a, \(\dfrac{6}{x-3}=\dfrac{9}{2x-7}\)

=> 6(2x-7) = 9(x-3)

=> 12x - 42 = 9x - 27

=> 12x - 9x = -27 + 42

=> 3x = 15 

=> x = 5

Vậy x = 5

b, \(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)

=> -7(x + 27) = 6(x + 1)

=> -7x - 189 = 6x + 6

=> -7x - 6x = 6 + 189

=> -13x = 195

=> x = -15

Vậy x = -15

a) Ta có: \(\dfrac{6}{x-3}=\dfrac{9}{2x-7}\)

\(\Leftrightarrow6\left(2x-7\right)=9\left(x-3\right)\)

\(\Leftrightarrow12x-42=9x-27\)

\(\Leftrightarrow12x-9x=-27+42\)

\(\Leftrightarrow3x=15\)

hay x=5

Vậy: x=5

b) Ta có: \(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)

\(\Leftrightarrow6\left(x+1\right)=-7\left(x+27\right)\)

\(\Leftrightarrow6x+6=-7x+189\)

\(\Leftrightarrow6x+7x=189-6\)

\(\Leftrightarrow13x=183\)

hay \(x=\dfrac{183}{13}\)

Vậy: \(x=\dfrac{183}{13}\)

a: =>2x-x=-5/2-1/3

=>x=-17/6

b: =>4(x-2)²=36

=>(x-2)²=9

=>x-2=3 hoặc x-2=-3

hay x=5 hoặc x=-1

c: =>2x+1/2=5/6

=>2x=1/3

hay x=1/6

a: =>2x-x=-5/2-1/3

=>x=-17/6

b: =>4(x-2)²=36

=>(x-2)²=9

=>x-2=3 hoặc x-2=-3

hay x=5 hoặc x=-1

c: =>2x+1/2=5/6

=>2x=1/3

hay x=1/6

Làm 1 câu thôi nhé lười lắm

1)

<=> 15 - x + 7 = -21

<=>     -x         = -21 - 15 - 7

<=>     -x         = -43

<=>      x         = 43

Các câu sau phá ngoặc đổi dấu, chuyển vế đổi dấu thôi kh có gì đâu

Bạn nhờ mình giúp cho

Câu cuối:

<=> -13 - 6 + x + 1 = 24

<=>               x        = 24 + 13 + 6 - 1

<=>               x        = 42

1: Ta có: 7x+6(3-x)=27-20+73

\(\Leftrightarrow7x+18-6x=80\)

\(\Leftrightarrow x=80-18=62\)

Vậy: x=62

2: Ta có: \(6x-5\left(x-7\right)=\left(27-514\right)-486-73\)

\(\Leftrightarrow6x-5x+35=27-514-486-73\)

\(\Leftrightarrow x+35=-1046\)

\(\Leftrightarrow x=-1081\)

Vậy: x=-1081

Bài 1:

a) 19/27+9/7+6/27+5/7+2/7

= (19/27+6/27) + (9/7+5/7+2/7)

=1 + 16/7

=23/7

b) 2/7 + 27/48 + 7/16 +45/63

= (2/7 + 45/63) + (27/48 +7/16)

= (2/7 + 5/7) + ( 9/16 + 7/16)

= 1 + 1

=2

c) Tính bằng cách bình thường

Bài 2:

6/11 + x = 1

           x = 1-6/11

           x = 5/11

Bài làm

a) \(\frac{19}{27}+\frac{9}{7}+\frac{6}{27}+\frac{5}{7}+\frac{2}{7}\)

\(=\frac{27}{27}+\frac{16}{7}\)

\(=1+\frac{16}{7}\)

\(=\frac{7}{7}+\frac{16}{7}\)

\(=\frac{23}{7}\)

b) \(\frac{2}{7}+\frac{27}{48}+\frac{7}{16}+\frac{45}{63}\)

\(=\left(\frac{18}{63}+\frac{45}{63}\right)+\left(\frac{27}{48}+\frac{21}{48}\right)\)

\(=\frac{63}{63}+\frac{48}{48}\)

\(=1+1\)

\(=2\)

c) \(\frac{17}{23}+\left(\frac{4}{7}-\frac{6}{13}\right)\)

\(=\frac{17}{23}+\left(\frac{52}{91}-\frac{42}{91}\right)\)

\(=\frac{17}{23}+\frac{10}{91}\)

\(=\frac{1547}{2093}+\frac{230}{2093}\)

\(=\frac{1770}{2093}\)

Bài 2

\(\frac{6}{11}+x=1\)

\(x=1-\frac{6}{11}\)

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

Vậy \(x=\frac{5}{11}\)

Cảm ơn 2 bn nhé

j mà hjhj thế nguyenthuy

j mà j mà hjhj thế nguyenthuy thế trieu dang

\(\frac{3^x}{27}=27\Rightarrow\frac{3^x}{27}=\frac{729}{27}=\frac{3^6}{27}\)

\(\Rightarrow3^x=3^6\Rightarrow x=6\)

Bui Bao Ngoc oi! phan a,la (x+1)/10 hay x+(1/10)?

cac phan khac nua!

Lời giải:

a. $121-3(x-5)=6$

$3(x-5)=121-6=115$

$x-5=115:3=\frac{115}{3}$

$x=\frac{115}{3}+5=\frac{130}{3}$

b.

$2x-138=2^3.3^2=72$

$2x=72+138=210$

$x=210:2=105$

c.

$x-3\vdots 7$

$\Rightarrow x-3\in\left\{0;7;14;21;28;35;42;49; 56;...\right\}$

Mà $10< x< 50$ nên $x\in\left\{14;21;28;35;42;49\right\}$

d.

$27\vdots x+1$

$\Rightarrow x+1\in\left\{\pm 1; \pm 3; \pm 9; \pm 27\right\}$

$\Rightarrow x\in\left\{0; -2; -4; 2; 8; -10; 26; -28\right\}$

a ) 121-3.(x - 5 ) = 6

3.(x-5) = 121 -6

3. (x-5)=115

x-5  = 115:3

x-5=35

x=35+5

x = 40

b) 2x - 138 = 2'3. 3'2

2x -138=8.9

2x-138=72

2x=72+138

2x=210

x=210:2

x=105

c) theo bài ra : x-3 ∈ B(7)

ta có B(7)=(0,7,14,21,28,35,49,56,...)

=) x-3 ∈ ( 0,7,14,21,28,35,49,56,...)

=) x ∈( 3 , 10,17,24,31,38,42,58,..)

mà 10 <x<50 nên x ∈ ( 17 , 24 ,31,38,42 )

vậy x ∈(17,24,31,38,42)

Lời giải:

a. $121-3(x-5)=6$

$3(x-5)=121-6=115$

x-5=115:3=\frac{115}{3}x−5=115:3=3115​

x=\frac{115}{3}+5=\frac{130}{3}x=3115​+5=3130​

b.

2x-138=2^3.3^2=722x−138=23.32=72

$2x=72+138=210$

$x=210:2=105$

c.

$x-3⋮7$

$\Rightarrow x-3\in \left\{0;7;14;21;28;35;42;49;56;...\right\}$

Mà $10<x<50$ nên $x\in \left\{14;21;28;35;42;49\right\}$

d.

$27⋮x+1$

$\Rightarrow x+1\in \left\{\pm 1;\pm 3;\pm 9;\pm 27\right\}$

$\Rightarrow x\in \left\{0;-2;-4;2;8;-10;26;-28\right\}$

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