\( 2\left( {x + \dfrac{3}{5}} \right) = 5 - \left( {\dfrac{{13}}{5} + x} \right)\\ \Leftrightarrow 2x + \dfrac{6}{5} = \dfrac{{12}}{5} - x\\ \Leftrightarrow 3x = \dfrac{6}{5}\\ \Leftrightarrow x = \dfrac{2}{5} \)
\( \dfrac{{10x + 3}}{{12}} = 1 + \dfrac{{6 + 8x}}{9}\\ \Leftrightarrow 3\left( {10x + 3} \right) = 36 + 4\left( {6 + 8x} \right)\\ \Leftrightarrow 30x + 9 = 36 + 24 + 32x\\ \Leftrightarrow - 2x = 51\\ \Leftrightarrow x = - \dfrac{{51}}{2} \)
bài 3,Giải PT
a,\(\frac{3}{2}.\left(x-\frac{5}{4}\right)-\frac{5}{8}=x\\ \)
b,\(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
c,\(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
a) \(\frac{5x-3}{50x^2-2}+\frac{5x-9}{12x-60x^2}+\frac{1}{12x}=\frac{8x-5}{80x^2+16x}\)
b) \(\frac{1}{x\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+9\right)}=\frac{1}{3}\left(27-\frac{1}{x+9}\right)\)
Bạn nào làm giúp mình với ạ!
#Cảm ơn nhiều! :)
1,Giải Pt
a,\(\frac{3x-7}{2}+\frac{x+1}{3}=-16\)
b,\(x-\frac{x+1}{3}=\frac{2x+1}{5}\)
c,\(\frac{7-3x}{12}+\frac{3}{4}=2\left(x-2\right)+\frac{5\left(5-2x\right)}{6}\)
e,\(\frac{3\left(x+3\right)}{4}+\frac{1}{2}=\frac{5x+9}{3}-\frac{7x-9}{4}\)
Câu 2: Giải phương trình:
a,\(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
b) 2x3 – 5x2 + 3x = 0
c) \(\frac{x}{2x-6}+\frac{x}{2x+2}-\frac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
Giaỉ PT
a) \(\left(\frac{x-1}{99}+x-99\right)+\left(\frac{x-3}{97}+\frac{x-7}{93}\right)+\left(\frac{x-5}{95}+\frac{x-95}{5}\right)=6\)
b) \(\left(4x-5\right)^2\left(2x-3\right)\left(x-1\right)=9\)
c) \(\frac{5}{x-8}+1=\frac{23}{x^2-5x-24}-\frac{2}{x+3}\)
Help !!
1) \(\frac{3x-1}{4}+\frac{2x-3}{3}=\frac{x-1}{2}\) Mc : 12
\(\Leftrightarrow\) \(\frac{3.\left(3x-1\right)}{12}+\frac{4.\left(2x-3\right)}{12}=\frac{6.\left(x-1\right)}{12}\)
\(\Leftrightarrow\) 9x - 3 + 8x - 12 = 6x - 6
\(\Leftrightarrow\) 9x + 8x - 6x = 3 + 12 - 6
\(\Leftrightarrow\) 11x = 9
\(\Leftrightarrow\) x = 0,8
Vậy S = {0,8}
2) \(\frac{x+1}{2}-\frac{x+3}{12}=3-\frac{5-3x}{3}\) Mc : 12
\(\Leftrightarrow\) \(\frac{6.\left(x+1\right)}{12}-\frac{x+3}{12}=\frac{12.3}{12}-\frac{4.\left(5-3x\right)}{12}\)
\(\Leftrightarrow\) 6x + 6 - x + 3 = 36 - 20 - 12x
\(\Leftrightarrow\) 6x - x + 12x = -6 - 3 + 36 - 20
\(\Leftrightarrow\) 17x = 7
\(\Leftrightarrow\) x = \(\frac{7}{17}\)
Vậy S = {\(\frac{7}{17}\)}
3) x - \(\frac{x+1}{3}\) = \(\frac{2x-1}{5}\) Mc : 15
\(\Leftrightarrow\) \(\frac{15.x}{15}-\frac{5.\left(x+1\right)}{15}=\frac{3.\left(2x-1\right)}{15}\)
\(\Leftrightarrow\) 15x - 5x - 5 = 6x - 3
\(\Leftrightarrow\) 15x - 5x - 6x = 5 - 3
\(\Leftrightarrow\) 4x = 2
\(\Leftrightarrow\) x = \(\frac{2}{4}=\frac{1}{2}\)
Vậy S = {\(\frac{1}{2}\)}
4) \(\frac{2x+7}{3}-\frac{x-2}{4}=-2\) Mc : 12
\(\Leftrightarrow\) \(\frac{4.\left(2x+7\right)}{12}-\frac{3.\left(x-2\right)}{12}=\frac{12.\left(-2\right)}{12}\)
\(\Leftrightarrow\) 8x + 28 -3x + 6 = -24
\(\Leftrightarrow\) 8x - 3x = -28 - 6 -24
\(\Leftrightarrow\) 5x = -58
\(\Leftrightarrow\) x = -11,6
Vậy S = {-11,6}
5) \(\frac{2x-3}{4}-\frac{4x-5}{3}=\frac{5-x}{6}\) Mc : 12
\(\Leftrightarrow\) \(\frac{3.\left(2x-3\right)}{12}-\frac{4.\left(4x-5\right)}{12}=\frac{2.\left(5-x\right)}{12}\)
\(\Leftrightarrow\) 6x - 9 - 16x + 20 = 10 - 2x
\(\Leftrightarrow\) 6x - 16x + 2x = 9 - 20 + 10
\(\Leftrightarrow\) -8x = -1
\(\Leftrightarrow\) x = \(\frac{1}{8}\)
Vậy S = {\(\frac{1}{8}\)}
6) \(\frac{12x+1}{4}=\frac{9x+1}{3}-\frac{3-5x}{12}\) Mc : 12
\(\Leftrightarrow\frac{3.\left(12x+1\right)}{12}=\frac{4.\left(9x+1\right)}{12}-\frac{3-5x}{12}\)
\(\Leftrightarrow\) 36x + 3 = 36x + 4 - 3 + 5x
\(\Leftrightarrow\) 36x - 36x - 5x = -3 + 4 - 3
\(\Leftrightarrow\) -5x = -2
\(\Leftrightarrow x=\frac{2}{5}\)
7) \(\frac{x+6}{4}\) - \(\frac{x-2}{6}-\frac{x+1}{3}=0\) Mc : 12
\(\Leftrightarrow\) \(\frac{3.\left(x+6\right)}{12}-\frac{2.\left(x-2\right)}{12}-\frac{4.\left(x+1\right)}{12}=0\)
\(\Leftrightarrow\) 3x + 18 - 2x + 4 - 4x - 4 = 0
\(\Leftrightarrow\) 3x - 2x - 4x = -18 - 4 + 4
\(\Leftrightarrow\) -3x = -18
\(\Leftrightarrow\) x = 6
Vậy S = {6}
8) x\(^2\) - x - 6 = 0
\(\Leftrightarrow\) x\(^2\) + 2x - 3x - 6 = 0
\(\Leftrightarrow\) x.(x + 2) - 3.(x + 2) = 0
\(\Leftrightarrow\) (x - 3).(x + 2) = 0
\(\Leftrightarrow\) x - 3 = 0 hoặc x + 2 = 0
\(\Leftrightarrow\) x = 3 hoặc x = -2
Vậy S = {3; -2}
GPT :
a, \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}..\right).503x=1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\)
b, \(\left(\frac{0,6+\frac{3}{7}-\frac{2}{11}}{1+\frac{5}{7}-\frac{5}{11}}+\frac{\frac{2}{3}-1,5+\frac{2}{9}}{\frac{5}{3}-3,75+\frac{5}{9}}\right)+93x=\left(\frac{3737}{4545}-\frac{954954}{975975}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)-7.\left(x-3\right)\)
Giải các phương trình sau
a) \(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
b) \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
c) \(\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2x-6}\)
d) \(\frac{3}{\left(x-1\right)\left(x-2\right)}+\frac{2}{\left(x-3\right)\left(x-1\right)}=\frac{1}{\left(x-2\right)\left(x-3\right)}\)
e) \(\frac{1}{x-2}+\frac{5}{x+1}=\frac{3}{2-x}\)
f) \(\frac{5x}{2x+2}+1=-\frac{6}{x+1}\)
g) \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4}{x^2-1}\)
h) \(\frac{3x}{x-2}-\frac{x}{x-5}=\frac{3x}{\left(x-2\right)\left(5-x\right)}\)
a, (x-1)3 - x(x-1)2 = 5(2-x) - 11(x+2)
b, (x-2)3 + (3x-1)(3x+1) = (x+1)3
c, \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{5}\)
d, \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}\)
e, \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
