\(\frac{x-2}{x+3}-\frac{x+1}{3-x}=\frac{2x^2+6}{x^2-9}\)
\(\Leftrightarrow\frac{x-2}{x+3}+\frac{x+1}{x-3}=\frac{x^2+6}{x^2-9}\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)+\left(x+1\right)\left(x+3\right)=x^2+6\)
\(\Leftrightarrow x^2+3x-3x-9+x^2+3x+3=2x^2+6^{ }\)
\(2x^2+3x-6-2x^2-6=0\)
\(\Leftrightarrow3x-12=0\)
\(\Leftrightarrow3x=12\)
\(\Leftrightarrow x=4\)
Vậy............
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
6) (-(x-3))/2-2=5(x+2)/4
7)2(2x+1)/5-(6+x)/3=(5-4x)/15
8) (7-3x)/2-(5+x)/5=1
9)(x-1)/2+3(x+1)/8=(11-5x)/3
10)(3+5x)/5-3=(9x-3)/4
1) x2 - x - (3x - 3) = 02) x(x - 6) - 7x + 42 = 0
3) x3 - 3x2 + 3x - 1 = 0
4) (2x - 5)2 - (x + 2) 2 = 0
5) x(2x - 9) = 3x(x - 5)
giải phương trình
[x - 6] [ 2x -5 ] [ 3x + 9 ] = 0
2x [ x - 3 ] + 5 [ x - 3 ] = 0
[ 4 - x ]2 - [ x - 2 ] [ 3 - 2x ] = 0
3x [x2 + 1 ] [ x - 2 ] =0
[ 3x + 5 ] [ x2 + x + 1 ] = 0
giải phương trình
1)\(2\left(x-3\right)+1=2\left(x+1\right)-9\)
2)\(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
3) \(\left(x-1\right)^2+\left(x+2\right)\left(x-2\right)=\left(2x+1\right)\left(x-3\right)\)
4)\(\left(x+5\right)\left(x-1\right)-\left(x+1\right)\left(x+2\right)=1\)
5) \(\dfrac{6x-1}{15}-\dfrac{x}{5}=\dfrac{2x}{3}\)
6)\(\dfrac{5\left(x-2\right)}{2}-\dfrac{x+5}{3}=1-\dfrac{4\left(x-3\right)}{5}\)
giải phương trình sau :
a/ x + 1 = 3x - 1
b/x - 4 +2x = 4 - x
c/\(\frac{x+3}{4}+1=\frac{1-x}{6}\)
\(\frac{2x+3}{4}=1+\frac{x-7}{5}\)
\(^{\left(x+1\right)^2-\left(x-1\right)^2=2x\left(x+1\right)-6}\)
\(\left(1+x\right)^3-\left(x-2\right)^2=9\left(x-1\right)^2\)
A, x/2x+6 - x/2x+2= 3x+2/(x+1)(x+3)
B, 5/x+7+8/2x+14= 2/3
C, x-1/x-1/x+1= 2x-1/x^2+x
1) Giải các pt sau:
a) \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)
b) \(\frac{3x-2}{6}-5=\frac{3-2\left(x+7\right)}{4}\)
c) \(\frac{x+8}{6}-\frac{2x-5}{5}=\frac{x-1}{3}-x+7\)
d) \(\frac{7x}{8}-5\left(x-9\right)=\frac{2x+1,5}{6}\)
e) \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\)
f) \(\frac{x+1}{3}+\frac{3\left(2x+1\right)}{4}=\frac{2x+3\left(x+1\right)}{6}+\frac{7+12x}{12}\)
a (2x-1)^3+6(3x-1)^2=2(x+1)^3+6(x+2)^3
b (x-2)^2+(3-2x)^2-(4x-4)(x-5)=(x+3)^2
c x-3+2(x-3)-1/3=3-x/4
ai giúp mik đc câu nào cũng đc ạ
Giải phương trình
1) 16-8x=0
2) 7x+14=0
3) 5-2x=0
4) 3x-5=7
5) 8-3x=6
6) 8=11x+6
7)-9+2x=0
8) 7x+2=0
9) 5x-6=6+2x
10) 10+2x=3x-7
11) 5x-3=16-8x
12)-7-5x=8+9x
13) 18-5x=7+3x
14) 9-7x=-4x+3
15) 11-11x=21-5x
16) 2(-7+3x)=5-(x+2)
17) 5(8+3x)+2(3x-8)=0
18) 3(2x-1)-3x+1=0
19)-4(x-3)=6x+(x-3)
20)-5-(x+3)=2-5x
