(x-3)(x+4)-2(x-2)=(x-4)2
⇔(x-3)(x+4)-2(x-2)-(x-4)2=0
⇔x2+x-12-2x+4-x2+8x-16=0
⇔7x-24=0
⇔7x=24
⇔x=\(\frac{24}{7}\)
Vậy x=\(\frac{24}{7}\)
Đúng 0
Bình luận (0)
Bài 3: Phương trình đưa được về dạng ax + b = 0
Các câu hỏi tương tự
Giải phương trình về dạng ax + b = 0
1. (3x - 2)/3 - 2 = (4x + 1)/4
2. (x - 3)/4 + ( 2x - 1 )/3 = (2 - x)/6
3. 1/2 (x + 1) + 1/4(x + 3) = 3 - 1/3 (x + 2)
4 (x + 4)/5 - x + 4 = x/3 - (x - 2)/2
5. (4 - 5x)/6 = 2 (-x + 1)/2
Giải phương trình về dạng ax + b = 0
1. (3x - 2)/3 - 2 = (4x + 1)/4
2. (x - 3)/4 + ( 2x - 1 )/3 = (2 - x)/6
3. 1/2 (x + 1) + 1/4(x + 3) = 3 - 1/3 (x + 2)
4 (x + 4)/5 - x + 4 = x/3 - (x - 2)/2
5. (4 - 5x)/6 = 2 (-x + 1)/2
a)5-(x-6)=4(3-2x)
b)2x(x+2)2-8x2=2(x-2)(x2+2x+4)
c)7-(2x+4)=-(x+4)
d)(x-2)3+(3x-1)(3x+1)=(x+1)3
e)(x+1)(2x-3)=(2x-1)(x+5)
f)(x-1)3-x(x+1)2=5x(2-x)-11(x+2)
g)(x-1)-(2x-1)=9-x
h)(x-3)(x+4)-2(3x-2)=(x+4)2
i)x(x+3)2-3x=(x+2)33+11
j)(x+1)(x2-x+1)-2x=x(x+1)(x-1)
a) (x-1)^3-x(x+1)^2=5x(2-x)-11(x+2)
b) (x-3)(x+4)-2(3x-2)=(x-4)^2
14, giải các PT sau.
1, 4-(x-5)=5(x-3x)
2, 32-4(0,5y-5)=3y+2
3, 19-2(x+11)=5(2x-3)-4(5x-7)
4, 4(x+3)-7x+17=8(5x-1)+166
5, 17-14(x+1)=13-4(x+1)-5(x-3)
6, 5(x+10)2+2x=5x2-3
7, (2x-1)2+5=(2x+3)(2x-3)-x
8, 3(x-2)2+2(x+3)(x-3)=5(x+1)2
a) (x-1)^3-x(x+1)^2=5x(2-x)-11(x+2)
b) (x-3)(x+4)-2(3x-2)=(x-4)^2
14, giải các PT sau.
1, 4-(x-5)=5(x-3x)
2, 32-4(0,5y-5)=3y+2
3, 19-2(x+11)=5(2x-3)-4(5x-7)
4, 4(x+3)-7x+17=8(5x-1)+166
5, 17-14(x+1)=13-4(x+1)-5(x-3)
6, 5(x+10)2+2x=5x2-3
7, (2x-1)2+5=(2x+3)(2x-3)-x
8, 3(x-2)2+2(x+3)(x-3)=5(x+1)2
1) dfrac{5x-2}{3} dfrac{5-3x}{2}2) dfrac{x+4}{5} - x + 4 dfrac{x}{3} - dfrac{x-2}{2}3) dfrac{10x+3}{12} 1 + dfrac{6+8x}{9}4) dfrac{x+1}{3}- dfrac{x-2}{6} dfrac{2x-1}{2}
Đọc tiếp
1) \(\dfrac{5x-2}{3}\)= \(\dfrac{5-3x}{2}\)
2) \(\dfrac{x+4}{5}\) - x + 4 = \(\dfrac{x}{3}\) - \(\dfrac{x-2}{2}\)
3) \(\dfrac{10x+3}{12}\)= 1 + \(\dfrac{6+8x}{9}\)
4) \(\dfrac{x+1}{3}\)- \(\dfrac{x-2}{6}\) = \(\dfrac{2x-1}{2}\)
BT: giải các pt sau
1. \(\frac{x+1}{2013}+\frac{x+2}{2012}=\frac{x+3}{2011}+\frac{x+4}{2010}\)
2. \(\frac{3x+2}{4}+\frac{x+3}{2}=\frac{x-1}{3}-\frac{-x-1}{12}\)
3.\(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2+3}{1-x^2}=0\)
4. \(\frac{1}{x+2}-\frac{3x}{x-2}=\frac{16}{x^2-4}\)