a.

Ta có:

(x+2)/327+(x+3)/326+(x+4)/325+(x+5)/324+(x+349)/5=0

<=>(x+2)/327+(x+3)/326+(x+4)/325+(x+5)/324+(x+329)-4   (giải thích: (x+349)/5=(x+329+20)/5=(x+329)/5+4)

<=>1+(x+2)/327+1+(x+3)/326+1+(x+4)/325+1+(x+5)324+(x+329)/5=0

<=>(x+329)/327+(x+329)/326+(x+329)/325+(x+329)/324+(x+329)/5=0

<=>x+329(1/327+1/326+1/325+1/324+1/5)=0

Vì (1/327+...+1/5) khác 0 => x+329=0

=>x=-329

\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)

\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)

\(\left(x+1\right)\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)

\(\)vì \(\dfrac{1}{3}>\dfrac{1}{6};\dfrac{1}{4}>\dfrac{1}{6};\dfrac{1}{5}>\dfrac{1}{6}=>\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}>0\)

\(=>x+1=0\)

\(=>x=-1\)

b,

\(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)

\(\left(\dfrac{x+1}{2020}+1\right)+\left(\dfrac{x+2}{2019}+1\right)=\left(\dfrac{x+3}{2018}+1\right)+\left(\dfrac{x+4}{2017}+1\right)\)

\(\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}=\dfrac{x+2021}{2018}+\dfrac{x+2021}{2017}\)

\(=>\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}-\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}=0\)

\(=>\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}\right)=0\)

Vì \(\dfrac{1}{2020}< \dfrac{1}{2018};\dfrac{1}{2019}< \dfrac{1}{2017}=>\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}< 0\)

\(=>x+2021=0\)

\(=>x=-2021\)

c,

\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)

\(\left(\dfrac{x+2}{327}+1\right)+\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}+1\right)+\left(\dfrac{x+349}{5}-4\right)=0\)

\(\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)

\(=>\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)

Vì \(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}>0\)

\(=>x+329=0\)

\(=>x=-329\)

nhìn thấy duy nhất một chữ rối!

(x+1/5)^2 =26/25-17/25

<=> (x +1/5)^2 =(3/5)^2

<=> x+1/5=3/5

=> x= 2/5

a) 
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)

\(\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\)

\(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)

\(\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(x+\frac{1}{5}=\frac{3}{5}\)

\(x=\frac{3}{5}-\frac{1}{5}\)

\(x=\frac{2}{5}\)
b) 
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)

\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+349}{5}=0\)
\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+4+\frac{x+329}{5}=0\)

\(\left(x+329\right). \left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+4+\frac{1}{5}\right)=0\)
\(\Rightarrow\left(x+329\right)=0\)
\(\Rightarrow x=-329\)

1.

(x + 1/5)² = 26/25 - 17/25 

(x + 1/5)² = 9/25 

Rút căn hai vế : 

|(x + 1/5)| = 3/5 

x = -4/5 
hoặc
x = 2/5

2.

(x + 2) / 327 + (x + 3) / 326 + (x + 4) / 325 + (x + 5) / 324 + (x + 349) / 5 = 0 
<=> (x + 2) / 327 +1+ (x + 3) / 326 +1+ (x + 4) / 325 +1+ (x + 5) / 324 +1+ (x + 349) / 5 -4 = 0 
<=> (x+ 329)/327 + (x+ 329)/326 + (x+ 329)/325 + (x+ 329)//324 + (x+ 329)/5 =0 
<=> (x+ 329).(1/327 + 1/ 326 + 1/325 + 1/324 +1/5) =0 
Do (1/327 + 1/ 326 + 1/325 + 1/324 +1/5) >0 nên x+ 329 =0 => x= -329 

Câu 1 chưa chắc đã đúng ( quên hết kiến thức lớp 6 rùi ) hihi

aaaaaaaa . chết rồi . cho mình sủa câu thứ nhất :

(x+1/5)² + 17/25=26/25

( x + 1/5 ) ² = 26/25 - 17/25

( x + 1/5 ) ² = 3/5²

x + 1/5 = 3/5

x = 2/5.

329

(x + 2) / 327 + (x + 3) / 326 + (x + 4) / 325 + (x + 5) / 324 + (x + 349) / 5 = 0 
<=> (x + 2) / 327 +1+ (x + 3) / 326 +1+ (x + 4) / 325 +1+ (x + 5) / 324 +1+ (x + 349) / 5 -4= 0 
<=> (x+ 329)/327 + (x+ 329)/326 + (x+ 329)/325 + (x+ 329)//324 + (x+ 329)/5 =0 
<=> (x+ 329).(1/327 + 1/ 326 + 1/325 + 1/324 +1/5) =0 
Do (1/327 + 1/ 326 + 1/325 + 1/324 +1/5) # 0 nên x+ 329 =0 => x= -329 

dấu # nghĩa là gì vậy bạn

nghĩa là khác Nguyễn Đăng Hoàng