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27 tháng 5 2020

\(\frac{11}{8}+\frac{13}{6}=\frac{85}{x}\)

\(\frac{85}{24}=\frac{85}{x}\)

\(x=24\)

\(2x-\frac{2}{11}=1\frac{1}{5}\)

\(2x-\frac{2}{11}=\frac{6}{5}\)

\(2x=\frac{76}{55}\)

\(x=\frac{38}{55}\)

27 tháng 5 2020

a. \(\frac{11}{8}\)\(\frac{13}{6}\)\(\frac{85}{x}\)

                    \(\frac{85}{24}\) =\(\frac{85}{x}\)

                             x   = \(\frac{24\cdot85}{85}\)

                             x   = 24

b. 2x - \(\frac{2}{11}\)=1\(\frac{1}{5}\)

     2x- \(\frac{2}{11}\)\(\frac{6}{5}\)

     2x             = \(\frac{6}{5}\)+\(\frac{2}{11}\)

       x             = \(\frac{76}{55}\): 2

       x             = \(\frac{38}{55}\)

Học tốt nha ~~~~~ ỌvỌ

18 tháng 3 2020

\( a)5\left( {x - 3} \right) - 4 = 2\left( {x - 1} \right) + 7\\ \Leftrightarrow 5x - 15 - 4 = 2x - 2 + 7\\ \Leftrightarrow 5x - 19 = 2x + 5\\ \Leftrightarrow 5x - 2x = 5 + 19\\ \Leftrightarrow 3x = 24\\ \Leftrightarrow x = 8\\ b)\dfrac{{8x - 3}}{4} - \dfrac{{3x - 2}}{2} = \dfrac{{2x - 1}}{2} + \dfrac{{x + 3}}{4}\\ \Leftrightarrow 8x - 3 - \left( {3x - 2} \right).2 = \left( {2x - 1} \right).2 + x + 3\\ \Leftrightarrow 8x - 3 - 6x + 4 = 4x - 2 + x + 3\\ \Leftrightarrow 2x + 1 = 5x + 1\\ \Leftrightarrow 2x - 5x = 0\\ \Leftrightarrow - 3x = 0\\ \Leftrightarrow x = 0 \)

18 tháng 3 2020

\( c)\dfrac{{2\left( {x + 5} \right)}}{3} + \dfrac{{x + 12}}{2} - \dfrac{{5\left( {x - 2} \right)}}{6} = \dfrac{x}{3} + 11\\ \Leftrightarrow 4\left( {x + 5} \right) + 3\left( {x + 12} \right) - \left[ {5\left( {x - 2} \right)} \right] = 2x + 66\\ \Leftrightarrow 4x + 20 + 3x + 36 - 5x + 10 = 2x + 66\\ \Leftrightarrow 2x + 66 = 2x + 66\\ \Leftrightarrow 0x = 0\left( {VSN} \right)\\ \Leftrightarrow x = 0 \)

\(d)\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}+\dfrac{x-4}{2000}+\dfrac{x-2}{2002}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}+\dfrac{x-1996}{8}+\dfrac{x-1994}{10}\\ \Leftrightarrow \dfrac{x-10}{1994}-1+\dfrac{x-8}{1996}-1+\dfrac{x-6}{1998}-1+\dfrac{x-4}{2000}-1+\dfrac{x-2}{2002}-1=\dfrac{x-2002}{2}-1+\dfrac{x-2000}{4}-1+\dfrac{x-1998}{6}-1+\dfrac{x-1996}{8}-1+\dfrac{x-1994}{10}-1\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}+\dfrac{x-2004}{8}+\dfrac{x-2004}{10}\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}-\dfrac{x-2004}{8}-\dfrac{x-2004}{10}=0\\ \Leftrightarrow \left(x-2004\right)\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}+\dfrac{1}{2000}+\dfrac{1}{2002}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}=0\right)\\ \Leftrightarrow x-2004=0\\ \Leftrightarrow x=2004\)

5 tháng 6 2019

#)Giải :

a) x + 2x + 3x + ... + 100x = - 213

=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213 

=> 100x + 5049 = - 213 

<=> 100x = - 5262

<=> x = - 52,62

5 tháng 6 2019

#)Giải :

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

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

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

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

\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)

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

1 tháng 4 2020

a) Đk: x \(\ne\)-2

Ta có: \(\frac{2}{x+2}-\frac{2x^2+16}{x^2+8}=\frac{5}{x^2-2x+4}\)

<=> \(\frac{2\left(x^2-2x+4\right)-\left(2x^2+16\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{5\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\)

<=> 2x2 - 4x + 8 - 2x2 - 16 = 5x + 10

<=> -4x - 8 = 5x + 10

<=> -4x - 5x = 10 + 8

<=> -9x = 18

<=> x = -2 (ktm)

=> pt vô nghiệm

b) Đk: x \(\ne\)2; x \(\ne\)-3

Ta có: \(\frac{1}{x-2}-\frac{6}{x+3}=\frac{5}{6-x^2-x}\)

<=> \(\frac{x+3}{\left(x-2\right)\left(x+3\right)}-\frac{6\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{5}{\left(x-2\right)\left(x+3\right)}\)

<=> x + 3 - 6x + 12 = -5

<=> -5x = -5 - 15

<=> -5x = -20

<=> x = 4 

vậy S = {4}

c) Đk: x \(\ne\)8; x \(\ne\)9; x \(\ne\)10; x \(\ne\)11

Ta có: \(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)

<=> \(\left(\frac{8}{x-8}+1\right)+\left(\frac{11}{x-11}+1\right)=\left(\frac{9}{x-9}+1\right)+\left(\frac{10}{x-10}+1\right)\)

<=> \(\frac{x}{x-8}+\frac{x}{x-11}-\frac{x}{x-9}-\frac{x}{x-10}=0\)

<=> \(x\left(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\right)=0\)

<=> x = 0 (vì \(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\ne0\)

Vậy S = {0}

3 tháng 8 2017

a Đ

b S

c S

d Đ

3 tháng 8 2017

a ) S 

b ) Đ

c ) S

d ) Đ

k cho mk nhé 

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!

2 tháng 8 2017

 mik ko chép lại đề, mik làm luôn: 

a)  x - \(\frac{31}{36}=\frac{-13}{38}\)

x = \(\frac{-13}{18}+\frac{31}{36}\)

\(x=\frac{5}{36}\)

b)\(2-x-\frac{3}{7}=\frac{9}{-21}\)

\(\frac{11}{7}-x=\frac{3}{7}\)

x = \(\frac{11}{7}-\frac{3}{7}\)

x = 8/7

c) x + 3/11 = 23/44

x = 23/44 - 3/11

x = 1/4

d) \(\frac{1}{12}-x=\frac{-11}{9}\)

x = \(\frac{1}{12}+\frac{11}{9}\)

x = 47/36

e) \(x-\frac{2}{3}=\frac{-17}{3}\)

x= -17/3 + 2/3

x = -5 

f) \(x-\frac{1}{2}=\frac{11}{4}.\frac{3}{11}\)

x - 1/2 = 3/4

x = 3/4 + 1/2 

x = 5/4

g) \(2x+\frac{3}{8}=\frac{-21}{32}.\frac{4}{7}\)

2x + 3/8 = -3 / 8

2x = -3/8 - 3/8 

2x = -9/8

x = -9/8.1/2 

x = -9/16

h) x - \(\frac{x}{3}=\frac{3}{57}.\frac{19}{12}\)

x  - \(\frac{x}{3}=\frac{1}{12}\)

x = \(\frac{1}{12}+\frac{x}{3}\)

x = \(\frac{1+4x}{12}\)

=> 12x = 1+4x

12x - 4x = 1

8x = 1

x = 1/8 

2 tháng 8 2017

Trả lời nhanh gọn lẹ nhé, mình k cho :)

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