Ta có: \(2x+\dfrac{3}{4}=1+\dfrac{x-7}{5}\)
\(\Leftrightarrow\dfrac{40x}{20}+\dfrac{15}{20}=\dfrac{20}{20}+\dfrac{4\left(x-7\right)}{20}\)
\(\Leftrightarrow40x+15=20+4x-28\)
\(\Leftrightarrow40x+15-4x+8=0\)
\(\Leftrightarrow36x+23=0\)
\(\Leftrightarrow36x=-23\)
\(\Leftrightarrow x=-\dfrac{23}{36}\)
Vậy: \(S=\left\{-\dfrac{23}{36}\right\}\)
x+1 phần 3 + 3( 2x + 1) phần 4 = 2x + 3 ( x+ 1) phần 6 +7 + 12x phần 12
1) Phương trình dạng ax+b=0
1) 2x+x+12=0
2) x-5=3-x
3)2x-(3-5x)=4(x+3)
4)2x+3/3=5-4x/2
5) x-3/5=6- 1-2x/3
6) 3x-2/6 -5=3-2(x+7)/4
7) 3x-7/2+ x+1/3= -16
8) x- x+1/3=2x+1/5
1) 3 - 2x + 4 + 6x = x + 7 + 3x
2) -6(1,5 - 2x) = 3(-15 + 2x)
3) 3(2x - 5) + 5(x -1) = 4(x + 1)
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
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
bài 2 giải các phương trình sau
b,2(x+3)-4=0
d,5(x-3)=3x-5
f,7(5-x)=11-5x
h,2(3x-1)-3x=10
j,3-2x=3.(x+1)-x-2
m,4(2x-3)-5=6(3-x)-7
4,\(\dfrac{x+1}{3}\)+\(\dfrac{3\left(2x+1\right)}{4}\)=\(\dfrac{2x+3\left(x+1\right)}{6}\)+\(\dfrac{7+12x}{12}\)
5,\(\dfrac{2x}{3}\)+\(\dfrac{2x-1}{6}\)=4-\(\dfrac{x}{3}\)
6,\(\dfrac{x-1}{2}\)+\(\dfrac{x-1}{4}\)=1-\(\dfrac{2\left(x-1\right)}{3}\)
Bài 1 : Giải các phương trình sau :
1) (x – 2)(x – 5) = (x – 3)(x – 4)
2) ( 6x + 2)(x – 2) = 2x(3x – 5)
3) (x – 2) 2 = (x – 3)(x + 2)
4) (x–1)(x +3) – (x+2)(x–3) = 0
5) (x–2)(x –5) – (x–3)(x–4) = 0
6) (3x – 2)(4x + 3) = 2x(6x – 1)
7) 4x 2 – (2x + 1)(2x – 1) = 0
8) (4x–5)(x+3) = (2x – 3)(7+2x)
9) (x + 3)(x – 2) = (x + 1) 2
10) (x+7)(x–7) + x 2 – 2 = 2(x 2 +5)
11) (x–1) 2 + (x+3) 2 = 2(x– 2)(x+2)
12) (x – 5) 2 = (x + 3) 2 + 2
13) (3x + 2) 2 – (3x – 2) 2 = 5x + 38
