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\(\left(\frac{x}{20}+1\right)+\left(\frac{x-1}{21}+1\right)=\left(\frac{x-2}{22}+1\right)+\left(\frac{x-3}{23}+1\right)\)
\(\frac{x+20}{20}+\frac{x+20}{21}-\frac{x+20}{22}-\frac{x+20}{23}=0\)
\(\left(x+20\right).\left(\frac{1}{20}+\frac{1}{21}-\frac{1}{22}-\frac{1}{23}\right)=0\)
mà \(\left(\frac{1}{20}+\frac{1}{21}-\frac{1}{22}-\frac{1}{23}\right)\ne0\)
=> x+20=0 => x=-20
vậy x=-20
\(\frac{x}{20}+\frac{x-1}{21}=\frac{x-2}{22}+\frac{x-3}{23}\)
\(1+\frac{x}{20}+1+\frac{x-1}{21}=1+\frac{x-2}{22}+1+\frac{x-3}{23}\)
\(\frac{x+20}{20}+\frac{21+x-1}{21}=\frac{22+x-2}{22}+\frac{23+x-3}{23}\)
\(\frac{x+20}{20}+\frac{x+20}{21}=\frac{x+20}{22}+\frac{x+20}{23}\)
\(\frac{x+20}{20}+\frac{x+20}{21}-\frac{x+20}{22}-\frac{x+20}{23}=0\)
\(\left(x+20\right)\left(\frac{1}{20}+\frac{1}{21}-\frac{1}{22}-\frac{1}{23}\right)=0\)
Mà \(\frac{1}{20}+\frac{1}{21}-\frac{1}{22}-\frac{1}{23}\ne0\)
\(\Rightarrow x+20=0\)
\(\Rightarrow x=-20\)
Vậy x = -20
Ta có :
\(\frac{x-20}{19}=\frac{y+35}{20}=\frac{z-15}{37}\left(1\right)\)
\(\frac{x-3}{2}+\frac{5x-6}{9}=1\left(2\right)\)
Giải phương trình 2 ,ta có :
\(\frac{19x-39}{18}=1\)
\(19x-39=18\)
\(19x=57\)
\(x=3\)
Thay x = 3 vào phương trình 1 ,ta có :
\(\frac{x-20}{19}=\frac{y+35}{20}=\frac{z-15}{37}\)
\(\frac{3-20}{19}=\frac{y+35}{20}=\frac{z-15}{37}\)
\(\frac{y-35}{20}=\frac{z-15}{37}=\frac{-17}{19}\)
\(\Rightarrow\hept{\begin{cases}\frac{y-35}{20}=\frac{-17}{19}\\\frac{z-15}{37}=\frac{-17}{19}\end{cases}}\Rightarrow\hept{\begin{cases}y=\frac{325}{19}\\z=\frac{-344}{19}\end{cases}}\)
a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\)
\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)
Nên x + 1 = 0
=> x = -1
\(\frac{x}{x+4}=\frac{5}{6}=>6x=5\left(x+4\right)=5x+20\)
\(=>6x-5x=20=>x=20\)
29-x/21 + 27-x/23 + 25-x/25 + 23-x/27 + 21-x/29 = -5
1 + 29-x/21 + 1 + 27-x/23 + 1 + 25-x/25 + 1 + 23-x/27 + 1 + 21-x/29 = 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 + 1/2 > 0
=> 50 - x = 0
x = 50
Vậy x = 50
\(\frac{-1}{3}+\frac{0,2-0,3+\frac{5}{11}}{-0,3+\frac{9}{16}-\frac{15}{12}}\)
\(=\frac{-1}{3}+\frac{\frac{2}{10}-\frac{3}{10}+\frac{5}{11}}{\frac{-3}{10}+\frac{9}{16}-\frac{15}{12}}\)
\(=\frac{-1}{3}+\frac{\frac{39}{110}}{\frac{-79}{80}}\)
\(=\frac{-1}{3}-\frac{312}{869}\)
\(=\frac{-1805}{2607}\)
a) 3x+3x+2=812
Suy ra 3x+3x.32=812
3x.(1+32) =812
3x.10 =812
3x =812:10
3x =406/5
Suy ra x ko có giá trị
b)4\(\frac{1}{3}\):\(\frac{x}{4}\)=6:0,3
suy ra \(\frac{13}{3}\):\(\frac{x}{4}\) =20
x/4 = 13/3:20
x/4 = 13/60
x = 13/15
c) I 2x + 0,5I=8,5
2x+0,5=8,5 hoặc 2x+0,5=-8,5
TH1:2x+0,5=8,5=>x=4
TH2:2x+0,5=-8,5=>x=-9/2
d) 8x: 2x =1635
=>(8:2)x=1635
=>4x =1635
=>4x =(42)35
=>4x =42.35 =>4x=470 =>x=70
Vậy x = 70
Đầy đủ và chính xác lắm đó.
Theo đề ta có :
\(\frac{x+3}{20}+\frac{x-15}{21}+\frac{x-35}{22}=66\)
\(\Rightarrow\left(\frac{x}{20}+\frac{3}{20}\right)+\left(\frac{x}{21}-\frac{15}{21}\right)+\left(\frac{x}{22}-\frac{35}{22}\right)=66\)
\(\Rightarrow\frac{x}{20}+\frac{3}{20}+\frac{x}{21}-\frac{5}{7}+\frac{x}{22}-\frac{35}{22}=66\)
\(\Rightarrow\left(\frac{x}{20}+\frac{x}{21}+\frac{x}{22}\right)+\left(\frac{3}{20}-\frac{5}{7}-\frac{35}{22}\right)=66\)
\(\Rightarrow x.\left(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}\right)+\frac{-3319}{1540}=66\)
\(\Rightarrow x.\frac{661}{4620}=66-\frac{-3319}{1540}\)
Tới đây lm đc r chứ nhưng mà hình như Akira Nishihiko , bn viết đề sai hay sao á?
THừa số 6 kìa