Áp dụng tính chất của tỉ lệ thức

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

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 

dễ mà bạn

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,\(\frac{4}{7}-x=\frac{1}{3}\Rightarrow x=\frac{4}{7}-\frac{1}{3}\)

\(\Rightarrow x=\frac{12}{21}-\frac{7}{21}\Rightarrow x=\frac{5}{21}\)

b,\(\left|x-5\right|=7\Rightarrow\orbr{\begin{cases}x-5=7\\x-5=-7\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=12\\x=-2\end{cases}}\)

c,\(x-\frac{2}{3}=-\frac{6}{7}\Rightarrow x=-\frac{6}{7}+\frac{2}{3}\Rightarrow x=-\frac{4}{21}\)

d,\(\frac{x}{3}=\frac{8}{12}\Rightarrow x=8.3:12=2\)

e,\(\frac{2}{3}-1\frac{4}{15}x=-\frac{3}{5}\Rightarrow1\frac{4}{15}x=\frac{2}{3}+\frac{3}{5}\Rightarrow\frac{19}{15}x=\frac{19}{15}\Rightarrow x=1\)

f,\(-2^3+0,5x=1,5\Rightarrow-8+0,5x=1,5\Rightarrow0,5x=1,5+8\)

\(\Rightarrow0,5x=9,5\Rightarrow x=19\)

g,\(2^{x-1}=16\Rightarrow x-1=4\Rightarrow x=5\)

Tìm x:

a)-2³+0,5x=1,5

-8+0,5x = 1,5

0,5x = 1,5-(-8) = 9,5

x = 9,5:0,5 = 19

b)(−3)x81=−27

(-3)^x:81 =-27

(-3)^x = -27.81 = -2187

(-3)^x = (-3)⁷

=> x=7

c)112.x−4=0,5

1,5.x = 0,5+4 = 4,5

x = 4,5:1,5 = 3

d)123:x4=6:0,3

\(\frac{5}{3}\):\(\frac{x}{4}\) = 20

\(\frac{x}{4}\) = \(\frac{5}{3}\):20 = \(\frac{1}{12}\)

=> x:4 = \(\frac{1}{12}\)

x = \(\frac{1}{12}\).4 = \(\frac{1}{3}\)

\(a)\) \(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\)

\(\Leftrightarrow\)\(\left(2x+1\right)\left(0,5x+2\right)=\left(x+1\right)\left(x+3\right)\)

\(\Leftrightarrow\)\(2x\left(0,5x+2\right)+0,5x+2=x\left(x+1\right)+3\left(x+1\right)\)

\(\Leftrightarrow\)\(x^2+4x+0,5+2=x^2+x+3x+3\)

\(\Leftrightarrow\)\(x^2+4x-x^2-4x=3-0,5-2\)

\(\Leftrightarrow\)\(0=0,5\) ( vô lí :'< )

Vậy không có x thoả mãn đề bài 

áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{x+1}{2x+1}=\frac{0,5+2}{x+3}=\frac{\left(x+1\right)-2.\left(0,5x+2\right)}{\left(2x+1\right)-2.\left(x+3\right)}=\frac{x+1-x-4}{2x+1-2x-6}=\frac{-3}{-5}=\frac{3}{5}\)

suy ra:

\(\frac{x+1}{2x+1}=\frac{3}{5}\Rightarrow5.\left(x+1\right)=3.\left(2x+1\right)\)

=>5x+5=6x+3

5x-6x=3-5

-x=-2

x=2

0,5x-2/3x=7/12

=>(1/2-2/3).x=7/12

=>-1/6x=7/12

=>x=7/12:1/6

=>x=7/2

b, x:4 1/3=-2,5

=>x:13/3=-2,5

=>x=-5/2x13/3

=>x=-13

c,5,5x=13/15

=>11/2x=13/15

=>x=13/15:11/2

=>x=26/165

d,(3x/7+1):-4=-1/28

=>3x/7+1=-1/28x(-4)

=>3x/7+1=-1/7

=>3x/7=-1/7-1

=>3x/7=-8/7

=>3x=-8/7x7

=>3x=8

=>x=8:3

=>x=8/3

\(0,5x-\frac{2}{3}x=\frac{7}{12}\)

\(\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)

\(\left(\frac{1}{2}-\frac{2}{3}\right)x=\frac{7}{12}\)

\(\frac{-1}{6}x=\frac{7}{12}\)

\(x=-21\)

b,c,d tương tự

câu trả lời b kết qua bằng -65/6