Bài 1: Tính nhanh 

A, 13 x 126 + 37 x 126 - 49 x 126

= 126 x ( 13 + 49 - 49 )

= 126 x 1

= 126

B, ( 1 + 2 + 4 + 8+.....+ 512 ) x ( 101 x 102 - 101 x 101 - 50 - 51 )

= ( 1 + 2 + 4 + 8+.....+ 512 ) x ( 101 x ( 102 - 101 - 1))

= ( 1 + 2 + 4 + 8+.....+ 512 ) x ( 101 x 0)

= ( 1 + 2 + 4 + 8+.....+ 512 ) x 0

= 0

\(=\left(1+2+4+...+408\right)\cdot\left[101\left(102-101-1\right)\right]=0\)

=0

bằng 0 nhé

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

=>x-105=0

=>x=105

\(\Leftrightarrow1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{101}{102}\)

\(\Leftrightarrow\dfrac{x+2-1}{x+2}=\dfrac{101}{102}\)

=>x+1=101

hay x=100

\(\dfrac{x+4}{104}+\dfrac{x+2}{102}=\dfrac{x+3}{103}+\dfrac{x+1}{101}\\ \Leftrightarrow\left(\dfrac{x+4}{104}-1\right)+\left(\dfrac{x+2}{102}-1\right)=\left(\dfrac{x+3}{103}-1\right)+\left(\dfrac{x+1}{101}-1\right)\\ \Leftrightarrow\dfrac{x-100}{104}+\dfrac{x-100}{102}-\dfrac{x-100}{103}-\dfrac{x-100}{101}=0\\ \Leftrightarrow\left(x-100\right)\left(\dfrac{1}{104}+\dfrac{1}{102}-\dfrac{1}{103}-\dfrac{1}{101}\right)=0\\ \Leftrightarrow x-100=0\left(vì.\dfrac{1}{104}+\dfrac{1}{102}-\dfrac{1}{103}-\dfrac{1}{101}\ne0\right)\\ \Leftrightarrow x=100\)

cộng cả 2 vế với -1

x=105

chỉ chi tiết giùm

bài này dạng lớp 6 nha

 1 x 3 x5 x.....x199=1.2.3.......200/2.4.6.....200=1.2.3....200/1.2.2.2.2.3....100.2=1.2.3......100.....200/1.2.3....100.2.2......2=101.....200/2.2...2

=101/2.102/2.....200/2

vậy ....

x + 1/100 + x + 2/101 = x + 3/102 - 1

<=> x + 1/100 - 1 + x + 2/101 - 1 = x + 3/102 - 1 - 2

<=> x - 99/100 + x - 99/101 = x - 99/102 - 2

<=> x - 99/100 + x - 99/101 - x - 99/102 = -2

<=> (x - 99)(1/100 + 1/101 - 1/102) = -2

<=> x - 99 = -2/1/100 + 1/101 - 1/102

<=> x = -2/1/100 + 1/101 - 1/102 + 99

Bạn chịu khó bấm máy hộ mình, số to quá

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 }