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\(\frac{x+1}{2017}+\frac{x+2}{2016}=\frac{x+3}{2015}+\frac{x+4}{2014}\)
\(\Leftrightarrow\frac{x+1}{2017}+1+\frac{x+2}{2016}+1=\frac{x+3}{2015}+1+\frac{x+4}{2014}+1\)
\(\Leftrightarrow\frac{x+2018}{2017}+\frac{x+2018}{2016}-\frac{x+2018}{2015}-\frac{x+2018}{2014}=0\)
\(\Leftrightarrow\left(x+2018\right)\left(\frac{1}{2017}+\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}\ne0\right)=0\Leftrightarrow x=-2018\)
\(\left(\frac{10}{99}+\frac{11}{199}-\frac{12}{299}\right)\times\left(\frac{1}{2}-\frac{1}{3}+-\frac{1}{6}\right)\)
\(=\left(\frac{10}{99}+\frac{11}{199}-\frac{12}{299}\right)\times\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\left(\frac{10}{99}+\frac{11}{199}-\frac{12}{299}\right)\times\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)
\(=\left(\frac{10}{99}+\frac{11}{199}-\frac{12}{299}\right)\times0\)
\(=0\)
A=[(3999/2+1)+(3998/3+1)+...+(1/4000+1)+1]/(1/2+1/3+...+1/4001)
A=(4001/2+4001/3+...+4001/4001)/(1/2+1/3+...+1/4001)
A=[4001(1/2+1/3+...+1/4001)]/(1/2+1/3+...+1/4001)
A=4001
Vậy A=4001
(x + 1) + (x + 2) + (x + 3 )+ ... + (x + 20) = 250 ( có 20 nhóm )
=> ( x + x + x +...+ x ) + ( 1 + 2 + 3 +...+ 20) = 250 ( có 20 x và 20 số hạng )
=> x . 20 + 20 . 21 : 2 = 250
=> x . 20 + 210 = 250
=> x . 20 = 250 - 210
=> x . 20 = 40
=> x = 40 : 20
x = 2
a, \(343\text{ : }\left(2^3-2^2+3^2\cdot x\right)=7\)
\(343\text{ : }\left(8-4+9\cdot x\right)=7\)
\(343\text{ : }\left(4+9\cdot x\right)=7\)
\(4+9\cdot x=343\text{ : }7\)
\(4+9\cdot x=49\)
\(9\cdot x=49-4\)
\(9\cdot x=45\)
\(x=45\text{ : }9\)
\(x=5\)
Đặt A=\(\frac{4000}{1}+\frac{3999}{2}+\frac{3998}{3}+........+\frac{1}{4000}\)
A=\(1+\left(1+\frac{3999}{2}\right)+\left(1+\frac{3998}{3}\right)+........+\left(1+\frac{1}{4000}\right)\)
A=\(\frac{4001}{4001}+\frac{4001}{2}+\frac{4001}{3}+...........+\frac{4001}{4000}\)
A=\(4001.\left(\frac{1}{2}+\frac{1}{3}+........+\frac{1}{4000}+\frac{1}{4001}\right)\)
=>\(y=\frac{4001.\left(\frac{1}{2}+\frac{1}{3}+........+\frac{1}{4001}\right)}{\frac{1}{2}+\frac{1}{3}+.........+\frac{1}{4001}}\)
=>\(y=4001\)
\(\left(2x-5\right)^3=8\)
\(\left(2x-5\right)^3=2^3\)
\(2x-5=2\)
\(2x=7\)
\(x=\frac{7}{2}\)
\(b,6⋮x-1\)
\(\Rightarrow x-1\inƯ6\left(\pm1,\pm2,\pm3\right)\)
+ x-1 =1 => x=2
+ x-1 =-1 =>x=0
+ x-1=2 =>x=3
+ x-1=-2 => x= -1
+ x-1 =3 =>x=4
+ x-1=-3 => x=-2
#Giải :
( 2x - 5 )3 = 8
( 2x - 5 )3 = 23
=> 2x - 5 = 2
2x = 7
x = 7/2
6 chia hết cho ( x - 1 )
=> (x - 1) € Ư (6)
Ư (6) = { 1 ; 2 ; 3 ; 6 }
Nếu x - 1 = 1 => x = 2
Nếu x - 1 = 2 => x = 3
Nếu x - 1 = 3 => x = 4
Nếu x - 1 = 6 => x = 7
[ P/S : € - thuộc ]
#By_Ami