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a, \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)

\(\frac{8x-3}{4}-\frac{6x-4}{4}=\frac{4x-2}{4}+\frac{x+3}{4}\)

\(8x-3-6x-4=4x-2+x+2\)

\(2x-7=5x\Leftrightarrow2x-5x+7=0\Leftrightarrow-3x=-7\Leftrightarrow x=\frac{7}{3}\)

b, \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)

\(\frac{3x+9}{6}-\frac{2x-2}{6}=\frac{x+5}{6}+\frac{6}{6}\)

\(3x+9-2x-2=x+5+6\)

\(x+7=x+11\Leftrightarrow-4\ne0\)

18 tháng 8 2020

1. \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)

\(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)

\(\Leftrightarrow35x-5+60x=96-6x\)

\(\Leftrightarrow95x-5=96-6x\)

\(\Leftrightarrow95x+6x=96+5\)

\(\Leftrightarrow101x=101\)

\(\Leftrightarrow x=1\)

2. \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\) 

\(\Leftrightarrow3\left(10x+3\right)=36+4\left(6+8x\right)\)

\(\Leftrightarrow30x+9=36+24+32x\)

\(\Leftrightarrow30x+9=32x+60\)

\(\Leftrightarrow30x-32x=60-9\)

\(\Leftrightarrow-2x=51\)

\(\Leftrightarrow x=-\frac{51}{2}\)

3. \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)

\(\Leftrightarrow8x-3-2\left(3x-2\right)=2\left(2x-1\right)+x+3\)

\(\Leftrightarrow8x-3-6x+4=4x-2+x+3\)

\(\Leftrightarrow2x+1=5x+1\)

\(\Leftrightarrow2x=5x\)

\(\Leftrightarrow x=0\)

19 tháng 8 2020

4) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)

=> \(\frac{9-3x}{8}+\frac{10-2x}{3}=\frac{1-x}{2}-\frac{2}{1}\)

=> \(\frac{3\left(9-3x\right)}{24}+\frac{8\left(10-2x\right)}{24}=\frac{12\left(1-x\right)}{24}-\frac{48}{24}\)

=> \(\frac{27-9x}{24}+\frac{80-16x}{24}=\frac{12-12x}{24}-\frac{48}{24}\)

=> \(\frac{27-9x+80-16x}{24}=\frac{12-12x-48}{24}\)

=> 27 - 9x + 80 - 16x = 12 - 12x - 48

=> 27 - 9x + 80 - 16x - 12 + 12x + 48 = 0

=> (27 + 80 - 12 + 48) + (-9x - 16x + 12x) = 0

=> 143 - 13x = 0

=> 13x = 143

=> x = 11

5) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)

=> \(\frac{2x-6}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)

=> \(\frac{3\left(2x-6\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)

=> \(\frac{6x-18}{21}+\frac{7x-35}{21}-\frac{13x+4}{21}=0\)

=> \(\frac{6x-18+7x-35-13x-4}{21}=0\)

=> 6x - 18 + 7x - 35 - 13x - 4 = 0

=> (6x + 7x - 13x) + (-18 - 35 - 4) = 0

=> -57 = 0(vô nghiệm)

6) \(\frac{6x+5}{2}-\left(2x+\frac{2x+1}{2}\right)=\frac{10x+3}{4}\)

=> \(\frac{6x+5}{2}-\frac{10x+3}{4}=2x+\frac{2x+1}{2}\)

=> \(\frac{2\left(6x+5\right)}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{2\left(2x+1\right)}{4}\)

=> \(\frac{12x+10}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{4x+2}{4}\)

=> \(\frac{12x+10-\left(10x+3\right)}{4}=\frac{8x+4x+2}{4}\)

=> \(\frac{12x+10-10x-3}{4}=\frac{12x+2}{4}\)

=> \(12x+10-10x-3=12x+2\)

=> \(2x+10-3=12x+2\)

=> 2x + 10 - 3 - 12x - 2 = 0

=> (2x - 12x) + (10 - 3 - 2) = 0

=> -10x + 5 = 0

=> -10x = -5

=> x = 1/2

7) \(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\)

=> \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{x+7}{15}=0\)

=> \(\frac{6x-3}{15}-\frac{5x-10}{15}-\frac{x+7}{15}=0\)

=> \(\frac{6x-3-\left(5x-10\right)-\left(x+7\right)}{15}=0\)

=> 6x - 3 - 5x + 10 - x - 7 = 0

=> (6x - 5x - x) + (-3 + 10 - 7) = 0

=> 0x + 0 = 0

=> 0x = 0

=> x tùy ý

Bài 8 tự làm nhé

22 tháng 3 2020

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 .

b) Ta có: \(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)

\(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)-\left(x+1\right)^3=0\)

\(x^3-6x^2+12x-8+9x^2-1-\left(x^3+3x^2+3x+1\right)=0\)

\(x^3+3x^2+12x-9-x^3-3x^2-3x-1=0\)

\(9x-10=0\)

hay 9x=10

\(x=\frac{10}{9}\)

Vậy: \(x=\frac{10}{9}\)

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

\(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{5}=0\)

\(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{3\left(x+7\right)}{15}=0\)

\(3\left(2x-1\right)-5\left(x-2\right)-3\left(x+7\right)=0\)

\(6x-3-5x+10-3x-21=0\)

\(-2x-14=0\)

\(-2x=14\)

hay x=-7

Vậy: x=-7

d) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}\)

\(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)

\(\frac{6\left(x-3\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)

\(6x-18+7x-35-13x-4=0\)

\(-21\ne0\)

Vậy: x∈∅

e) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)

\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}-\frac{\left(x+10\right)\left(x-2\right)}{3}=0\)

\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{3\left(x+4\right)\left(2-x\right)}{12}-\frac{4\left(x+10\right)\left(x-2\right)}{12}=0\)

\(x^2+14x+40-\left(3x+12\right)\left(2-x\right)-\left(4x+40\right)\left(x-2\right)=0\)

\(x^2+14x+40-\left(24-6x-3x^2\right)-\left(4x^2+32x-80\right)=0\)

\(x^2+14x+40-24+6x+3x^2-4x^2-32x+80=0\)

\(-12x+96=0\)

\(-12x=-96\)

hay x=8

Vậy: x=8

29 tháng 4 2020

1) \(\frac{x-3}{2}+\frac{4x+1}{3}=\frac{2x-7}{6}\)

<=> 3(x - 3) + 2(4x + 1) = 2x - 7

<=> 3x - 9 + 8x + 2 = 2x - 7

<=> 11x - 7 = 2x - 7

<=> 11x - 7 - 2x = -7

<=> 9x - 7 = -7

<=> 9x = -7 + 7

<=> 9x = 0

<=> x = 0

25 tháng 2 2020

giup minh voi cac bạn

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,...
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

26 tháng 3 2020

a)

\(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)

\(\Leftrightarrow\frac{4x-10x-15x}{12}=\frac{3x-60}{12}\)

\(\Leftrightarrow\frac{-10x-11}{12}=\frac{3x-60}{12}\)

\(\Leftrightarrow\frac{-10x-11-3x+60}{12}=0\)

\(\Leftrightarrow\frac{49-13x}{12}=0\)

\(\Rightarrow49-13x=0\)

\(\Rightarrow x=\frac{-49}{13}\)

26 tháng 3 2020

b)

\(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)

\(\Leftrightarrow\frac{8x-3-6x+4}{4}=\frac{4x-2+x+3}{4}\)

\(\Leftrightarrow\frac{2x+1}{4}=\frac{5x+1}{4}\)

\(\Leftrightarrow\frac{2x+1-5x-1}{4}=0\)

\(\Leftrightarrow\frac{-3x}{4}=0\)

\(\Rightarrow-3x=0\)

\(\Rightarrow x=0\)