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ĐK : x \(\ne\)0 ; \(x\ne-4\)
Ta có : \(\frac{1}{5}+\frac{1}{45}+\frac{1}{117}+...+\frac{1}{x\left(x+4\right)}=\frac{53}{216}\)
\(\Rightarrow\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{x\left(x+4\right)}=\frac{53}{216}\)
=> \(\frac{1}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{x\left(x+4\right)}\right)=\frac{53}{216}\)
=> \(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{x}-\frac{1}{x+4}=\frac{53}{216}:\frac{1}{4}\)
=> \(1-\frac{1}{x+4}=\frac{53}{54}\)
=> \(\frac{1}{x+4}=\frac{1}{54}\)
=> x + 4 = 54
=> x = 50 (tm)
Vậy x = 50
-3x+(-9)+5x-5=-10
(-3x+5x)+(-9-5)=-10
-2x+(-14)=-10
-2x=-10-(-14)
-2x=24
x=24:(-2)
x=-12. chúc bạn học tối nha
Ta có : (x + 3) + (x + 5) + (x + 7) + ..... + (x + 21) = 335
=> x + x + x + x + (3 + 5 + 7 + ..... + 21) = 335
=> 10x + 120 = 335
=> 10x = 335 - 120
=> 10x = 215
=> x = 21,5
ta có : (x + 3) + (x + 5) + (x + 7) + ..... + (x + 21) = 335
<=> x + x + x + x + (3 + 5 + 7 + ..... + 21) = 335
<=> 10x + 120 = 335
=> 10x = 335 - 120
=> 10x = 215
vậy x = 21,5
a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{-5}{12}\)
b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}\)
c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)
\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-77}{120}\)
d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{-7}{20}\)
e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-59}{105}\)
g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-13}{12}\)
a) x+45-[90+(-20)+5-(-45)]
=x+45-120
=x+-75
b) x+(294+13)+(94-13)
=x+307+81
=x+388
a) x+45-[90+(-20)+5-(-45)]
=x+45-[(90-20)+(5+45)]
=x+45-[70+50]
=x+45-120
=x+(45-120)
=x-75
b) x+(294+13)+(94-13)
=x+307+81
=x+(307+81)
=x+388
a) \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{-1}{3}\)
\(\Rightarrow x=\dfrac{-1}{3}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{-2}{3}\)
b)\(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)
\(\Rightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{-11}{20}\)
c) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{3}{35}-\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{-1}{5}\)
\(\Rightarrow x=\dfrac{-1}{5}-\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{-4}{5}\)
d)\(\dfrac{2}{3}.x=\dfrac{4}{27}\)
\(\Rightarrow x=\dfrac{4}{27}:\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{2}{9}\)
e) \(\dfrac{-3}{5}.x=\dfrac{21}{10}\)
\(\Rightarrow x=\dfrac{21}{10}:\dfrac{-3}{5}\)
\(\Rightarrow x=\dfrac{-7}{2}\)
a) 31 - 2 ( x + 3 ) = 21
2 ( x + 3 ) = 31 - 21
2 ( x + 3 ) = 10
x + 3 = 10 : 2
x + 3 = 5
x = 5 - 3
x = 2
b) ( x - 11 ) : 7 + 5 = 7
( x - 11 ) : 7 = 7 - 5
( x - 11 ) : 7 = 2
( x - 11 ) = 2 x 7
x - 11 = 14
x = 14 + 11
x = 25
a) x + 45 -[ 90 + (-20) + 5 - (-45)] +3x
= x +45 -( 90 - 20 + 5 + 45) + 3x
= x +45 -120 +3x
= -75 +4x
b) x +(-294 +13 -2x) +(94 -13) +9x
= x -281 -2x +81 +9x
= -200 +8x
\(A=\left(x+1\right)+\left(x+\dfrac{5}{45}\right)+\left(x+\dfrac{5}{117}\right)+\left(x+221\right)=10\\ \Rightarrow x+1+x+\dfrac{1}{9}+x+\dfrac{5}{117}+x+221=10\\ \Rightarrow4x+\left(1+\dfrac{1}{9}+\dfrac{5}{117}+221\right)=10\\ \Rightarrow4x+\dfrac{2888}{13}=10\\ \Rightarrow4x=10-\dfrac{2888}{13}\\ \Rightarrow4x=-\dfrac{2758}{13}\\ \Rightarrow x=-\dfrac{1379}{26}\)
Cho hỏi 1/9 ở đâu v ak