a) |5/3 - x| - |-5/6| = |-5/9|

=> |5/3 - x| - 5/6 = 5/9

=> |5/3 - x| = 5/9 + 5/6

=> |5/3 - x| = 25/18

=> \(\orbr{\begin{cases}\frac{5}{3}-x=\frac{25}{18}\\\frac{5}{3}-x=-\frac{25}{18}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{5}{18}\\x=\frac{55}{18}\end{cases}}\)

a, \(\left|\frac{5}{3}-x\right|-\left|-\frac{5}{6}\right|=\left|-\frac{5}{9}\right|\)

\(\Leftrightarrow\left|\frac{5}{3}-x\right|-\frac{5}{6}=\frac{5}{9}\Rightarrow\left|\frac{5}{3}-x\right|=\frac{5}{9}+\frac{5}{6}=\frac{25}{18}\)

\(\Rightarrow\orbr{\begin{cases}\frac{5}{3}-x=\frac{25}{18}\\\frac{5}{3}-x=-\frac{25}{18}\end{cases}\Rightarrow}x.\)

b, \(\left|x+\frac{1}{102}\right|+\left|x+\frac{2}{102}\right|+...+\left|x+\frac{100}{102}\right|\ge0\)

\(\Rightarrow102x\ge0\Leftrightarrow x\ge0\)=> Ta có thể phá dấu GTTĐ

\(\Rightarrow x+\frac{1}{102}+x+\frac{2}{102}+...+x+\frac{100}{102}=102x\)

\(\Rightarrow100x+\frac{1+2+3+...+100}{102}=102x\Rightarrow2x\Rightarrow x.\)

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

=>x-105=0

=>x=105

Lời giải:

$75813=7.10000+5.1000+8.100+1.10+3$

$=7.10^4+5.10^3+8.10^2+1.10+3$

\(\left(-\frac{1}{5}\right)^{300}=-\frac{1^{300}}{5^{300}}=-\frac{1}{5^{300}}\)

\(\left(-\frac{1}{5}\right)^{500}=-\frac{1^{500}}{5^{500}}=-\frac{1}{5^{500}}\)

Ta có : 

\(5^{300}< 5^{500}\)

\(\Rightarrow-5^{300}>-5^{500}\)

\(\Rightarrow-\frac{1}{5^{300}}>-\frac{1}{5^{500}}\)

\(\Rightarrow\left(-\frac{1}{5}\right)^{300}>\left(-\frac{1}{5}\right)^{500}\)

cộng cả 2 vế với -1

x=105

chỉ chi tiết giùm

Chọn A

NHO CHO MINH NHA !!!

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 = \(\left\{105\right\}\)

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