\(\left(\frac{x}{^{x^2-25}}-\frac{x-5}{x^2+5x}\right)\):\(\left(\fra...">
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8 tháng 1 2020

1.

\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)

\(MC:12\)

Quy đồng :

\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)

\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)

\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)

\(\Leftrightarrow6x+9-3x=-4-9+16\)

\(\Leftrightarrow-7x=3\)

\(\Leftrightarrow x=\frac{-3}{7}\)

2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)

\(MC:20\)

Quy đồng :

\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)

\(\Leftrightarrow30x+15-20=15x-2\)

\(\Leftrightarrow15x=3\)

\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)

18 tháng 3 2019

a,  (2x+5)mũ 2=(x+2) mũ 2

=.> (2x+5) mũ 2-(x+2) mũ 2=0

=> (2x+5+x+2)x(2x+5-x-2)=0

=>(3x+7)x(x+3)=0

=>3x+7=0 hoặc x+3=0

3x+7=0=>x=-7/3

x+3=0 =>x=-3

vậy x=-7/3 hoặc x=-3

hok tot

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,...
Đọc tiếp

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, \(\frac{2x-\frac{4-3x}{5}}{15}=\frac{7x\frac{x-3}{2}}{5}-x+1\)

17

Đây là những bài cơ bản mà bạn!

29 tháng 3 2020

bạn ấy muốn thách xem bạn nào đủ kiên nhẫn đánh hết chỗ này

4 tháng 2 2020

a/ \(7x-5=13-5x\)

\(\Leftrightarrow7x+5x=13+5\)

\(\Leftrightarrow12x=18\)

\(\Leftrightarrow x=\frac{3}{2}\)

b/\(5\left(2x-3\right)-4\left(5x-7\right)=19-2\left(x+11\right)\)

\(\Leftrightarrow10x-15-20x+28=19-2x-22\)

\(\Leftrightarrow10x-20x+2x=19-22-28+15\)

\(\Leftrightarrow-8x=-16\)

\(\Leftrightarrow x=2\)

c/ \(\frac{2x-1}{3}-\frac{5x+2}{7}=x+13\)

\(\Leftrightarrow\frac{7\left(2x-1\right)-3\left(5x+2\right)-21\left(x+13\right)}{21}=0\)

\(\Leftrightarrow14x-7-15x-6-21x-273=0\)

\(\Leftrightarrow-22x-286=0\)

\(\Leftrightarrow x=-13\)

e/ \(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x+2\right)}\)

\(\Leftrightarrow\frac{2}{x+1}-\frac{1}{x-2}-\frac{3x-11}{\left(x+1\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{2\left(x-2\right)\left(x+2\right)-\left(x+1\right)\left(x+2\right)-\left(3x-11\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{2\left(x^2-4\right)-\left(x^2+3x+2\right)-\left(3x^2-17x+22\right)}{\left(x+1\right)\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow2x^2-8-x^2-3x-2-3x^2+17x-22=0\)

\(\Leftrightarrow-2x^2+14x-32=0\)

\(\Leftrightarrow x^2-7x+16=0\)

\(\Leftrightarrow x=\frac{-\left(-7\right)\pm\sqrt{\left(-7\right)^2-4.1.16}}{2}\)

\(\Leftrightarrow x=\frac{7\pm\sqrt{-15}}{2}\left(ktm\right)\)

\(\Leftrightarrow x\in\varnothing\)

4 tháng 2 2020

Bài 1:

a) \(7x-5=13-5x\)

\(\Leftrightarrow7x+5x=13+5\)

\(\Leftrightarrow12x=18\)

\(\Leftrightarrow x=18:12\)

\(\Leftrightarrow x=\frac{3}{2}.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{\frac{3}{2}\right\}.\)

b) \(5.\left(2x-3\right)-4.\left(5x-7\right)=19-2.\left(x+11\right)\)

\(\Leftrightarrow10x-15-\left(20x-28\right)=19-\left(2x+22\right)\)

\(\Leftrightarrow10x-15-20x+28=19-2x-22\)

\(\Leftrightarrow13-10x=-3-2x\)

\(\Leftrightarrow13+3=-2x+10x\)

\(\Leftrightarrow16=8x\)

\(\Leftrightarrow x=16:8\)

\(\Leftrightarrow x=2.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2\right\}.\)

c) \(\frac{2x-1}{3}-\frac{5x+2}{7}=x+13\)

\(\Leftrightarrow\frac{7.\left(2x-1\right)}{7.3}-\frac{3.\left(5x+2\right)}{3.7}=\frac{21.\left(x+13\right)}{21}\)

\(\Leftrightarrow\frac{14x-7}{21}-\frac{15x+6}{21}=\frac{21x+273}{21}\)

\(\Leftrightarrow14x-7-\left(15x+6\right)=21x+273\)

\(\Leftrightarrow14x-7-15x-6=21x+273\)

\(\Leftrightarrow-x-13=21x+273\)

\(\Leftrightarrow-x-21x=273+13\)

\(\Leftrightarrow-22x=286\)

\(\Leftrightarrow x=286:\left(-22\right)\)

\(\Leftrightarrow x=-13.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{-13\right\}.\)

Chúc bạn học tốt!

14 tháng 2 2020

Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\) 

 \(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)

\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\) 

 \(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)

\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\) 

\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)

\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)

\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)