2+(-4)+6+(-8)+.....+x= 2016
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\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{2014.2016}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+......+\frac{1}{2014}-\frac{1}{2016}\)
\(=\frac{1}{2}-\frac{1}{2016}\)
\(=\frac{1008}{2016}-\frac{1}{2016}=\frac{1007}{2016}\)
\(\frac{2}{2\times4}+\frac{2}{4\times6}+...+\frac{2}{2014\times2016}\)
=\(\left(\frac{2}{2}-\frac{2}{4}\right)+\left(\frac{2}{4}-\frac{2}{6}\right)+...+\left(\frac{2}{2014}-\frac{2}{2016}\right)\)
=\(\frac{2}{2}-\frac{2}{4}+\frac{2}{4}-\frac{2}{6}+...+\frac{2}{2014}-\frac{2}{2016}\)
=\(\frac{2}{2}-\frac{2}{2016}=\frac{1007}{1008}\)
`#3107.101107`
A,
\(2\times32\times12+4\times6\times41+8\times27\times3\\ =24\times32+24\times41+24\times27\\ =24\times\left(32+41+27\right)\\ =24\times100\\ =2400\)
B,
\(\left(2006\times2005^{2016}-2005^{2016}\right)\div2005^{2017}\\ =\left[2005^{2016}\times\left(2006-1\right)\right]\div2005^{2017}\\ =\left(2005^{2016}\times2005\right)\div2005^{2017}\\ =2005^{2017}\div2005^{2017}\\ =1\)
(1+2+3+4+5+6+7+8+9+...............................+2016+2025) x (24,2 - 24,2) = (1 + 2 +3+4+5+6+7+8+9+...............................+2016+2025) x 0 = 0
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 $x=0$
a) 612 : (x - 7) + 8 = 59
612 : (x - 7) = 51
x - 7 = 612 : 51 = 12
x = 12 + 7
x = 19
b) 2016 - ( 2+4+6+8....+x)= 376
2 + 4 + 6+ 8 + ... +x = 2016 - 376 = 1640
Áp dụng công thức tính dãy số ta có :
\(\frac{\left[\left(x-2\right):2+1\right].\left(x+2\right)}{2}=1640\)
\(\Rightarrow\left(\frac{x-2}{2}+1\right).\left(x+2\right)=3280\)
\(\Rightarrow\frac{x}{2}.\left(x+2\right)=3280\)
\(\Rightarrow\frac{x^2}{2}+x=3280\)
\(\Rightarrow\frac{x^2}{2}+\frac{2x}{2}=3280\)
\(\Rightarrow x^2+2x=6560\)
=> x . (x+2) = 6560
=> x = 80
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