Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
a) \(\left(x-2\right)\left(x^2+2x+7\right)+2\left(x^2-4\right)-5\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+2x+7\right)+2\left(x-2\right)\left(x+2\right)-5\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left[x^2+2x+7+2\left(x+2\right)-5\right]=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+2x+7+2x+4-5\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+4x+6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x^2+4x+6=0\end{matrix}\right.\)
Ta có:
\(x^2+4x+6\)
\(=x^2+2.x.2+4+2\)
\(=\left(x+2\right)^2+2\)
Vì \(\left(x+2\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x+2\right)^2+2\ge2\) với mọi x
\(\Rightarrow x^2+4x+6\) vô nghiệm
\(\Rightarrow x-2=0\)
\(\Rightarrow x=2\)
b) \(3x\left(x-1\right)+\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(3x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
c) \(2\left(x+3\right)x^2-3x=0\)
\(\Rightarrow x\left[2\left(x+3\right)x-3\right]=0\)
\(\Rightarrow x\left(2x^2+6x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x^2+6x-3=0\end{matrix}\right.\)
Ta có:
\(2x^2+6x-3\)
\(=2\left(x^2+3x-\dfrac{3}{2}\right)\)
\(=2\left(x^2+2.x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{9}{4}-\dfrac{3}{2}\right)\)
\(=2\left(x+\dfrac{3}{2}\right)^2-\dfrac{15}{2}\)
Vì \(2\left(x+\dfrac{3}{2}\right)^2\ge0\) với mọi x
\(\Rightarrow2\left(x+\dfrac{3}{2}\right)^2-\dfrac{15}{2}\ge-\dfrac{15}{2}\) với mọi x
\(\Rightarrow2x^2+6x-3\) vô nghiệm
\(\Rightarrow x=0\)
1. Tìm x :
a) \(( x+2 )^2 - 2( x - 3 ) = ( x + 1 ) ^ 2\) (1)
\(\Leftrightarrow x^2+4x+4-2x+6=x^2+2x+1\)
\(\Leftrightarrow x^2+2x-x^2-2x=1-10\)
\(\Leftrightarrow x\in\varnothing\)
Vậy phương trình (1) vô nghiệm.
b. \(( x - 1 ) ^ 2 + ( x -2 ) ^ 2 = 2( x + 4 ) ^ 2 - ( 22x + 27 )\) (2)
\(\Leftrightarrow x^2-2x+1+x^2-4x+4=2\left(x^2+8x+16\right)-22x-27\)
\(\Leftrightarrow2x^2-6x+5=2x^2+16x+32-22x-27\)
\(\Leftrightarrow2x^2-6x-2x^2-16x+22x=32-27-5\)
\(\Leftrightarrow x\in\varnothing\)
Vậy phương trình (2) vô nghiệm.
a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
Sorry mình nhầm câu a
a) (2x - 1)2 + (x + 3)2 - 5(x + 7)(x - 7) = 0
b) (x + 2)(x2 - 2x + 4) - x(x2 + 2) = 15
c) (x + 3)3 - x(3x + 1)2 + (2x - 1)(4x2 - 2x + 1) = 28
d) (x2 - 1)3 - (x4 + x2 + 1)(x2 - 1) = 0
Giải:
a) (2x - 1)2 + (x + 3)2 - 5(x + 7)(x - 7) = 0
\(\Leftrightarrow\) 4x2 - 4x + 1 + x2 + 6x + 9 - 5(x2 - 49) = 0
\(\Leftrightarrow\) 4x2 - 4x + 1 + x2 + 6x + 9 - 5x2 + 245 = 0
\(\Leftrightarrow\) 2x + 255 = 0
\(\Leftrightarrow\) 2x = - 255
\(\Leftrightarrow\) x = - 255 : 2
\(\Leftrightarrow\) x = \(-\frac{255}{2}\)
Vậy x = \(-\frac{255}{2}\)
b) (x + 2)(x2 - 2x + 4) - x(x2 + 2) = 15
\(\Leftrightarrow\) x3 + 8 - x3 - 2x = 15
\(\Leftrightarrow\) 8 - 2x = 15
\(\Leftrightarrow\) 2x = 8 - 1
\(\Leftrightarrow\) 2x = - 7
\(\Leftrightarrow\) x = - 7 : 2
\(\Leftrightarrow\) x = \(-\frac{7}{2}\)
Vậy x = \(-\frac{7}{2}\)
c) (x + 3)3 - x(3x + 1)2 + (2x - 1)(4x2 - 2x + 1) = 28
\(\Leftrightarrow\) x3 + 6x2 + 27x + 27 - x(9x2 + 6x + 1) + 8x3 - 1 = 28
\(\Leftrightarrow\) x3 + 6x2 + 27x + 27 - 9x3 - 6x2 - x + 8x3 - 1 = 28
\(\Leftrightarrow\) 26x + 26 = 28
\(\Leftrightarrow\) 26x = 28 - 26
\(\Leftrightarrow\) 26x = 2
\(\Leftrightarrow\) x = 2 : 26
\(\Leftrightarrow\) x = \(\frac{1}{13}\)
Vậy x = \(\frac{1}{13}\)
d) (x2 - 1)3 - (x4 + x2 + 1)(x2 - 1) = 0
\(\Leftrightarrow\) x6 - 2x2 + 1 - (x6 - 1) = 0
\(\Leftrightarrow\) x6 - 2x2 + 1 - x6 + 1 = 0
\(\Leftrightarrow\) -2x2 + 2 = 0
\(\Leftrightarrow\) -2x2 = - 2
\(\Leftrightarrow\) x2 = - 2 : (- 2)
\(\Leftrightarrow\) x2 = 1
\(\Leftrightarrow\) x = 1 hoặc x = - 1
Vậy x \(\in\) {1; - 1}
