\(3^x:3^2-12=69\)

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10 tháng 10 2019

<=>3x-2=69+12

<=>3x-2=81

<=>3x-2=34

<=>x-2=4

<=>x=6

21 tháng 3 2017

1:

a)10/20-15/20+16/20=-5/20+16/20

=11/20.

b)2/3+8/3:8/5=2/3+5/3

=7/3.

2:a)Ta có:

2x=1/4=3/4

2x=4/4=1

x=1:2

x=0,5

b)x:(2/12-1/12)=-3/8.

x:1/12=-3/8.

x=-3/8x1/12.

x=-1/32.

23 tháng 1 2020

a) x + 15 = 36 - 2x

x + 15 = 36 - (x + x )

15 =36 - ( x + x) - x

15 = 36 - x - x - x

15 = 36 - 3x

3x = 36 - 15

3x = 21

x = 21 : 3

=> x = 7

b) (x - 7) - (2x +5) = -14

x - 7 -( 2x + 5) = -14

x - (2x + 5) = -14 + 7 = -7

x - 2x - 5 = -7

x - 2x = -7 + 5 = -2

x - x + x = 2

x = 2 (-x + x cũng bằng chính nó)

=> x = 2

c) (x - 12) - 15 = (-7 + 20) - (18+x)

(x - 12) - 15 = 13 - (18 + x)

(x - 12) - 15 = 13 - 18 - x

(x - 12) - 15 = -5 - x

15 = (x - 12 ) - (-5 - x)

15 = x - 12 + 5 + x

15 = x + (-12) + 5 + x

15 = 2x + [(-12) + 5]

15 = 2x + -7

2x = -7 + 15

2x = 8

x = 8 : 2

=> x = 4

..................

16 tháng 7 2018

\(a)\dfrac{1}{3}x+\dfrac{2}{5}\left(x+1\right)=0\)

\(\Leftrightarrow\dfrac{1}{3}x+\dfrac{2}{5}x+\dfrac{2}{5}=0\)

\(\Leftrightarrow x\left(\dfrac{5}{15}+\dfrac{6}{15}\right)=\dfrac{-2}{5}\)

\(\Leftrightarrow x.\dfrac{11}{15}=\dfrac{-2}{5}\)

\(\Leftrightarrow x=\dfrac{-2}{5}.\dfrac{15}{11}\)

\(\Leftrightarrow x=\dfrac{-6}{11}\)

25 tháng 3 2019

(-6,17 +3+5/9-2-36/97)*(1/3-1/4-1/12)=(-6,17+3+5/9-2-36/97)*(4/12-3/12-1/12)=(-6,17+3+5/9-2-36/97)*0=0

2 tháng 5 2017

Bài 1:

a)

\(\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\ \Leftrightarrow x-1=24\\ x=24+1\\ x=25\)

b)

\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\ \dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\ \dfrac{3x}{7}+1=\dfrac{1}{2}\\ \dfrac{3x}{7}=\dfrac{1}{2}-1\\ \dfrac{3x}{7}=\dfrac{-1}{2}\\ 3x=\dfrac{-1}{2}\cdot7\\ 3x=\dfrac{-7}{2}\\ x=\dfrac{-7}{2}:3\\ x=\dfrac{-7}{6}\)

c)

\(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\ x+\dfrac{7}{12}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{12}\\ x=\dfrac{1}{4}\)

d)

\(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\ \dfrac{-1}{6}x=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{-1}{6}\\ x=\dfrac{-7}{2}\)

e)

\(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\ \dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\ \dfrac{277}{690}-x=\dfrac{7}{46}\\ x=\dfrac{277}{690}-\dfrac{7}{46}\\ x=\dfrac{86}{345}\)

f)

\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\ \left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\ x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\ x-\dfrac{1}{12}=\dfrac{7}{24}\\ x=\dfrac{7}{24}+\dfrac{1}{12}\\ x=\dfrac{3}{8}\)

g)

\(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\ \dfrac{13}{21}+x=\dfrac{2}{7}\\ x=\dfrac{2}{7}-\dfrac{13}{21}\\ x=\dfrac{-1}{3}\)

h)

\(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{29}{24}\\ x=\dfrac{29}{24}:\dfrac{1}{2}\\ x=\dfrac{29}{12}\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\)

i)

\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)