Bài 2: Phương trình bậc nhất một ẩn và cách giải
Các câu hỏi tương tự
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) (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) Giải các pt sau:
a) frac{x-3}{5}6-frac{1-2x}{3}
b) frac{3x-2}{6}-5frac{3-2left(x+7right)}{4}
c) frac{x+8}{6}-frac{2x-5}{5}frac{x-1}{3}-x+7
d) frac{7x}{8}-5left(x-9right)frac{2x+1,5}{6}
e) frac{5left(x-1right)+2}{6}-frac{7x-1}{4}frac{2left(2x+1right)}{7}-5
f) frac{x+1}{3}+frac{3left(2x+1right)}{4}frac{2x+3left(x+1right)}{6}+frac{7+12x}{12}
Đọc tiếp
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
x-2/x+3-x+1/3-x=2x^2+6/x^2-9
giải phương trình 1)dfrac{1-6x}{x-2}+dfrac{9x+4}{x+2}dfrac{xleft(3x-2right)+1}{x^2-4}2) dfrac{3x+2}{3x-2}-dfrac{6}{2+3x}dfrac{9x^2}{9x^2-4}3) dfrac{x+5}{3x-6}-dfrac{1}{2}dfrac{2x-3}{2x-4}4) dfrac{x-1}{x}+dfrac{1}{x+1}dfrac{2x-1}{2x^2+2}5) dfrac{2}{x+1}+dfrac{3x+1}{x+1}dfrac{1}{left(x+1right)left(x-2right)}
Đọc tiếp
giải phương trình 1)\(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)2) \(\dfrac{3x+2}{3x-2}-\dfrac{6}{2+3x}=\dfrac{9x^2}{9x^2-4}\)3) \(\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4}\)4) \(\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x-1}{2x^2+2}\)5) \(\dfrac{2}{x+1}+\dfrac{3x+1}{x+1}=\dfrac{1}{\left(x+1\right)\left(x-2\right)}\)
Giải phương trình:a)
\(\dfrac{1}{x-1}+\dfrac{1}{x^2+x+1}=\dfrac{1}{x^3-1}\)
b)
\(\dfrac{1}{x^2-2x+2}+\dfrac{2}{x^2-2x+3}=\dfrac{6}{x^2-2x+4}\)
giải phương trình :
a) \(\dfrac{1}{x-1}+\dfrac{1}{x^2+x+1}=\dfrac{1}{x^3-1}\)
b) \(\dfrac{1}{x^2-2x+2}+\dfrac{2}{x^2-2x+3}=\dfrac{6}{x^2-2x+4}\)
a. 1+2x-5/6 + 3- x/4
b. x+1/2 - x-2/3
c. 2x-1/3 +x =x+4
d. x+5/4 -2x-3/3 -6x-1/8 + 2x-1/12
Mọi người giúp em với đề này khó quá
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)