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\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)
\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}\right)=\frac{20}{41}\)
\(\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{\left(x+2\right)-x}{x\left(x+2\right)}\right)=\frac{20}{41}\)
\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(1-\frac{1}{x+2}=\frac{20}{41}:\frac{1}{2}\)
\(1-\frac{1}{x+2}=\frac{40}{41}\)
\(\frac{1}{x+2}=1-\frac{40}{41}\)
\(\frac{1}{x+2}=\frac{1}{41}\)
\(x+2=41\)
\(x=41-2\)
\(x=39\)
Tìm x
a) \(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{x\times\left(x+2\right)}=\frac{20}{41}\)
\(\Rightarrow\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{x\times\left(x+1\right)}\right)=\frac{20}{41}\)
\(\Rightarrow\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{x\times\left(x+2\right)}=\frac{20}{41}:\frac{1}{2}\)
\(\Rightarrow\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{x\times\left(x+2\right)}=\frac{40}{41}\)
\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{\left(x+2\right)}=\frac{40}{41}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{40}{41}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{40}{41}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{41}\)
\(\Rightarrow x+2=41\)
\(\Rightarrow x=41-2\)
\(\Rightarrow x=39\)
Vậy x = 39
2^x = 4 . 128 = 512
2^x = 2^9
=> x = 9
b) 2^x . 3^x = 4.9
=> 2^x . 3^x = 36
=> x = 2
\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)
\(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x.\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(x=305\)
(a-b+c-d)-9a+b+c+d)=a-b+c-d-a-b-c-d
=(a-a)-(b+b)+(c-c)-(d-d)
=0-2b+0-2d
=-2b-2d
(-a+b-c)+(a-b)-(a-b+c)
=-a+b-c+a-b-a+b-c
=(-a+a-a)+(b-b+b)-(c+c)
=-a+b-2c
-(a-b-c)+(b-c+d)-(a+b+d)
=-a+b+c+c-c+d-a-b-d
=(-a-a)+(b-b)+(c+c-c)+(d-d)
=-2a+0+c+0
=-2a+c
x+x+x+82=-2-x
3x+82=-2-x
3x+x=-2-82
4x=-84
x=-84:4=-21
5.(-40).x=-100
-200.x=-100
x=-100:(-200)
x=0,5
-1.(-3).(-6).x=36
-18.x=36
x=36:(-18)
x=-2
|1-4x|=7
Xảy ra hai trường hợp:1-4x=7=>4x=1-7=-6=>x=-6:4=-1,5
1-4x=-7=>4x=1+7=8=>x=8:4=2
Vì x là số nguyên nên ta chỉ có một đáp án đó là:2
x.(x-2)=0+> x=0 hoặc x=2
x.(x-2)>0=>x=-1;-2;-3;-4;...
A=chia hết cho 2 ko chia hết cho 5
B=chia hết cho cả 2 và 5
C=chia hết cho 2 ko chia hết cho 5