Thay \(a=t\) nhé ,t nhìn trất's hơn

\(M=\left(t+1\right)\left(t+2\right)\left(t+3\right)\left(t+4\right)+1\)

\(M=\left[\left(t+1\right)\left(t+4\right)\right]\left[\left(t+2\right)\left(t+3\right)\right]+1\)

\(M=\left[t\left(t+4\right)+1\left(t+4\right)\right]\left[t\left(t+3\right)+2\left(t+3\right)\right]+1\)

\(M=\left(t^2+4t+t+4\right)\left(t^2+3t+2t+6\right)+1\)

\(M=\left(t^2+5t+4\right)\left(t^2+5t+6\right)+1\)

\(M=\left(t^2+5t+5-1\right)\left(t^2+5t+5+1\right)+1\)

\(M=\left(t^2+5t+5\right)^2-1+1\)

\(M=\left(t^2+5t+5\right)^2\)

Vì \(t\in Z\Rightarrow\left(t^2+5t+5\right)\in Z\)

Nên \(M\) là bình phương của 1 số nguyên (đpcm)

Sửa đề:

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{b+c+1}=\dfrac{b}{a+c+1}=\dfrac{c}{a+b-2}=\dfrac{a+b+c}{b+c+1+a+c+1+a+b+-2}=\dfrac{a+b+c}{\left(b+c+a+c+a+b\right)+\left(1+1-2\right)}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)

Tương đương với:

\(\left\{{}\begin{matrix}\dfrac{a}{b+c+1}=\dfrac{1}{2}\\\dfrac{b}{a+c+1}=\dfrac{1}{2}\\\dfrac{c}{a+c-2}=\dfrac{1}{2}\\a+b+c=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b+c+1=2a\\a+c+1=2b\\a+c-2=2c\\a+b+c=\dfrac{1}{2}\end{matrix}\right.\)

\(\circledast\) Từ \(a+b+c=\dfrac{1}{2}\Leftrightarrow b+c=\dfrac{1}{2}-a\)

Nên \(\dfrac{1}{2}-a+1=2a\)(tự tìm a nhé dễ lắm)

\(\circledast\) Từ \(a+b+c=\dfrac{1}{2}\Leftrightarrow a+c=\dfrac{1}{2}-b\)

Nên \(\dfrac{1}{2}-b+1=2b\)(tự tính b)

\(\circledast\) Từ \(a+b+c=\dfrac{1}{2}\Leftrightarrow a+b=\dfrac{1}{2}-c\)

Nên\(\dfrac{1}{2}-c-2=2c\)(tự tính c)

Vậy...

Gọi 3 số tự nhiên liên tiếp đó là \(t;t+1;t+2\)

Theo đề bài ta có:

\(t^2+\left(t+1\right)^2=\left(t+2\right)^2\)

\(\Rightarrow t^2+t^2+2t+1=t^2+4t+4\)

\(\Rightarrow2t^2+2t+1=t^2+4t+4\)

\(\Rightarrow\left(2t^2+2t+1\right)-\left(t^2+2t\right)=\left(t^2+4t+4\right)-\left(t^2+2t\right)\)

\(\Rightarrow t^2+1=2t+4\)

\(\Rightarrow t^2+1-2t-4=0\)

\(\Rightarrow t^2-3-2t=0\)

\(\Rightarrow t^2-3t+t-3=0\)

\(\Rightarrow t\left(t+1\right)-3\left(t+1\right)=0\)

\(\Rightarrow\left(t-3\right)\left(t+1\right)=0\)

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

Vì \(t\in N\Rightarrow t=3\)

\(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)

\(\Leftrightarrow\dfrac{ab}{cd}=\dfrac{a^2-2ab+b^2}{c^2-2cd+d^2}\)

\(\Rightarrow ab\left(c^2-2cd+d^2\right)=cd\left(a^2-2ab+b^2\right)\)

\(\Rightarrow c^2ab-2abcd+d^2ab=a^2cd-2abcd+b^2cd\)

\(\Rightarrow c^2ab-d^2ab=a^2cd-b^2cd\)

Ko hiểu đề ,làm tới đây thui :)

\(\left\{{}\begin{matrix}\left(-\dfrac{1}{2}\right)^{19}< 0\\\left(\dfrac{1}{4}\right)^9>0\end{matrix}\right.\)

Suy ra:\(\left(-\dfrac{1}{2}\right)^{19}< \left(\dfrac{1}{4}\right)^9\)

\(\left(x-2\right)\left(x+\dfrac{2}{3}\right)< 0\)

Với mọi \(x\in R\) thì:

\(x-2< x+\dfrac{2}{3}\) Nên \(\left\{{}\begin{matrix}x-2< 0\\x+\dfrac{2}{3}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 2\\x>-\dfrac{2}{3}\end{matrix}\right.\)

Suy ra\(-\dfrac{2}{3}< x< 2\)

\(\left(x+1\right)\left(x-2\right)< 0\)

Với mọi \(x\in R\) thì:

\(x-2< x+1\) nên \(\left\{{}\begin{matrix}x-2< 0\\x+1>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 2\\x>-1\end{matrix}\right.\)

Suy ra \(-1< x< 2\)

Áp dụng bđt AM-GM cho 2 số dương:

\(a^3+b^3+c^3\ge3abc\)

Dấu "=" xảy ra khi:

\(a=b=c\)

Khi đó:

\(\left\{{}\begin{matrix}\dfrac{a}{b}=1\\\dfrac{b}{c}=1\\\dfrac{a}{c}=1\end{matrix}\right.\) \(\Leftrightarrow\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{a}{c}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

\(A=\left|x-2009\right|+\left|x-3\right|\)

\(A=\left|x-2009\right|+\left|3-x\right|\)

\(A\ge\left|x-2009+3-x\right|\)

\(A\ge2006\)

Dấu "=" xảy ra khi:

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2009\ge0\Rightarrow x\ge2009\\3-x\ge0\Rightarrow x\le3\end{matrix}\right.\\\left\{{}\begin{matrix}x-2009\le0\Rightarrow x\le2009\\3-x\le0\Rightarrow x\ge3\end{matrix}\right.\end{matrix}\right.\)

Suy ra \(3\le x\le2009\)

ko có công thức đâu bạn 

mk học rồi ko có quy luật đâu