5 ( năm )

a, <=> (x-5/100) -1 +(x-4/101) -1 +(x-3/102) -1= (x-100/5) -1+(x-101/4) -1 +(x-102/3) -1
<=> (x-105)(1/100 +1/101 +1/102)= (x-105)(1/5+1/4+1/3)
<=> (x-105)(1/100+1/101+1/102-1/5-1/4-1/3)=0
vì 1/100+1/101+1/102-1/5-1/4-1/3 khác 0 <=> x-105=0
<=> x=105

b, 29-x/21 +1+27-x/23 +1+25-x/25 +1+23-x/27 +1+21-x/29 +1=0
<=> 50-x/21 +50-x/23 +50-x/25 +50-x/27 +50-x/29=0
<=> (50-x)(1/21 +1/23 +1/25 +1/27 +1/29)=0
vì 1/21+1/23+1/25+1/27+1/29 lớn hơn 0
nên 50-x=0
<=> x=50

\(3x^2+x+11=0\)

\(x^2+x+\frac{1}{4}+2x^2+\frac{43}{4}=0\) 

\(\left(x+\frac{1}{2}\right)^2+2x^2+\frac{43}{4}=0\) 

Mà \(\left(x+\frac{1}{2}\right)^2+2x^2+\frac{43}{4}\ge\frac{43}{4}\forall x\)

=> PT vô nghiêm

\(3x^2+x+11=0\)

\(\Leftrightarrow x^2+\frac{1}{3}x+\frac{11}{3}=0\)

\(\Leftrightarrow x^2+2\frac{1}{3}.\frac{1}{2}x+\frac{1}{36}+\frac{131}{36}=0\)

\(\Leftrightarrow\left(x+\frac{1}{6}\right)^2=-\frac{131}{36}\left(voly\right)\)

=> Phương Trình Vô Nghiệm

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 } 

\(\frac{x-101}{86}+\frac{x-103}{84}+\frac{x-101}{82}=3\)

-> Đề đúng chứ bạn?