Giải:

Gọi 3 cạnh của tam giác là a, b, c ( a, b, c > 0 )

Ta có: \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\) và a + b + c = 45

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

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{a+b+c}{2+3+4}=\dfrac{45}{9}=5\)

\(\Rightarrow\left\{{}\begin{matrix}a=10\\b=15\\c=20\end{matrix}\right.\)

Vậy 3 cạnh của tam giác lần lượt là 10, 15, 20

\(\left(x-3\right)\left(x-4\right)< 0\)

\(\Rightarrow\left\{{}\begin{matrix}x-3>0\\x-4< 0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x-3< 0\\x-4>0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x>3\\x< 4\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x< 3\\x>4\end{matrix}\right.\)

\(\Rightarrow3< x< 4\)

Vậy 3 < x < 4

\(\left(2x+y\right)\left(2z+y\right)+\left(x-y\right)\left(y-z\right)\)

\(=4xz+2xy+2yz+y^2+xy-xz-y^2+yz\)

\(=3xz+3xy+3yz\)

\(=3\left(xz+xy+yz\right)\)

Ta có: \(VT=a^3+b^3+c^3\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b\right)-3abc+3abc\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc\right)-3ab\left(a+b+c\right)+3abc\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc-3ab\right)+3abc\)

\(=3abc=VP\) ( do a + b + c = 0 )

\(\Rightarrowđpcm\)

\(A=3^2+3^3+...+3^{2006}\)

\(\Rightarrow3A=3^3+3^4+...+3^{2007}\)

\(\Rightarrow3A-A=\left(3^3+3^4+...+3^{2007}\right)-\left(3^2+3^3+...+3^{2006}\right)\)

\(\Rightarrow2A=3^{2007}-3^2\)

Do \(2A+3=3x\)

\(\Rightarrow3^{2007}-3^2+3=3x\)

\(\Rightarrow3^{2007}-6=3x\)

\(\Rightarrow x=3^{2006}-2\)

Vậy...

pho mat lam tu pho mai 

 l i k e nha

\(P=\dfrac{14^5.9^4-6^9.49^2}{2^{10}.49^3.3^8+6^8.7^5.13}\)

\(=\dfrac{2^5.7^5.3^8-2^9.3^9.7^4}{2^{10}.7^6.3^8+2^8.3^8.7^5.13}\)

\(=\dfrac{2^5.7^4.3^8\left(7-2^4.3\right)}{2^8.3^8.7^5\left(2^2.7+13\right)}\)

\(=\dfrac{-41}{2^3.7.41}\)

\(=\dfrac{-1}{56}\)

\(\left\{{}\begin{matrix}xy=\dfrac{3}{5}\\yz=\dfrac{4}{5}\\zx=\dfrac{3}{4}\end{matrix}\right.\Rightarrow x^2y^2z^2=\dfrac{3}{5}.\dfrac{4}{5}.\dfrac{3}{4}=\dfrac{9}{25}\)

\(\Rightarrow xyz=\pm\dfrac{3}{5}\)

+) \(xyz=\dfrac{3}{5}\Rightarrow\left\{{}\begin{matrix}z=1\\x=\dfrac{3}{4}\\y=\dfrac{4}{5}\end{matrix}\right.\)

+) \(xyz=\dfrac{-3}{5}\Rightarrow\left\{{}\begin{matrix}z=-1\\x=\dfrac{-3}{4}\\y=\dfrac{-4}{5}\end{matrix}\right.\)

Vậy...

\(\dfrac{x+1}{2011}+\dfrac{x+2}{2010}+\dfrac{x+3}{2009}+\dfrac{x+4}{2008}=-4\)

\(\Rightarrow\dfrac{x+1}{2011}+1+\dfrac{x+2}{2010}+1+\dfrac{x+3}{2009}+1+\dfrac{x+4}{2008}+1=0\)

\(\Rightarrow\dfrac{x+2012}{2011}+\dfrac{x+2012}{2010}+\dfrac{x+2012}{2009}+\dfrac{x+2012}{2008}=0\)

\(\Rightarrow\left(x+2012\right)\left(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}\right)=0\)

Mà \(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}\ne0\)

\(\Rightarrow x+2012=0\Rightarrow x=-2012\)

Vậy x = -2012

a, \(4x^3y^2-8x^2y^3+12x^3y^2\)

\(=4x^2y^2\left(x-2y+3x\right)\)

\(=4x^2y^2\left(4x-2y\right)\)

\(=8x^2y^2\left(2x-y\right)\)

b, \(x\left(2-x\right)^2-\left(2-x\right)^3\)

\(=\left(2-x\right)^2\left(x-2+x\right)\)

\(=\left(2-x\right)^2\left(2x-2\right)\)

\(=2\left(2-x\right)^2\left(x-2\right)\)

\(=2\left(x-2\right)^3\)