tìm x a)3. (x-1)+ 2x-2x2 =0 b) x2 + 8x + 15 =0
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\(a,\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)
\(\left(x-1\right)\left(5x+3-3x+8\right)=0\)
\(\left(x-1\right)\left(2x+11\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\2x=-11\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=-\frac{11}{2}\end{cases}}}\)
\(b,3x\left(25x+15\right)-35\left(5x+3\right)=0\)
\(15x\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\left(5x+3\right).5\left(3x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x+3=0\\5\left(3x-7\right)=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=-3\\3x-7=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\x=\frac{7}{3}\end{cases}}}\)
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
a: \(\left(x-2\right)^2-\left(2x+3\right)^2=0\)
=>(x-2-2x-3)(x-2+2x+3)=0
=>(-x-5)(3x+1)=0
=>(x+5)(3x+1)=0
=>\(\left[\begin{array}{l}x+5=0\\ 3x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-5\\ x=-\frac13\end{array}\right.\)
b: \(9\left(2x+1\right)^2-4\left(x+1\right)^2=0\)
=>\(\left\lbrack3\left(2x+1\right)\right\rbrack^2-\left\lbrack2\left(x+1\right)\right\rbrack^2=0\)
=>\(\left(6x+3\right)^2-\left(2x+2\right)^2=0\)
=>(6x+3+2x+2)(6x+3-2x-2)=0
=>(8x+5)(4x+1)=0
=>\(\left[\begin{array}{l}8x+5=0\\ 4x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac58\\ x=-\frac14\end{array}\right.\)
c: \(x^3-6x^2+9x=0\)
=>\(x\left(x^2-6x+9\right)=0\)
=>\(x\left(x-3\right)^2=0\)
=>\(\left[\begin{array}{l}x=0\\ \left(x-3\right)^2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=3\end{array}\right.\)
d: \(x^2\left(x+1\right)-x\left(x+1\right)+x\left(x-1\right)=0\)
=>\(\left(x+1\right)\left(x^2-x\right)+x\left(x-1\right)=0\)
=>x(x+1)(x-1)+x(x-1)=0
=>x(x-1)(x+1+1)=0
=>x(x-1)(x+2)=0
=>\(\left[\begin{array}{l}x=0\\ x-1=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\\ x=-2\end{array}\right.\)
e: \(\left(x-2\right)^2-\left(x-2\right)\left(x+2\right)=0\)
=>(x-2)(x-2-x-2)=0
=>-4(x-2)=0
=>x-2=0
=>x=2
g: \(x^4-2x^2+1=0\)
=>\(\left(x^2-1\right)^2=0\)
=>\(x^2-1=0\)
=>\(x^2=1\)
=>\(\left[\begin{array}{l}x=1\\ x=-1\end{array}\right.\)
h: \(4x^2+y^2-20x-2y+26=0\)
=>\(4x^2-20x+25+y^2-2y+1=0\)
=>\(\left(2x-5\right)^2+\left(y-1\right)^2=0\)
=>\(\begin{cases}2x-5=0\\ y-1=0\end{cases}\Rightarrow\begin{cases}x=\frac52\\ y=1\end{cases}\)
i: \(x^2-2x+5+y^2-4y=0\)
=>\(x^2-2x+1+y^2-4y+4=0\)
=>\(\left(x-1\right)^2+\left(y-2\right)^2=0\)
=>\(\begin{cases}x-1=0\\ y-2=0\end{cases}\Rightarrow\begin{cases}x=1\\ y=2\end{cases}\)
a) \(3\left(x-1\right)+2x-2x^2=0\)
\(\Leftrightarrow3x-3+2x-2x^2=0\)
\(\Leftrightarrow-2x^2+5x-3=0\)
\(\Leftrightarrow-2x^2+2x+3x-3=0\)
\(\Leftrightarrow-2x\left(x-1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\-2x+3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy..
b) \(x^2+8x+15=0\)
\(\Leftrightarrow x^2+3x+5x+15=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)
Vậy..
Tìm x :
a) 3(x - 1 ) + 2x - 2x2 = 0
\(\Leftrightarrow3\left(x-1\right)-2x^2+2x=0\)
\(\Leftrightarrow\) 3\(\left(x-1\right)-2x\left(x-1\right)=0\)
\(\Leftrightarrow\) (x - 1 )( 3-2x) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\-2x=-3\Rightarrow x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy....
b) x2 + 8x + 15 = 0
\(\Leftrightarrow x^2+3x+5x+15=0\)
\(\Leftrightarrow\) (x2 + 3x ) + ( 5x + 15 ) =0
\(\Leftrightarrow x\left(x+3\right)+5\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)
Vậy....