(2x-5)(x+1)=0
⇔\(\left\{{}\begin{matrix}2x-5=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\x=-1\end{matrix}\right.\)
Giai phường trình sau:
a, \(3x^2+2x-1=0\) e, \(4x^2-12x+5=0\) i,\(2x^2+5x-3=0\)
b,\(x^2-5x+6=0\) f, \(2x^2+5x+3=0\) j,\(x^2+6x-16=0\)
c,\(x^2-3x+2=0\) g,\(x^2+x-2=0\)
d,\(2x^2-6x+1=0\) h, \(x^2-4x+3=0\)
Bài 3: Giải các phương trình sau:
a) 2x – 5 = 0 b) 12 – 3x = 3( 4-x) c) 2(x – 3) = 2x-1
5(x-3)+7=8x-(2x+9)
4x(x-1)+2(x-1)=0
Giải phương trình
1) 5(x - 3) - 4 = 2(x - 1)
2) 5(x - 3) - 2(x - 5) = x-2
3) 3(x - 2) - 14x = 2(3-) + 1
4) (x + 1)²+ 2x = x(x + 1) + 6
5) 3 - 4x(3 - 2x) = 8x² + x - 30
6) x²-x(5 - x) = 8
7) (x - 1)² - 36 = 0
8) (3x - 1)(4x - 3) + 2x(6x - 1) = 2(2x + 7)
9) (x - 2)² + 4(x - 3) =(x² + x - 3)
10) (x - 2)² – 2(x + 1) = (x - 1)(x - 2)
11) (x - 2)² + 3(x - 5) = x² + 3x - 3
12)(x - 3)² + (x + 3)² = 2 (x² +9)
13) (3x - 1)2 + (3x +1)² = 2(9x² + 4) + 1
14) (x - 1)(x - 2) + (2x + 1) = 5x²
giải phương trình, tiếp
\(\left(x+1\right)^2=4\left(x^2-2x+1\right)^2\)
\(\left(2x+7\right)^2=9\left(x+2\right)^2\)
\(4\left(2x+7\right)^2=9\left(x+3\right)^2\)
\(\frac{1}{9}\left(x-3\right)^2-\frac{1}{25}\left(x+5\right)^2=0\)
\(2x^2-6x+1=0\)
\(3x^2+12x-66=0\)
\(9x^2-30x+225=0\)
\(3x^2-7x+1=0\)
\(3x^2-7x+8=0\)
\(x^2-4x+1=0\)
\(2x^2-6x+1=0\)
Câu2:giải các phương trình sau:
a)5(3x+2) =4x+1
b)\(\frac{4x-5}{x-1}=2+\frac{X}{X-1}\) c)\(2x^3+4x^2+2x=0\)
4(3x-2)-3(x-4)=7x-10
(3x-1/2)(-2/3x+1)=0
2x/x+1+3(x-1)/x=5
a)\(x^4+\left(x-1\right)\left(3^{x^2}+2x-2\right)=0\)
b)\(\dfrac{x^2-3x+5}{x^2-4x+5}-\dfrac{x^2-5x+5}{x^2-6x+5}=\dfrac{-1}{4}\)
(x2-25) +(x-5)(2x-11)=0
