Ta có:

B. ( x - 21.13 ) : 11 = 30

=> x = 30 x 11 + 21 x 13

=> x = 330 + 273

=> x = 603

C. ( 25 - x ) + ( 29 - x ) + ( 33 - x ) + ... + ( 101 - x ) = 128

=> ( 25 + 29 + 33 + ... + 101 ) - ( x + x + x +...+ x ) = 128

=> 1260 - 20x = 128

=> 20x = 1260 - 128

=> x = 1132 : 20

=> x = 283/5

B. ( x - 21.13 ) : 11 = 30

=> x = 30 x 11 + 21 x 13

=> x = 330 + 273

=> x = 603

C. ( 25 - x ) + ( 29 - x ) + ( 33 - x ) + ... + ( 101 - x ) = 128

=> ( 25 + 29 + 33 + ... + 101 ) - ( x + x + x +...+ x ) = 128

=> 1260 - 20x = 128

=> 20x = 1260 - 128

=> x = 1132 : 20

=> x = 283/5

283/5

=> 25 - x + 29 - x + 33 - x + ....+ 101 - x = 12 

=> ( 25 + 29 + 33 +....+ 101 ) - ( x + x + .....+ x ) = 12 

=> \(\frac{\left(25+101\right).20}{2}-20x=12\)

=> 1260 - 20x = 12

=> x = 62,4 

ủa vậy bài này

là sao vậy bn 

dấu này / là đấu

gì vậy

\(\left(25-x\right)+\left(29-x\right)+...+\left(101-x\right)=108\)

\(\Leftrightarrow\left(25+29+...+101\right)-\left(x+x+....+x\right)=108\)

\(\Leftrightarrow1260-20x=180\)

\(\Leftrightarrow20x=1080\)

\(\Leftrightarrow x=54\)

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

ai làm ơn trả lời hộ mình câu này với

a) \(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
\(\Leftrightarrow\left(\frac{x-5}{100}-1\right)+\left(\frac{x-4}{101}-1\right)+\left(\frac{x-3}{102}-1\right)=\left(\frac{x-100}{5}-1\right)+\left(\frac{x-101}{4}-1\right)+\left(\frac{x-102}{3}-1\right)\)
\(\Leftrightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)
\(\Leftrightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
\(\Leftrightarrow x=105\)
b) \(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5\)
\(\Leftrightarrow\left(\frac{29-x}{21}+1\right)+\left(\frac{27-x}{23}+1\right)+\left(\frac{25-x}{25}+1\right)+\left(\frac{23-x}{27}+1\right)+\left(\frac{21-x}{29}+1\right)=0\)
\(\Leftrightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)
\(\Leftrightarrow x=50\)