tim x
x:15,x:20
15:x,20:x
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VẾ TRÁI LÀ:
A=1/2.3+1/3.2+1/4.5+...+1/x[x+1]
A=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/n-1/n+1
A=1-1/n+1
1-1/n+1=2015/2016
1/n+1=1-2015/2016
1/n+1=1/2016
n=2016-1
n=2015
Vế trái là
S=1/2+1/6+1/12+1/20+..+1/x(x+1)
S=1/1.2+1/2.3+1/3.4+1/4.5+...+1/x.(x+1)
S=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/x.(x+1)
S=1-1/(x+1)=Vế phải=2015/2016
<=>1-1/(x+1)=2015/2016
1/(x+1)=1/2016
=>x+1=2016
x=2015
Ủng hộ mk mha Chí Tiến
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
\(\left(4.x-15\right)^{2016}=\left(4.x-15\right)^{2015}\)
=>\(\left(4.x-15\right)^{2016}-\left(4.x-15\right)^{2015}=0\)
=>\(\left(4.x-15\right)^{2015}.\left[\left(4.x-15\right)-1\right]=0\)
=>\(\hept{\begin{cases}\left(4.x-15\right)^{2015}=0\\\left(4.x-15\right)-1=0\end{cases}}\)=>\(\hept{\begin{cases}4x-15=0\\4.x-15-1=0\end{cases}}\)=>\(\hept{\begin{cases}4x=15\\4x=16\end{cases}}\)
=>\(\hept{\begin{cases}x=\frac{15}{4}\\x=4\end{cases}}\)
Vậy.................
(4x - 15)2016 = (4x - 15)2015
(4x - 5)2016 - (4x - 15)2015 = 0
(4x - 5)2015.(4x - 15) - (4x - 15)2015 = 0
(4x - 5)2015.[(4x - 15) - 1] = 0
=> \(\orbr{\begin{cases}\left(4x-5\right)^{2015}=0\\\left(4x-15\right)-1=0\end{cases}}\)=>\(\orbr{\begin{cases}4x-15=0\\4x-15=1\end{cases}}\)
=> \(\orbr{\begin{cases}4x=15\\4x=16\end{cases}}\)=>\(\orbr{\begin{cases}x=\frac{15}{4}\\x=4\end{cases}}\)
Ta có : A=20/11×13 + 20/13×15 +20/15×17+...+20/53×55
A = 10 ×( 2/11×13+2/13×15+...12/53×55)
A = 10 ×(1/11-1/13+1/13-1/15+1/15-1/17+...+1/53-1/55)
A = 10 × (1/11-1/55)
A =10 × 4/55
A = 8/11
\(\left|x-1\right|+\left(y+2\right)^{20}=0\)
=>x-1=0 và y+2=0
=>x=1 và y=-2
Thay x=1 và y=-2 vào X, ta được:
\(X=2\cdot1^5-5\cdot\left(-2\right)^3+2015\)
\(=2017+40=2057\)
180 - ( x + 15 ) : x x 20 = 100
( x + 15 ) : x x 20 = 80
( x + 15 ) : x = 4
x : x + 15/x = 4
1 + 15/x = 4
15/x = 3
x = 5
\(2015-\left|x-2015\right|=x\)
\(\Leftrightarrow\left|x-2015\right|=2015-x\)
\(\Leftrightarrow\orbr{\begin{cases}x-2015=2015-x\\x-2015=-\left(2015-x\right)\end{cases}}\)
\( TH1:x-2015=2015-x\)
\(\Leftrightarrow2x=4030\)
\(\Leftrightarrow x=\frac{4030}{2}\)
\(\Leftrightarrow x=2015\)
\(TH2:x-2015=-\left(2015-x\right)\)
\(\Leftrightarrow x-2015=x-2015\)
Vậy x = 2015
a) x E B(15) = {0;15;30;45;60;75;90;....}
x E B(20) = {0;20;40;60;80;100;120;...}
b) x E Ư(15) = {1;3;5;15}
x E Ư(20) = {1;2;4;5;10;20}