Tìm x, biết
1.\(\frac{2x+4}{7}=\frac{4x-2}{15}\)
2. \(\frac{12-7x}{-13}=\frac{4-3x}{-5}\)
3. \(\frac{11x-2}{7x+5}=\frac{11}{8}\)
Tìm x:\(a,\frac{6x-5}{-7}=\frac{5x-3}{-5}\\ b,\frac{12-7x}{-13}=\frac{4-3x}{-5}\\ c,\frac{2x+4}{7}=\frac{4x-2}{15}\)
a) \(\frac{6x-5}{-7}=\frac{5x-3}{-5}\)
=> -5(6x - 5) = -7(5x - 3)
=> -30x + 25 = -35x + 21
=> -30x + 25 + 35x - 21 = 0
=> (-30x + 35x) + (25 - 21) = 0
=> 5x + 4 = 0
=> 5x = -4
=> x = -4/5
b) \(\frac{12-7x}{-13}=\frac{4-3x}{-5}\)
=> -5(12 - 7x) = -13(4 - 3x)
=> -60 + 35x = -52 + 39x
=> -60 + 35x + 52 - 39x = 0
=> (-60 + 52) + (35x - 39x) = 0
=> -8 - 4x = 0
=> -8 = 4x
=> x = -2
c) \(\frac{2x+4}{7}=\frac{4x-2}{15}\)
=> 15(2x + 4) = 7(4x - 2)
=> 30x + 60 = 28x - 14
=> 30x + 60 - 28x + 14 = 0
=> 2x + 74 = 0
=> 2x = -74
=> x = -37
Tìm x biết:
\(\frac{2x+3}{7}=\frac{4x-1}{15}\)
\(\frac{11x-2}{7x+5}=\frac{11}{8}\)
Câu 3: Giải các phương trình sau bằng cách đưa về dạng ax+b=0
1. a, \(\frac{5x-2}{3}=\frac{5-3x}{2}\); b, \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
c, \(2\left(x+\frac{3}{5}\right)=5-\left(\frac{13}{5}+x\right)\); d, \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
e, \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\); f, 4 (0,5-1,5x)=\(\frac{5x-6}{3}\)
g, \(\frac{3x+2}{2}-\frac{3x+1}{6}=\frac{5}{3}+2x\); h, \(\frac{x+4}{5}.x+4=\frac{x}{3}-\frac{x-2}{2}\)
i, \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\); k, \(\frac{5x+2}{6}-\frac{8x-1}{3}=\frac{4x+2}{5}-5\)
m, \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\); n, \(\frac{1}{4}\left(x+3\right)=3-\frac{1}{2}\left(x+1\right).\frac{1}{3}\left(x+2\right)\)
p, \(\frac{x}{3}-\frac{2x+1}{6}=\frac{x}{6}-x\); q, \(\frac{2+x}{5}-0,5x=\frac{1-2x}{4}+0,25\)
r, \(\frac{3x-11}{11}-\frac{x}{3}=\frac{3x-5}{7}-\frac{5x-3}{9}\); s, \(\frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{6}-\frac{5\left(0,4-2x\right)}{6}\)
t, \(\frac{2x-8}{6}.\frac{3x+1}{4}=\frac{9x-2}{8}+\frac{3x-1}{12}\); u, \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{3}+\frac{2x-1}{12}\)
v, \(\frac{5x-1}{10}+\frac{2x+3}{6}=\frac{x-8}{15}-\frac{x}{30}\); w, \(\frac{2x-\frac{4-3x}{5}}{15}=\frac{7x\frac{x-3}{2}}{5}-x+1\)
Đây là những bài cơ bản mà bạn!
\(\frac{5x-2}{3}=\frac{5-3x}{2}\)
\(< =>\frac{\left(5x-2\right).2}{6}=\frac{\left(5-3x\right).3}{6}\)
\(< =>\left(5x-2\right).2=\left(5-3x\right).3\)
\(< =>10x-4=15-9x\)
\(< =>10x+9x=15+4\)
\(< =>19x=19< =>x=1\)
\(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
\(< =>\frac{\left(10x+3\right).3}{36}=\frac{36}{36}+\frac{\left(6+8x\right).4}{36}\)
\(< =>\left(10x+3\right).3=36+\left(6+8x\right).4\)
\(< =>30x+9=36+24+32x\)
\(< =>32x-30x=9-36-24\)
\(< =>2x=9-60=-51< =>x=-\frac{51}{2}\)
1, |x-5|=|2x-9|
2, ||2x-11|-x|=8
3, |4x-7|+\(\left|\frac{7-4x}{4x-7}\right|\)=9
4, \(\frac{\left|7x^2-9x+2\right|}{5x+4}\)=2-7x
5, \(\frac{x^2-8x+15}{\left|2x^2-9x-5\right|}\)=3x-9
tìm x biết
\(\frac{2x}{3}=\frac{7x-3}{15}\)
\(\frac{11x-2}{7x+5}=\frac{11}{8}\)
\(\frac{2x+3}{3x+1}=\frac{3}{4}\)
giup mk mk se tk
\(\frac{2x}{3}=\frac{7x-3}{15}\)
\(\Rightarrow30x=3\left(7x-3\right)\)
\(30x=21x-9\)
\(9x=-9\)
\(x=-1\)
\(\frac{11x-2}{7x+5}=\frac{11}{8}\)
\(\Rightarrow8\left(11x-2\right)=11\left(7x+5\right)\)
\(88x-16=77x+55\)
\(11x=71\)
\(x=\frac{71}{11}\)
\(\frac{2x+3}{3x+1}=\frac{3}{4}\)
\(\Rightarrow4\left(2x+3\right)=3\left(3x+1\right)\)
\(8x+12=9x+3\)
\(-x=-9\)
\(x=9\)
mk lộn đề câu a
bn giải lại giùm mk nha
\(\frac{2x+3}{6}=\frac{7x-3}{15}\)
giải các phương trình và hệ phương trình sau
1 , 4 ( x + 5 ) ( x + 6 ) ( x + 10 ) ( x + 12 ) = 3x2
2 , ( 2x - 1 ) ( 4x + 5 ) ( 8x + 3 ) ( 16x - 15 ) = 99x2
3 ,( x - 1 ) ( x - 2 ) ( x - 4 ) ( x - 8 ) =\(\frac{10}{9}\) x2
4, \(\frac{3x}{x^2-x+4}\) + \(\frac{x}{2x^2-6x+8}\) = 1
5 , \(\frac{3x}{x^2-4x+1}\) - \(\frac{2x}{x^2+x+1}\) = \(\frac{8}{3}\)
6, \(\frac{3x}{x^2-3x+1}\) + \(\frac{7x}{x^2+x+1}\) = -4
7, \(\frac{4x}{4x^2-8x+7}\) + \(\frac{3x}{4x^2-10x+7}\)= 1
8, \(\frac{2x}{x^2-3x+1}\) + \(\frac{7x}{x^2+x+1}\) = 6
9, \(\frac{x^2-10x+15}{x^2-6x+15}\) - \(\frac{4x}{x^2-12x+15}\)= 2
\(\frac{x-3}{3xy}\)+ \(\frac{5x+3}{3xy}\)
\(\frac{5x-7}{2x-3}+\frac{4-3x}{2x-3}\)
\(\frac{3x+5}{7x-1}-\frac{6-4x}{7x-1}\)
\(\frac{11x-7}{3-5x}-\frac{6x+4}{5x-3}\)
\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
\(\frac{1}{2x-10}+\frac{2x}{3x^2-15x}\)
1/ \(\frac{x-3}{3xy}\)+\(\frac{5x+3}{3xy}\)= \(\frac{6x}{3xy}\)=\(\frac{3}{y}\)
2/\(\frac{5x-7}{2x-3}\)+\(\frac{4-3x}{2x-3}\)=\(\frac{2x-3}{2x-3}\)=1
3/\(\frac{11x-7}{3-5x}\)-\(\frac{6x+4}{5x-3}\)=\(\frac{11x-7}{3-5x}\)+\(\frac{6x+4}{3-5x}\)=\(\frac{17x-3}{3-5x}\)
4/\(\frac{3}{2x+6}\)-\(\frac{x-6}{2x^2+6x}\)=\(\frac{3x}{x\left(2x+6\right)}\)-\(\frac{x-6}{x\left(2x+6\right)}\)=\(\frac{2x-6}{x\left(2x+6\right)}\)
5/\(\frac{1}{2x-10}\)+\(\frac{2x}{3x^2-15x}\)=\(\frac{1}{2\left(x-5\right)}\)+\(\frac{2x}{3x\left(x-5\right)}\)=\(\frac{3x}{6x \left(x-5\right)}\)+\(\frac{4x}{6x\left(x-5\right)}\)
=\(\frac{7x}{6x\left(x-5\right)}\)=\(\frac{7}{6\left(x-5\right)}\)
Giải các phương trình sau:
a,\(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
b,\(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
c,\(\frac{2\left(x+5\right)}{3}+\frac{x+12}{2}-\frac{5\left(x-2\right)}{6}=\frac{x}{3}+11\)
d,\(\frac{x-4}{5}+\frac{3x-2}{10}-x=\frac{2x-5}{3}-\frac{7x+2}{6}\)
e,\(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}\)
f,\(\frac{3x-1}{2}-\left(x-\frac{1}{4}\right)=\frac{4x-9}{8}\)
a, Ta có : \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
=> \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}=\frac{x+7}{15}\)
=> \(3\left(2x-1\right)-5\left(x-2\right)=x+7\)
=> \(6x-3-5x+10-x-7=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
b, Ta có : \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
=> \(\frac{3\left(x+3\right)}{6}-\frac{2\left(x-1\right)}{6}=\frac{x+5}{6}+\frac{6}{6}\)
=> \(3\left(x+3\right)-2\left(x-1\right)=x+5+6\)
=> \(3x+9-2x+2-x-5-6=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
c, Ta có : \(\frac{2\left(x+5\right)}{3}+\frac{x+12}{2}-\frac{5\left(x-2\right)}{6}=\frac{x}{3}+11\)
=> \(\frac{4\left(x+5\right)}{6}+\frac{3\left(x+12\right)}{6}-\frac{5\left(x-2\right)}{6}=\frac{2x}{6}+\frac{66}{6}\)
=> \(4\left(x+5\right)+3\left(x+12\right)-5\left(x-2\right)=2x+66\)
=> \(4x+20+3x+36-5x+10-2x-66=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
\(a,\frac{x-3}{2}+\frac{4x+1}{3}=\frac{2x-7}{6}\)
\(b,\frac{x-4}{5}+\frac{3x-2}{10}-x=\frac{2x-5}{3}-\frac{7x+5}{6}\)
\(c,\frac{2\left(x-3\right)}{4}+\frac{x-5}{3}=\frac{13x+4}{12}\)