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a. \(0,5x-\frac{1}{3}x=\frac{1}{3}\)

 \(\left(0,5-\frac{1}{3}\right)x=\frac{1}{3}\)

                   \(\frac{1}{6}x=\frac{1}{3}\)

                         \(x=\frac{1}{3}:\frac{1}{6}\)

                        \(x=2\)

b. \(x:3\frac{1}{2}=-1,5\)

           \(x=-1,5\times3\frac{1}{2}\)

           \(x=-5,25\)

c. \(1,5x=\frac{11}{5}\)

          \(x=\frac{22}{15}\)

28 tháng 3 2017

A. 0,5x - 1/3x = 1/3

    (0,5 - 1/3)x = 1/3

              1/6x = 1/3

                  x = 2

B. x : 3 1/2 = -1,5

              x = -1,5 x 3 1/2

              x = -5,25

C. 1,5x = 11/5

        x = 22/15

28 tháng 3 2020

a, x( x - 3) - ( x + 2)( x - 1) = 3

<=> x2 - 3x - x2 + x - 2x + 2 = 3

<=>  x2 - 3x - x2 + x - 2x = 3 - 2

<=> -4x = 1

<=> x =\(-\frac{1}{4}\)

Vậy_

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

\(\Leftrightarrow\frac{30x}{30}-\frac{\left(x-1\right)6}{30}+\frac{\left(x+2\right)5}{30}=\frac{120}{30}+\frac{\left(x+1\right)10}{30}\)

=> 30x - 6x + 6 + 5x + 10 = 120 + 10x + 10

<=> 30x - 6x + 5x - 10x = 120 + 10 - 6 - 10

<=> 19x = 114

<=> x = 6

Vậy _

6 tháng 7 2018

\(\hept{\begin{cases}\text{|}0,5x\text{|}=0,5x\\\sqrt{\left(0,5x\right)^2}=0,5x\\\left(0,5x\right)^2=\left(0,5x\right)^2\end{cases}}\)

2, tương tự

\(\hept{\begin{cases}\text{|}-\frac{2}{3}x\text{|}=\frac{2}{3}x\\\sqrt{\left(-\frac{2}{3}x\right)^2}=\frac{2}{3}x\\\left(-\frac{2}{3}x\right)^2=\left(\frac{2}{3}x\right)^2\end{cases}}\)

4, tương tự 

2 tháng 5 2015

a.x=3

b.x=-21/2

c.-7/2

d.112/81

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

HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023

a)

\(\begin{array}{l}x:{\left( {\frac{{ - 1}}{2}} \right)^3} =  - \frac{1}{2}\\x =  - \frac{1}{2}.{\left( {\frac{{ - 1}}{2}} \right)^3}\\x = {\left( {\frac{{ - 1}}{2}} \right)^4}\\x = \frac{1}{{16}}\end{array}\)              

Vậy \(x = \frac{1}{{16}}\).

 b)

\(\begin{array}{l}x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9}\\x = {\left( {\frac{3}{5}} \right)^9}:{\left( {\frac{3}{5}} \right)^7}\\x = {\left( {\frac{3}{5}} \right)^2}\\x = \frac{9}{{25}}\end{array}\)

Vậy \(x = \frac{9}{{25}}\).

c)

\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^{11}}:{\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^2}\\x = \frac{4}{9}.\end{array}\)         

Vậy \(x = \frac{4}{9}\).

d)

\(\begin{array}{l}x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)

Vậy \(x = \frac{1}{{16}}\).

14 tháng 12 2017

dễ mà bạn

14 tháng 12 2017

a) x=4/7 - 1/3=19/21

b) /x-5/=7 -->x-5=7 hoặc x-5=-7

--> x=12 hoặc x= -2

a) Ta có: \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)

\(\Leftrightarrow\frac{7x}{8}-5x+45-\frac{20x+1,5}{6}=0\)

\(\Leftrightarrow\frac{21x}{24}-\frac{120x}{24}+\frac{1080}{24}-\frac{4\left(20x+1,5\right)}{24}=0\)

\(\Leftrightarrow-99x+1080-4\left(20x+1,5\right)=0\)

\(\Leftrightarrow-99x+1080-80x-6=0\)

\(\Leftrightarrow1074-179x=0\)

\(\Leftrightarrow179x=1074\)

hay x=6

Vậy: x=6

b) Ta có: \(4\left(0,5-1,5x\right)=-\frac{5x-6}{3}\)

\(\Leftrightarrow2-6x=\frac{6-5x}{3}\)

\(\Leftrightarrow\frac{3\left(2-6x\right)}{3}-\frac{6-5x}{3}=0\)

\(\Leftrightarrow6-18x-6+5x=0\)

\(\Leftrightarrow-13x=0\)

mà -13≠0

nên x=0

Vậy: x=0

c) Ta có: \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)

\(\Leftrightarrow\frac{6\left(x+4\right)}{30}+\frac{30\left(-x+4\right)}{30}-\frac{10x}{30}+\frac{15\left(x-2\right)}{30}=0\)

\(\Leftrightarrow6\left(x+4\right)+30\left(4-x\right)-10x+15\left(x-2\right)=0\)

\(\Leftrightarrow6x+24+120-30x-10x+15x-30=0\)

\(\Leftrightarrow-19x+114=0\)

\(\Leftrightarrow-19x=-114\)

hay x=6

Vậy: x=6

d) Ta có: \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)

\(\Leftrightarrow\frac{21\left(4x+3\right)}{105}-\frac{15\left(6x-2\right)}{105}-\frac{35\left(5x+4\right)}{105}-\frac{315}{105}=0\)

\(\Leftrightarrow84x+63-90x+30-175x-140-315=0\)

\(\Leftrightarrow-181x-362=0\)

\(\Leftrightarrow-181x=362\)

hay x=-2

Vậy: x=-2

e) Ta có: \(\frac{1}{4}\left(x+3\right)=3-\frac{1}{2}\left(x+1\right)-\frac{1}{3}\left(x+2\right)\)

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

\(\Leftrightarrow\frac{3\left(x+3\right)}{12}-\frac{36}{12}+\frac{6\left(x+1\right)}{12}+\frac{4\left(x+2\right)}{12}=0\)

\(\Leftrightarrow3x+9-36+6x+6+4x+8=0\)

\(\Leftrightarrow13x-13=0\)

\(\Leftrightarrow13x=13\)

hay x=1

Vậy: x=1