Cần gấp ^^

\(\Leftrightarrow\left(\dfrac{x-11}{111}+1\right)+\left(\dfrac{x-12}{112}+1\right)=\left(\dfrac{x-23}{123}+1\right)+\left(\dfrac{x-24}{124}+1\right)\)

=>x+100=0

=>x=-100

a: =>\(\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+8}{92}+1\right)+\left(\dfrac{x+3}{97}+1\right)=0\)

=>x+100=0

=>x=-100

b: =>(x-11/111+1)+(x-12/112+1)=(x-23/123+1)+(x-24/124+1)

=>x+100=0

=>x=-100

a) 12 + ( 5 + x ) = 20 

⇔ 5 + x = 8

⇔ x = 3

Vậy x = 3

b) 120 + ( 50 + x ) = 180

⇔ 50 + x = 60

⇔ x = 10

Vậy x = 10

c) 175 + (  30 - x ) = 200

⇔ 30 - x = 25

⇔ x = 5

Vậy x = 5

d) 130 - ( 100 + x ) = 25

⇔ 100 + x = 105

⇔ x = 5

Vậy x = 5

e) 145 - ( 125 + x ) = 12

⇔ 125 + x = 133

⇔ x = 8

Vậy x = 8

j) 5 ( x + 12 ) + 22 = 92

⇔ 5x + 60 + 22 = 92

⇔ 5x + 82 = 92

⇔ 5x = 10

⇔ x = 2

Vậy x = 2

g) 3 ( x + 23 ) + 6 = 96

⇔ 3x + 69 + 6 = 96

⇔ 3x + 75 = 96

⇔ 3x = 21

⇔ x = 7

Vậy x = 7

Tìm x : 

a) 12 + ( 5 + x ) = 20

5 + x =20 - 12

5 +  x = 8

x = 3 

b) 120 + ( 50 + x ) = 180

50 + x = 180-120

50 + x = 60

x = 60 - 50

x = 10

c) 175 + ( 30 - x ) = 200

30 -x = 200 - 175

30 - x = 25

x = 30-25

x=5

d) 130- ( 100 +x ) =25

100 + x =130 - 25

100+x= 105

x=105-100

x=5

e) 145 - (125+x) = 12

125+x = 145-12

125+x=133

x= 133-125

x=8

j) 5 (x+12) +22 = 92 

5 ( x+12) = 92 -22

5 (x+12)=70

x+12=70 : 5

x + 12 = 14

x = 14-12

x=2

g) 2 (x+23) + 6 =96

2 (x+23) = 96 - 6

2 (x+23)= 90

x+23=90:2

x+23=45

x= 45-23

x=22

CHÚC BẠN HỌC TỐT

a) 12 + ( 5 + x ) = 20

=> 5 + x = 20 - 12

=> 5 + x = 8

=> x = 8 - 5

tìm x :

a) 12 + ( 5 + x ) = 20 

b) 120 + ( 50 + x ) = 180

c) 175 + (  30 - x ) = 200

d) 130 - ( 100 + x ) = 25

e) 145 - ( 125 + x ) = 12

j) 5 ( x + 12 ) + 22 = 92

g) 3 ( x + 23 ) + 6 = 96

giúp mk nha mn mk cần gấp

=> x = 3

b) 120 + ( 50 + x ) = 180

=> 50 + x = 180 - 120

=> 50 + x = 60

=> x = 60 - 50

=> x = 10

c) 175 + ( 30 - x ) = 200

=> 30 - x = 200 - 175

=> 30 - x = 25

=> x = 30 - 25

=> x = 5

d) 130 - ( 100 + x ) = 25

=> 100 + x  = 130 - 25

=> 100 + x = 105

=> x = 105 - 100 

=> x = 5

a) \(\Leftrightarrow\frac{x+2}{98}+\frac{x+4}{96}+2=\frac{x+6}{94}+\frac{x+8}{92}+2\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}+\frac{1}{94}+\frac{1}{92}\right)=0\)

\(\Leftrightarrow x+100=0\Leftrightarrow x=-100\)

b)\(\Leftrightarrow\frac{x-12}{77}+\frac{x-11}{78}-2=\frac{x-74}{15}+\frac{x-73}{16}-2\)

\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}-\frac{x-89}{15}-\frac{x-89}{16}=0\)

\(\Leftrightarrow\left(x-89\right)\left(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right)=0\)

\(\Leftrightarrow x-89=0\Leftrightarrow x=89\)

Lời giải:

a)

PT \(\Leftrightarrow \frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{92}+1\)

\(\Leftrightarrow \frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)

\(\Leftrightarrow (x+100)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)

Dễ thấy \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}<0\) nên $x+100=0$

$\Rightarrow x=-100$

b)

PT \(\Leftrightarrow \frac{x-12}{77}-1+\frac{x-11}{78}-1=\frac{x-74}{15}-1+\frac{x-73}{16}-1\)

\(\Leftrightarrow \frac{x-89}{77}+\frac{x-89}{78}=\frac{x-89}{15}+\frac{x-89}{16}\)

\(\Leftrightarrow (x-89)\left(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right)=0\)

Dễ thấy \(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}< 0\)

\(\Rightarrow x-89=0\Rightarrow x=89\)

Ta có : \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+2}{98}\)

\(\Rightarrow\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+2}{98}+1\)

\(\Rightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)

=> \(\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}\right)=\left(x+100\right).\left(\frac{1}{94}+\frac{1}{92}\right)\)

=> \(\left(x+100\right).\left(\frac{1}{98}+\frac{1}{96}\right)-\left(x+100\right)\left(\frac{1}{94}+\frac{1}{92}\right)=0\)

=> \(\left(x+100\right).\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\Rightarrow x+100=0\left(\text{vì }\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\right)\)

=> x = - 100 

Vậy x = - 100

\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)

\(\Leftrightarrow\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{92}+1\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)

Vì \(\frac{1}{92}>\frac{1}{94}>\frac{1}{96}>\frac{1}{98}\)nên \(\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)< 0\)

\(\Rightarrow x+100=0\Leftrightarrow x=-100\)

\(\Leftrightarrow\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0.Ma:\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}< 0\Rightarrow x=-100\)

\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)

\(\Leftrightarrow\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)

Vì : \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\)

Do đó : \(x+100=0\Leftrightarrow x=-100\)

Vậy phương trình có nghiệm : \(x=-100\)

Chúc bạn học tốt !!!

\(x^2+7x+12\)

\(=x^2+3x+4x+12\)

\(=x\left(x+3\right)+4\left(x+3\right)\)

\(=\left(x+3\right)\left(x+4\right)\)

\(a^{10}+a^5+1\)

\(=\left(a^{10}-a\right)+\left(a^5-a^2\right)+\left(a^2+a+1\right)\)

\(=a\left(a^9-1\right)+a^2\left(a^3-1\right)+\left(a^2+a+1\right)\)

\(=a\left(a^3-1\right)\left(a^3+1\right)+a^2\left(a^3-1\right)+\left(a^2+a+1\right)\)

\(=\left(a^4+a\right)\left(a^2+a+1\right)\left(a-1\right)+a^2\left(a-1\right)\left(a^2+a+1\right)+\left(a^2+a+1\right)\)

\(=\left(a^2+a+1\right)\left(a^5-a^4+a^2-a\right)+\left(a^3-a^2\right)\left(a^2+a+1\right)+\left(a^2+a+1\right)\)

\(=\left(a^2+a+1\right)\left(a^5-a^4+a^2-a+a^3-a^2+1\right)\)

\(=\left(a^2+a+1\right)\left(a^5-a^4+a^3-a+1\right)\)

Bài 2 :
\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)

<=> \(\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)\)

<=> \(\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)

Vì \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\)

Do đó x + 100 = 0 <=> x = -100

Vậy phương trình có nghiệm x = -100