=>\(\dfrac{x-5}{100}-1+\dfrac{x-4}{101}-1+\dfrac{x-3}{102}-1=\dfrac{x-100}{5}-1+\dfrac{x-101}{4}-1+\dfrac{x-102}{3}-1\)

=>x-105=0

=>x=105

cộng cả 2 vế với -1

x=105

chỉ chi tiết giùm

a) \(\left(x-15\right)-35=58\)

\(x-15=58+35\)

\(x-15=93\)

\(x=93+15\)

\(x=108\)

b) \(x-7100=1100+1\)

\(x-7100=1101\)

\(x=1101+7100\)

\(x=8201\)

help mik zới mn ưi

100.1=100

100.2=200

100.3=300

100.4=400

100.5=500

100 x 1 = 100

100 x 2 = 200

100 x 3 = 300

100 x 4 = 400

100 x 5 = 500

 aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

\(\text{-12(x-5)+7(3-x)=5 }\)

\(-12x+60+21-7x=5\)

\(-12x-7x=5-21-60\)

\(-19x=-76\)

\(x=-76:\left(-19\right)\)

\(x=4\)

\(\text{ 30(x+2)-6(x-5)-24x=100}\)

\(30x+60-6x+30-24x=100\)

\(30x-6x-24x=100-30-60\)

\(0=10\)

\(\Rightarrow x\)ko tồn tại

\(\text{(x+1-5)+7(3-x)=5 }\)

\(x+1-5+21-7x=5\)

\(x-7x=5-21+5-1\)

\(-6x=-12\)

\(x=\left(-12\right):\left(-6\right)\)

\(x=2\)

\(\text{(x+1)+(x+3)+(x+5)+........+(x+99)=0}\)

\(\text{(x+1)+(x+3)+...+(x+99)=0}\)
tổng các số hang là\(\frac{\left(99+1\right)}{2}=50\)(số hạng)
=>\(\text{(x+1)+(x+3)+...+(x+99)=0}\)<=> \(\text{50.x+(1+3+5+..+99)=0}\)
<=>\(\text{50.x+(99+1)}\)\(.\frac{50}{2}=0\)<=> \(\text{50.x+2500=0=}\)>\(x=\frac{-2500}{50}=-50\)

chúc bạn học tốt

a) 2 x 2 = 4              3 x 3 = 9

2 x 4 = 8              3 x 5 = 15

2 x 6 = 12              3 x 7 = 21

2 x 8 = 16              3 x 9 = 27

4 x 4 = 16              5 x 5 = 25

4 x 2 = 8              5 x 7 = 35

4 x 6 = 24              5 x 9 = 45

4 x 8 = 32              5 x 3 = 15

b) 200 x 4 = 800              300 x 2 = 600

200 x 2 = 400             300 x 3 = 900

400 x 2 = 800             500 x 1 = 500

100 x 4 = 400             100 x 3 = 300

Trả lời 1 câu thôi nhé em 100 nhân 4 bằng 400 nhá em

a) 2 x 2 =4 ......              3 x 3 = ..9....

2 x 4 = ....8..              3 x 5 = ...15...

2 x 6 = ....12..              3 x 7 = .21.....

2 x 8 = ..16....              3 x 9 = ..27....

4 x 4 = ...16...              5 x 5 = ...25...

4 x 2 = ...8...              5 x 7 = ..35....

4 x 6 = .....24.              5 x 9 = .45.....

4 x 8 = ..32....              5 x 3 = .15.....

b) 200 x 4 = 800......              300 x 2 =600 ......

200 x 2 = ..400....             300 x 3 = 900......

400 x 2 = .....800.             500 x 1 = ..500....

100 x 4 = ....400..             100 x 3 = .300.....

cộng cả 2 vế với -1

x=105

Ta có :\(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)

<=> \(\left(\frac{x-5}{100}-1\right)+\left(\frac{x-4}{101}-1\right)+\left(\frac{x-3}{102}-1\right)=\left(\frac{x-100}{5}-1\right)+\left(\frac{x-101}{4}-1\right)+\left(\frac{x-102}{3}-1\right)\)

<=> \(\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)

<=> \(\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}\right)=\left(x-105\right)\left(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}\right)\)

<=> \(\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)

<=> x - 105 = 0 (Vì \(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\))

<=> x = 105

Vậy nghiệm phương trình là x = 105

#muon roi ma sao con

\(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)

\(\Leftrightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}-\frac{x-105}{5}-\frac{x-101}{4}-\frac{x-102}{3}=0\)

( cả 2 vế trừ đi 3 và từng phân thức trừ đi 1 ) 

\(\Leftrightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\right)=0\Leftrightarrow x=105\)

Vậy tập nghiệm của pt là S = { 105 } 

\(< =>\left(\dfrac{x-5}{100}-1\right)+\left(\dfrac{x-4}{101}-1\right)+\left(\dfrac{x-3}{102}-1\right)+3=\left(\dfrac{x-100}{5}-1\right)+\left(\dfrac{x-101}{4}-1\right)+\left(\dfrac{x-102}{3}-1\right)+3\)\(< =>\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}=\dfrac{x-105}{5}+\dfrac{x-105}{4}+\dfrac{x-105}{3}\)

\(< =>\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}\right)\) = 0

<=> x - 105 = 0

<=> x = 105

Vậy tập nghiệm của phương trình là S = \(\left\{105\right\}\)

Bạn tham khảo tại link : https://olm.vn/hoi-dap/detail/205275532692.html