Violympic toán 7

Lớn nhất nhé bạn :D

\(A=\dfrac{1}{\left|x-y\right|+\left|y-z\right|+\left|z-x\right|+2016}\)

\(\left\{{}\begin{matrix}\left|x-y\right|\ge0\\\left|y-z\right|\ge0\\\left|z-x\right|\ge0\end{matrix}\right.\) \(\Rightarrow\left|x-y\right|+\left|y-z\right|+\left|z-x\right|\ge0\)

\(\Rightarrow\left|x-y\right|+\left|y-z\right|+\left|z-x\right|+2016\ge2016\)

\(A=\dfrac{1}{\left|x-y\right|+\left|y-z\right|+\left|z-x\right|+2016}\le\dfrac{1}{2016}\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|x-y\right|=0\\\left|y-z\right|=0\\\left|z-x\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=z\\z=x\end{matrix}\right.\)

\(\Rightarrow x=y=z\)

\(\left|x-3\right|+\left|x-5\right|=\left|x-3\right|+\left|5-x\right|\)

Áp dụng bđt:

\(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)

\(\left|x-3\right|+\left|5-x\right|\ge\left|x-3+5-x\right|=2\)
Dấu "=" xảy ra khi:

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\Rightarrow x\ge3\\5-x\ge0\Rightarrow x\le5\end{matrix}\right.\\\left\{{}\begin{matrix}x-3< 0\Rightarrow x< 3\\5-x< 0\Rightarrow x>5\end{matrix}\right.\end{matrix}\right.\)

Vậy \(3\le x\le5\)

Lập bảng xét dấu nhé bạn !

\(A=0,5-\left|x-3,5\right|\)

Với mọi \(x\in R\) thì:

\(\left|x-3,5\right|\ge0\)

\(A=0,5-\left|x-3,5\right|\le0,5\)

Dấu "=" xảy ra khi:

\(\left|x-3,5\right|=0\Rightarrow x=3,5\)

Bạn A sẽ bốc một lúc hết 2014 viên thế là chẳng còn viên nào cho bạn B bốc! Chắc chắn chiến thắng! :D

Bạn A sẽ bốc sao cho:2014 sẽ chia hết cho số bi mà bạn bốc

\(A=\dfrac{\sqrt{\dfrac{9}{4}-3^{-1}+2018^0}}{25\%+1\dfrac{1}{4}-1,3}-\dfrac{\left(-\dfrac{1}{2}\right)^2-\sqrt{\dfrac{4}{9}}+0,4}{0,6-\dfrac{2}{3}.\left(-\dfrac{1}{4}-\dfrac{1}{2}\right)}\)

\(A=\dfrac{\sqrt{\dfrac{9}{4}-\dfrac{1}{3}+1}}{\dfrac{1}{4}+\dfrac{5}{4}-\dfrac{13}{10}}-\dfrac{\dfrac{1}{4}-\dfrac{2}{3}+\dfrac{2}{5}}{\dfrac{3}{5}-\dfrac{2}{3}\left(-\dfrac{1}{4}-\dfrac{1}{2}\right)}\)

\(A=\dfrac{\sqrt{\dfrac{35}{12}}}{\dfrac{1}{5}}-\dfrac{-\dfrac{1}{60}}{\dfrac{11}{10}}\)

\(A=\dfrac{5\sqrt{105}}{6}+\dfrac{11}{66}\)

\(A=\dfrac{55\sqrt{105}}{66}+\dfrac{11}{66}\)

\(A=\dfrac{55\sqrt{105}+11}{66}\)

\(9\left(\dfrac{1}{3}\right)^3:\left[\left(\dfrac{-2}{3}\right)^2+0,5-1\dfrac{1}{2}\right]\)

\(=9.\dfrac{1}{27}:\left[\dfrac{2}{3}+0,5-1,5\right]\)

\(=\dfrac{1}{3}:\left[\dfrac{2}{3}+-1\right]\)

\(=\dfrac{1}{3}:\dfrac{-1}{3}=\dfrac{-1}{9}\)

\(A=9\left(\dfrac{1}{3}\right)^3:\left[\left(-\dfrac{2}{3}\right)^2+0,5-1\dfrac{1}{2}\right]\)

\(A=9.\dfrac{1}{3}:\left(\dfrac{4}{9}+\dfrac{1}{2}-\dfrac{3}{2}\right)\)

\(A=3:\left(\dfrac{8}{18}+\dfrac{9}{18}-\dfrac{27}{18}\right)\)

\(A=3:-\dfrac{5}{9}\)

\(A=-\dfrac{27}{5}\)

| x - y - 2 | + | y + 3 | = 0

x-y-2+y+3=0

x-(y-y)-(2-3)=0

x-0+1=0

x+1=0

x=-1

Theo như sách đã nói hiền phải tick đúng cho Khải

\(\left|x-y-2\right|+\left|y+3\right|=0\)

Với mọi \(x;y\in R\) thì:

\(\left\{{}\begin{matrix}\left|x-y-2\right|\ge0\\\left|y+3\right|\ge0\end{matrix}\right.\) \(\Rightarrow\left|x-y-2\right|+\left|y+3\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|x-y-2\right|=0\\\left|y+3\right|=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\)

\(A=1+11+....+11^7+11^8+11^9\)

\(11A=11\left(1+11+.....+11^7+11^8+11^9\right)\)

\(\)\(11A=11+11^2+....+11^8+11^9+11^{10}\)

\(11A-A=\left(11+11^2+....+11^8+11^9+11^{10}\right)-\left(1+11+....+11^7+11^8+11^9\right)\)\(10A-11^{10}-1\)

\(A=\dfrac{11^{10}-1}{10}\)

Được biết:\(11^n=\overline{....1}\)

Nên: \(11^{10}-1=\overline{....1}-1=\overline{....0}\)

\(A=\dfrac{11^{10}-1}{10}=\dfrac{\overline{....0}}{10}=\overline{...0}⋮5\)

\(D=11^9+11^8+11^7+...+11+1\)
Đặt 11D= \(11^{10}+11^9+11^8+...+11^2+11\)
=> 11D-D= \(\left(11^{10}+11^9+11^8+...+11^2+11\right)-\left(11^9+11^8+11^7+...+11+1\right)\)=> 10D= \(11^{10}-1\)
=> D= \(11^{10}-1:10\)
Ta thấy: \(11^{10}\) có tận cùng là 1 mà \(11^{10}-1\) sẽ có tận cùng là 0
Mà 0 chia hết cho 5 =>\(11^{10}-1:10\) chia hết cho 5
Vậy....(đpcm)

Ta có :

\(D=11^9+11^8+.........+11+1\)

\(\Leftrightarrow11A=11^{10}+11^9+..........+11\)

\(\Leftrightarrow11A-A=\left(11^{10}+11^9+......+11\right)-\left(11^9+11^8+......+1\right)\)

\(\Leftrightarrow10A=11^{10}-1\)

\(\Leftrightarrow A=\dfrac{11^{10}-1}{10}\)

Ta thấy \(A=\dfrac{11^{10}-1}{10}=\dfrac{\left(....1\right)-1}{10}=\dfrac{\left(...0\right)}{10}=\left(...0\right)⋮5\)

\(\Leftrightarrowđpcm\)

\(S_n=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+....+\dfrac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)

\(S_n=\dfrac{1}{3}\left(\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+....+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}-\dfrac{1}{n\left(n+2\right)\left(n+3\right)}\right)\)\(S_n=\dfrac{1}{3}\left(\dfrac{1}{2.3.4}-\dfrac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\right)\)

\(S_n=\dfrac{1}{3}\left(\dfrac{1}{24}-\dfrac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\right)\)

\(S_n=\dfrac{1}{72}-\dfrac{1}{3\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)

\(\left(2x+3\right)^2=\dfrac{9}{121}\)

\(\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)

\(\left(2x+3\right)=\dfrac{3}{11}\)

\(2x=\dfrac{3}{11}-3\)

\(2x=-\dfrac{14}{11}\)

\(x=-\dfrac{7}{11}\)

\(\left(2x+3\right)^2=\dfrac{9}{121}\)

\(\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)

\(\Rightarrow2x+3=\dfrac{3}{11}\)

=> 2x = \(\dfrac{3}{11}-3\)

=> 2x = \(-\dfrac{30}{11}\)

=> x = \(-\dfrac{30}{11}\div2=-\dfrac{15}{11}\)

\(\left(2x+3\right)^2=\dfrac{9}{121}\)

\(\Rightarrow\left(2x+3\right)^2=\left(\pm\dfrac{3}{11}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}2x+3=\dfrac{3}{11}\\2x+3=-\dfrac{3}{11}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=-\dfrac{30}{11}\\2x=-\dfrac{36}{11}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{15}{11}\\x=-\dfrac{18}{11}\end{matrix}\right.\)