a) \(x\cdot25=5^3\)

\(x\cdot25=125\)

\(x=\frac{125}{25}\)

\(x=5\)

b) \(6^x:6^2=6^4\)

\(6^x=6^4\cdot6^2\)

\(6^x=6^6\)

\(x=6\)

c) \(3x=27\)

\(x=27:3\)

\(x=9\)

d) \(5^x+1=125\)

\(5^x=124\)

...

e) \(x\cdot36=6^8\)

\(x=6^8:6^2\)

\(x=6^6\)

\(x=46656\)

g) \(7^x+1:7^2=7^9\)

\(7^x=7^9-\frac{1}{7^2}\)

......

a, x =  125: 25 =5

b , 6 ^x =6^4 . 6^2

6^x =6^6

=> x =6

c, 3x = 27 

=> 3x = 3 .9

=> x = 9 ( vì 3 =3 )

d, 5^x +1 =125

5 ^ x+1 =5^3

=>x +1 =3

=>x=2

e, x. 36 =6 ^8

x . 6^2 = 6^8

x = 6^8 : 6^2

x= 6^6

g , 7^x+1 : 7^2 = 7^9

7^x+1 = 7^2+9

7^x +1=7^ 11

=>x +1 =11

=> x =10

Bài 1.

Bài 1:

a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)

b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)

c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)

d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)

hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)

e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)

hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)

dad

dái wa      

Giúp mình đi mình đang cần gấp

tìm x biết:

(3x-1) [- 1/2x+5]=0

1/4+1/3:(2x-1)=-5

[2x+3/5]² - 9/25=0

-5(x+1/5)-1/2(x-2/3)=3/2x - 5 /6

[x+1/2]x [2/3-2x]=0

17/2-|2x-3/4|=-7/4

2/3x-1/2x =5/12

(x+1/5)²+17/25=26/25

[x.44/7+3/7].11/5-3/7=-2

3[3x-1/2]+1/9=0

Toán lớp 6Tìm x

 Trả lời  Câu hỏi tương tự

Chưa có ai trả lời câu hỏi này,bạn hãy là người đâu tiên giúp nguyenvanhoang giải bài toán này !

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

a)\(\left(5x+1\right)^2=\frac{36}{49}\\ \left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\\ \Rightarrow\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=\frac{-6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{matrix}\right.\)

vậy...

2.

a) \(\left(5x+1\right)^2=\frac{36}{49}\)

⇒ \(5x+1=\pm\frac{6}{7}\)

⇒ \(\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=-\frac{6}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}5x=\frac{6}{7}-1=-\frac{1}{7}\\5x=\left(-\frac{6}{7}\right)-1=-\frac{13}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\left(-\frac{1}{7}\right):5\\x=\left(-\frac{13}{7}\right):5\end{matrix}\right.\)

⇒ \(\left[{}\begin{matrix}x=-\frac{1}{35}\\x=-\frac{13}{35}\end{matrix}\right.\)

Vậy \(x\in\left\{-\frac{1}{35};-\frac{13}{35}\right\}.\)

Chúc bạn học tốt!

a) x=7

b)x=2/3

c)x=-13

d)x=3

e)x=-1

\(x\) \(\times\) \(\dfrac{4}{5}\) = 15

\(x\)         = 15 : \(\dfrac{4}{5}\)

\(x\)         = \(\dfrac{75}{4}\) 

\(\dfrac{12}{25}\) \(\times\) \(x\) = \(\dfrac{4}{10}\)

          \(x\) = \(\dfrac{4}{10}\) : \(\dfrac{12}{25}\)

          \(x\) = \(\dfrac{5}{6}\) 

1, \(x^2\) - 9 = 0

 (\(x\) - 3)(\(x\) + 3) = 0

 \(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

 vậy \(x\) \(\in\) {-3; 3}

5, 4\(x^2\) - 36 = 0

    4.(\(x^2\) - 9) = 0

       \(x^2\) - 9 = 0

       (\(x\) - 3)(\(x\) + 3) = 0

        \(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\)

        \(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

Vậy \(x\) \(\in\) {-3; 3}

\(a,5x-16=40-1\)

<=>\(5x=40-1+16\)

<=>\(5x=55\)

<=>\(x=11\)

\(b,x-10=-25\)

<=>\(x=\left(-25\right)-10\)

<=>\(x=-35\)

\(c,-12+x=-30\)

<=>\(x=\left(-30\right)+12\)

<=>\(x=-18\)

a) 5x-16=40-1

5x-16=39

5x=39+16

5x=55

x=55:5

x=11

b)x-10=(-25)

x=(-25)+10

x=(-15)

c) -12+x= -30

           x= -30-(-12)

           x= -30+12

           x= -18

d) 2x +12 = -40+ 6

2x+12=(-34)

2x=(-34)-12

2x=(-46)

x=(-46):2

x=(-23)

e)125:(3x-13)=25

3x-13=125:25

3x-13=5

3x=5+13

3x=18

x=18:3

x=6

f)541+(218-x)=735

218-x=735-541

218-x=194

x=218-194

x=  24

g) 3(2x+1)-19=14

3(2x+1)=14+19

3(2x+1)=33

2x+1=33:3

2x+1=11

2x=11-1

2x=10

x=10:2

x=5

h)175-5(x+3)=85

5(x+3)=175-85

5(x+3)=90

x+3=90:5

x+3=18

x=18-3

x=15

i)8x+(-3)=39

8x=39-(-3)

8x=42

x=42:8

x=42/8

L)2x+4x=36+(-6)

6x=30

x=30:6

x=5