - 3( x - 4 ) = 0
( x + 2 )( x^2 - 2x + 4 ) + x( x + 5 )( x - 5 )
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
5 tháng 11 2023
2: \(3x\left(x-4\right)+2x-8=0\)
=>\(3x\left(x-4\right)+2\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(3x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)
3: 4x(x-3)+x2-9=0
=>\(4x\left(x-3\right)+\left(x+3\right)\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(4x+x+3\right)=0\)
=>\(\left(x-3\right)\left(5x+3\right)=0\)
=>\(\left[{}\begin{matrix}x-3=0\\5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{5}\end{matrix}\right.\)
4: \(x\left(x-1\right)-x^2+3x=0\)
=>\(x^2-x-x^2+3x=0\)
=>2x=0
=>x=0
5: \(x\left(2x-1\right)-2x^2+5x=16\)
=>\(2x^2-x-2x^2+5x=16\)
=>4x=16
=>x=4
29 tháng 8 2021
\(a,x^4-2x^3+5x^2-10x=0\\ \Leftrightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Leftrightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x^2+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x\in\varnothing\left(x^2+5>0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(b,\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Leftrightarrow\left(3x+5\right)^2-\left(2x-2\right)^2=0\\ \Leftrightarrow\left(3x+5+2x-2\right)\left(3x+5-2x+2\right)=0\\ \Leftrightarrow\left(5x+3\right)\left(x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-7\end{matrix}\right.\)
\(c,x^3-2x^2+x=0\\ \Leftrightarrow x\left(x-1\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(d,x^2\left(x-1\right)-4x^2+8x-4=0\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
29 tháng 8 2021
a) \(x^4-2x^3+5x^2-10x=0\\ \Rightarrow\left(x^4-2x^3\right)+\left(5x^2-10x\right)=0\\ \Rightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Rightarrow\left(x^3+5x\right)\left(x-2\right)=0\\ \Rightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2+5=0\\x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{5}\\x=2\end{matrix}\right.\)
Vậy \(x=\left\{-\sqrt{5};0;\sqrt{5};2\right\}\)
b) \(\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Rightarrow\left[{}\begin{matrix}3x+5=2x-2\\3x+5=-2x+2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\)
c) \(x^3-2x^2+x=0\\ \Rightarrow x\left(x^2-2x+1\right)=0\\ \Rightarrow x\left(x-1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
vậy ...
d) \(x^2\left(x-1\right)-4x^2+8x-4=0\\ x^2\left(x-1\right)-\left(4x^2-8x+4\right)=0\\ x^2\left(x-1\right)-\left(2x-2\right)^2=0\\ \Rightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left[x^2-4\left(x-1\right)\right]=0\\ \Rightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Rightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
6 tháng 6 2017
a) (x + 2)(x + 3) - (x - 2)(x + 5) = 0
<=> x2 + 3x + 2x + 6 - (x2 + 5x - 2x - 10) = 0
<=> x2 + 3x + 2x + 6 - x2 - 5x + 2x + 10 = 0
<=> 2x + 16 = 0
<=> 2x = -16
<=> x = -8
b) (2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)
<=> (2x + 3)(x - 4) + (x - 5)(x - 2) - (3x - 5)(x - 4) = 0
<=> 2x2 - 8x + 3x - 12 + x2 - 2x - 5x + 10 - (3x2 - 12x - 5x + 20) = 0
<=> 2x2 - 8x + 3x - 12 + x2 - 2x - 5x + 10 - 3x2 + 12x + 5x - 20 = 0
<=> 5x = 12 - 10 + 20
<=> 5x = 22
<=> x = 22/5
c) (8 - 5x)(x + 2) + 4(x - 2)(x + 1) + 2(x - 2)(x + 2) = 0
<=> 8x + 16 - 5x2 - 10x + (4x - 8)(x + 1) + 2(x2 - 4) = 0
<=> 8x + 16 - 5x2 - 10x + 4x2 + 4x - 8x - 8 + 2x2 - 8 = 0
<=> x2 - 6x = 0
<=> x(x - 6) = 0
<=> x = 0 hay x - 6 = 0
I<=> x = 6
d) (8x - 3)(3x + 2) - (4x + 7)(x + 4) = (2x + 1)(5x - 1) - 33
<=> 24x2 + 16x - 9x - 6 - (4x2 + 16x + 7x + 28) = 10x2 - 2x + 5x - 1 - 33
<=> 24x2 + 16x - 9x - 6 - 4x2 - 16x - 7x - 28 - 10x2 + 2x - 5x + 1 + 33 = 0
<=> 10x2 - 19x = 0
<=> x(10x - 19) = 0
<=> x = 0 hay 10x - 19 = 0
I <=> 10x = 19
I <=> x = 19/10
LN
15 tháng 4 2020
1) (x+6)(3x-1)+x+6=0
⇔(x+6)(3x-1)+(x+6)=0
⇔(x+6)(3x-1+1)=0
⇔3x(x+6)=0
2) (x+4)(5x+9)-x-4=0
⇔(x+4)(5x+9)-(x+4)=0
⇔(x+4)(5x+9-1)=0
⇔(x+4)(5x+8)=0
3)(1-x)(5x+3)÷(3x-7)(x-1)
=\(\frac{\left(1-x\right)\left(5x+3\right)}{\left(3x-7\right)\left(x-1\right)}=\frac{\left(1-x\right)\left(5x+3\right)}{\left(7-3x\right)\left(1-x\right)}=\frac{\left(5x+3\right)}{\left(7-3x\right)}\)
16 tháng 11 2021
a: \(x\in\left\{0;25\right\}\)
c: \(x\in\left\{0;5\right\}\)
12 tháng 7 2023
a: \(\dfrac{x+5}{x-1}+\dfrac{8}{x^2-4x+3}=\dfrac{x+1}{x-3}\)
=>(x+5)(x-3)+8=x^2-1
=>x^2+2x-15+8=x^2-1
=>2x-7=-1
=>x=3(loại)
b: \(\dfrac{x-4}{x-1}-\dfrac{x^2+3}{1-x^2}+\dfrac{5}{x+1}=0\)
=>(x-4)(x+1)+x^2+3+5(x-1)=0
=>x^2-3x-4+x^2+3+5x-5=0
=>2x^2+2x-6=0
=>x^2+x-3=0
=>\(x=\dfrac{-1\pm\sqrt{13}}{2}\)
e: =>x^2-2x+1+2x+2=5x+5
=>x^2+3=5x+5
=>x^2-5x-2=0
=>\(x=\dfrac{5\pm\sqrt{33}}{2}\)
g: (x-3)(x+4)*x=0
=>x=0 hoặc x-3=0 hoặc x+4=0
=>x=0;x=3;x=-4
Lần sau bạn ghi đúng lớp ạ ~
( x - 4 )( 5x - 2 ) - 3( x - 4 ) = 0
⇔ 5x2 - 2x - 20x + 8 - 3x + 12 = 0
⇔ 5x2 - 25x + 20 = 0
⇔ 5x2 - 5x - 20x + 20 = 0
⇔ 5x( x - 1 ) - 20( x - 1 ) = 0
⇔ ( x - 1 )( 5x - 20 ) = 0
⇔ \(\orbr{\begin{cases}x-1=0\\5x-20=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=4\end{cases}}\)
( x + 2 )( x2 - 2x + 4 ) + x( x + 5 )( x - 5 )
= x3 + 23 + x( x2 - 25 )
= x3 + 8 + x3 - 25x
= 2x3 - 25x + 8