1/6 x 1/3 + 17/6 x 1/3 + 2015/2016 - 1 = ?
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{6}.\frac{1}{3}+\frac{17}{6}.\frac{1}{3}+\frac{2015}{2016}-1\)
\(=\frac{1}{3}\left(\frac{1}{6}+\frac{17}{6}\right)+\frac{2015}{2016}-1\)
\(=\frac{1}{3}.3+\frac{2015}{2016}-1\)
\(=1-1+\frac{2015}{2016}=\frac{2015}{2016}\)
\(\frac{1}{6}\times\frac{1}{3}+\frac{17}{6}\times\frac{1}{3}+\frac{2015}{2016}-1\)
\(=\left(\frac{1}{6}+\frac{17}{6}\right)\times\frac{1}{3}+\frac{2015}{2016}-1\)
\(=3\times\frac{1}{3}+\frac{2015}{2016}-1\)
\(=1+\frac{2015}{2016}-1\)
\(=0+\frac{2015}{2016}=\frac{2015}{2016}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
câu 1. tìm x nguyên để \(\frac{-35}{6}\)<x<\(\frac{-18}{5}\)
<=> -4,375<x<-3,6
mà x\(\in\)Z nên x={-4}
câu 2. A=\(\frac{2015}{2016}\)+\(\frac{2016}{2017}\)
B=\(\frac{2015+2016}{2016+2017}\)=\(\frac{2015}{2016+2017}\)+\(\frac{2016}{2016+2017}\)
Vì \(\frac{2015}{2016+2017}\)<\(\frac{2015}{2016}\); \(\frac{2016}{2016+2017}\)<\(\frac{2016}{2017}\)
Vậy B<A
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{2015}{2016}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2015}{2016}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}:2\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{4032}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4032}=\frac{1}{4032}\)
=> x+1=4032
=> x = 4032 - 1
=> x = 4031
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
ĐKXĐ: $x\geq 1; y\geq 2; z\geq 3$
PT \(\Leftrightarrow x+y+z+8-2\sqrt{x-1}-4\sqrt{y-2}-6\sqrt{z-3}=0\)
\(\Leftrightarrow [(x-1)-2\sqrt{x-1}+1]+[(y-2)-4\sqrt{y-2}+4]+[(z-3)-6\sqrt{z-3}+9]=0\)
\(\Leftrightarrow (\sqrt{x-1}-1)^2+(\sqrt{y-2}-2)^2+(\sqrt{z-3}-3)^2=0\)
\(\Rightarrow \sqrt{x-1}-1=\sqrt{y-2}-2=\sqrt{z-3}-3=0\)
\(\Leftrightarrow \left\{\begin{matrix} x=2\\ y=6\\ z=12\end{matrix}\right.\)
2.
ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow \sqrt{x+1}=1-\sqrt{x}$
$\Rightarrow x+1=(1-\sqrt{x})^2=x+1-2\sqrt{x}$
$\Leftrightarrow 2\sqrt{x}=0$
$\Leftrightarrow x=0$
Thử lại thấy thỏa mãn
Vậy $x=0$
![](https://rs.olm.vn/images/avt/0.png?1311)
1/6x 1/3 + 17/6 x 1/3 + 2015/2016-1
= ( 1/6 + 17/6 ) x 1/3 + 2015/2016-1
= 3 x 1/3 + 2015/2016-1
= 1 +2015/2016-1
= 1-1+2015/2016
= 2015/2016