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3/ Chu vi hình chữ nhật:
\(\left(\dfrac{1}{4}+\dfrac{3}{10}\right)\cdot2=\dfrac{11}{10}\) (chưa biết đơn vị)
Diện tích hình chữ nhật:
\(\dfrac{1}{4}\cdot\dfrac{3}{10}=\dfrac{11}{20}\) (chưa biết đơn vị)
(sữa đề tìm \(x\) nguyên )
\(2^x+3+2^x=144\Leftrightarrow2^x+2^x=141\)
ta có : \(2^x+2^x\) là số chẳn
mà \(141\) là số lẽ \(\Rightarrow\) phương trình vô nghiệm
\(\dfrac{2n-1}{n+1}=\dfrac{2\left(n+1\right)-3}{n+1}\)
Để \(\dfrac{2\left(n+1\right)-3}{n+1}\in Z\Rightarrow3⋮n+1\)
\(\Rightarrow n+1\inƯ\left(3\right)=\left\{-1;-3;1;3\right\}\)
\(n+1=-1\Rightarrow n=-2\)
\(n+1=-3\Rightarrow n=-4\)
\(n+1=1\Rightarrow n=0\)
\(n+1=3\Rightarrow n=2\)
\(\dfrac{x-7}{y-6}=\dfrac{7}{6}\)
\(\Leftrightarrow6\left(x-7\right)=7\left(y-6\right)\)
\(6x-42=7y-42\)
\(6x=7y\Leftrightarrow x=\dfrac{7}{6}y\)
\(x=-4:\left(7-6\right).7=-28\)
\(y=-28-4=-24\)
b tương tự
Giải:b)
\(\dfrac{x-7}{y-6}=\dfrac{7}{6}\) nên \(6\left(x-7\right)=7\left(y-6\right)\)
Do đó \(6x-42=7y-42\) nên \(6x=7y\)
Suy ra \(6x-6y=y\) hay \(6\left(x-y\right)=y\)
Nên 6.(-4) = y
Vậy y = -24, x = \(\dfrac{7.\left(-24\right)}{6}\)= -28
c)
\(\dfrac{x+3}{y+5}=\dfrac{3}{5}\) nên \(5\left(x+3\right)=3\left(y+5\right)\)
Do đó \(5x+15=3y+15\) nên \(5x=3y\)
Suy ra \(5x+5y=3y+5y\)
\(5\left(x+y\right)=8y\)
\(5.16=8y\)
Nên \(y=\dfrac{5.16}{8}=\dfrac{80}{8}=10\)
Vậy y = 10, x = 16 - 10 =6
)
Gọi \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}\)là \(S\)
\(S=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}\\ S>\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{100\cdot101}\\ S>\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{100}-\dfrac{1}{101}\\ S>\dfrac{1}{5}-\dfrac{1}{101}>\dfrac{1}{5}\)
Vậy \(S>\dfrac{1}{5}\)(đpcm)
Đề sai, tớ sửa lại
Ta có :
\(A=2+2^2+..............+2^{60}\)
\(\Leftrightarrow A=\left(2+2^2\right)+\left(2^3+2^4\right)+...........+\left(2^{59}+2^{60}\right)\)
\(\Leftrightarrow A=2\left(1+2\right)+2^3\left(1+2\right)+.........+2^{59}\left(1+2\right)\)
\(\Leftrightarrow A=2.3+2^3.3+...........+2^{59}.3\)
\(\Leftrightarrow A=3\left(2+2^2+..........+2^{59}\right)\)
\(\Leftrightarrow A⋮3\rightarrowđpcm\)
Lại có :
\(A=2+2^2+2^3+............+2^{60}\)
\(\Leftrightarrow A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+..........+\left(2^{58}+2^{59}+2^{60}\right)\)
\(\Leftrightarrow A=2\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+..........+2^{59}\left(1+2+2^2\right)\)
\(\Leftrightarrow A=2.7+2^4.7+............+2^{58}.7\)
\(\Leftrightarrow A=7\left(2+2^3+..........+2^{58}\right)\)
\(\Leftrightarrow A⋮7\rightarrowđpcm\)
Ta tiếp tục có :
\(A=2+2^2+2^3+............+2^{60}\)
\(\Leftrightarrow A=\left(2+2^2+2^3+2^4\right)+..............+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(\Leftrightarrow A=2\left(1+2+2^2+2^3\right)+.............+2^{57}\left(1+2+2^2+2^3\right)\)
\(\Leftrightarrow A=2.15+............+2^{57}.15\)
\(\Leftrightarrow A=15\left(2+.........+2^{57}\right)\)
\(\Leftrightarrow A⋮15\rightarrowđpcm\)
Ta có: ( x + 2)( x - 5) = -12
=> \(x+2\inƯ\left(-12\right);x-5\inƯ\left(-12\right)\)
mà Ư (-12) = \(\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}x+2\in\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\\x-5\in\left\{"....."\right\}\end{matrix}\right.\)
Xét các t/h:
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2012^2}+\dfrac{1}{2013^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2011.2012}+\dfrac{1}{2012.2013}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2012}-\dfrac{1}{2013}\)
\(=1-\dfrac{1}{2013}\)
\(\Rightarrow A< 1-\dfrac{1}{2013}\)
\(\Rightarrow A< 1\) ( đpcm )
mình gợi ý nè :
Chứng minh A <\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(\dfrac{2}{3^2}+\dfrac{2}{4^2}+\dfrac{2}{5^2}+....\dfrac{2}{2016^2}\)
Ta thấy: \(\dfrac{2}{3^2}< \dfrac{2}{2.3}\)
\(\dfrac{2}{4^2}< \dfrac{2}{3.4}\)
...\(\dfrac{2}{2016^2}< \dfrac{2}{2015.2016}\)
Đặt:A=\(\dfrac{2}{3^2}+\dfrac{2}{4^2}+\dfrac{2}{5^2}+...+\dfrac{2}{2016^2}\)
=>\(A< \dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{2015.2016}\)
=>\(A< \dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+\dfrac{2}{4}-\dfrac{2}{5}+...+\dfrac{2}{2015}-\dfrac{2}{2016}\)
=>A<\(\dfrac{2}{2}-\dfrac{2}{2016}\)
=>A<\(\dfrac{1007}{1008}\) mà \(\dfrac{1007}{1008}\) < 1
=>A<1
Vậy \(\dfrac{2}{3^2}+\dfrac{2}{4^2}+\dfrac{2}{5^2}+...+\dfrac{2}{2016^2}\)<1 (\(đpcm\))
\(\dfrac{2}{3^2}+\dfrac{2}{4^2}+...+\dfrac{2}{2016^2}=2\left(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}\right)\)
Ta có: \(\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{2016^2}< \dfrac{1}{2015.2016}\)
\(\Rightarrow2\left(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}\right)< 2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2015.2016}\right)\)
\(\Rightarrow2\left(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}\right)< 2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)
\(\Rightarrow2\left(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}\right)< 2\left(\dfrac{1}{2}-\dfrac{1}{2017}\right)=1-\dfrac{2}{2017}< 1\)
=> đpcm