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NT
1
14 tháng 7 2021
(4x-3)(2x-5) +(3-4x)(x-1)=0
(4x-3)(2x-5)-(4x-3)(x-1)=0
(4x-3)(2x-5-x+1)=0
(4x-3)(x-4)=0
4x-3=0 hoặc x-4=0
x=\(\frac{3}{4}\)hoặc x=4
TP
1
KN
18 tháng 10 2019
a) \(3x^2-2x=0\)
Phương trình này xác định với mọi x
b)\(\frac{1}{x-1}=3\)
pt xác định \(\Leftrightarrow x-1\ne0\Leftrightarrow x\ne1\)
c) \(\frac{2}{x-1}=\frac{x}{2x-4}\)
pt xác định\(\Leftrightarrow\hept{\begin{cases}x-1\ne0\\2x-4\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne2\end{cases}}\)
d) \(\frac{2x}{x^2-9}=\frac{1}{x+3}\)
pt xác định\(\Leftrightarrow\hept{\begin{cases}x^2-9\ne0\\x+3\ne0\end{cases}}\Leftrightarrow x\ne\pm3\)
e) \(2x=\frac{1}{x^2-2x+1}\)
pt xác định\(\Leftrightarrow x^2-2x+1\ne0\Leftrightarrow\left(x-1\right)^2\ne0\)
\(\Leftrightarrow x-1\ne0\Leftrightarrow x\ne1\)
f) \(\frac{1}{x-2}=\frac{2x}{x^2-5x+6}\)
\(\Leftrightarrow\frac{1}{x-2}=\frac{2x}{\left(x-3\right)\left(x-2\right)}\)
pt xác định\(\Leftrightarrow\hept{\begin{cases}x-2\ne0\\\left(x-2\right)\left(x-3\right)\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne2\\x\ne3\end{cases}}\)
29 tháng 6 2017
Ta có : x2 - 2x - (x + 3)2 = 6
<=> x2 - 2x - x2 - 6x - 9 = 6
<=> -8x - 9 = 6
=> -8x = 15
=> x = \(\frac{15}{-8}\)
ND
21
4 tháng 9
Bài 4:
a: \(2x^4+18x^2=0\)
=>\(2x^2\left(x^2+9\right)=0\)
=>\(x^2=0\) (Vì \(2\left(x^2+9\right)=2x^2+18\ge18>0\forall x\) )
=>x=0
b: (x-5)(x+5)-15x+75=0
=>(x-5)(x+5)-15(x-5)=0
=>(x-5)(x+5-15)=0
=>(x-5)(x-10)=0
=>\(\left[\begin{array}{l}x-5=0\\ x-10=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=10\end{array}\right.\)
c: \(x^4=x^2\)
=>\(x^4-x^2=0\)
=>\(x^2\left(x^2-1\right)=0\)
=>\(\left[\begin{array}{l}x^2=0\\ x^2-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x^2=0\\ x^2=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\\ x=-1\end{array}\right.\)
d: \(12x\left(6x-1\right)-24x^2=0\)
=>12x(6x-1-2x)=0
=>x(4x-1)=0
=>\(\left[\begin{array}{l}x=0\\ 4x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=\frac14\end{array}\right.\)
Bài 2:
a: 4x-16+3y(4-x)
=4(x-4)-3y(x-4)
=(x-4)(4-3y)
b: \(9y^2-6y+1=\left(3y\right)^2-2\cdot3y\cdot1+1^2=\left(3y-1\right)^2\)
c: \(25x^2-4=\left(5x\right)^2-2^2=\left(5x-2\right)\left(5x+2\right)\)
d: \(x^2-12x+36=x^2-2\cdot x\cdot6+6^2=\left(x-6\right)^2\)
e: \(8x^3+36x^2+54x+27\)
\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2+3^3\)
\(=\left(2x+3\right)^3\)
f: \(\left(2x-5\right)^2-\left(2x+y\right)^2\)
=(2x-5-2x-y)(2x-5+2x+y)
=(-y-5)(4x+y-5)
g: \(\left(2x-y\right)^3+\left(2x+y\right)^3\)
\(=8x^3-12x^2y+6xy^2-y^3+8x^3+12x^2y+6xy^2+y^3\)
\(=16x^3+12xy^2=4x\left(4x^2+3y^2\right)\)
Câu 1:
a: \(6x^2-72x=0\)
=>\(6\left(x^2-12x\right)=0\)
=>\(x^2-12x=0\)
=>x(x-12)=0
=>\(\left[\begin{array}{l}x=0\\ x-12=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=12\end{array}\right.\)
b: \(-2x^4+16x=0\)
=>\(-2x\left(x^3-8\right)=0\)
=>\(x\left(x^3-8\right)=0\)
=>\(\left[\begin{array}{l}x=0\\ x^3-8=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^3=8\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=2\end{array}\right.\)
c: \(\left(2x-1\right)^3-8x\left(x-3\right)\cdot\left(x+3\right)=-1\)
=>\(8x^3-12x^2+6x-1-8x\cdot\left(x^2-9\right)=-1\)
=>\(8x^3-12x^2+6x-1-8x^3+72x=-1\)
=>\(-12x^2+78x=0\)
=>-6x(2x-13)=0
=>x(2x-13)=0
=>\(\left[\begin{array}{l}x=0\\ 2x-13=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=\frac{13}{2}\end{array}\right.\)
d: \(x\left(x-5\right)-\left(x-3\right)^2=0\)
=>\(x^2-5x-\left(x^2-6x+9\right)=0\)
=>\(x^2-5x-x^2+6x-9=0\)
=>x-9=0
=>x=9
e: \(x\left(x-5\right)+3\left(x-5\right)=0\)
=>(x-5)(x+3)=0
=>\(\left[\begin{array}{l}x-5=0\\ x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=-3\end{array}\right.\)
f: 2x(x-8)-5(8-x)=0
=>2x(x-8)+5(x-8)=0
=>(x-8)(2x+5)=0
=>\(\left[\begin{array}{l}x-8=0\\ 2x+5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=-\frac52\end{array}\right.\)
g: \(30x-15x^2=0\)
=>15x(2-x)=0
=>x(2-x)=0
=>\(\left[\begin{array}{l}x=0\\ 2-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=2\end{array}\right.\)
h: \(-4x^3-12x=0\)
=>\(-4x\left(x^2+3\right)=0\)
=>x=0
\(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)
\(\left(2x+3\right)\left(x-1\right)-\left(2x-3\right)\left(x-1\right)=0\)
\(\left(x-1\right)\left(2x+3-2x+3\right)=0\)
\(\left(x-1\right)\cdot6=0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
(2x+3).(x-1) + (2x-3).(1-x) = 0
(2x+3).(x-1) - (2x+3).(1-x) = 0
(2x+3).[(x-1) - (1-x)] = 0
(2x+3).( x - 1 -1 + x) = 0
(2x+1). ( 2x - 2) = 0
(2x+1).2.(x-1) = 0
=> 2x+1 = 0 => 2x = -1 => x = -1/2
x-1=0 => x = 1