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cau noi tren S

\(\left(2+1\right)\left(2^2+1\right)...\left(2^{16}+1\right)\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1\)

a, \(VT=\left(a+b\right)^2-\left(a-b\right)^2\)

\(=a^2+2ab+b^2-a^2+2ab-b^2\)

\(=4ab=VP\)

\(\Rightarrowđpcm\)

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

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

\(=a^3+b^3=VT\)

\(\Rightarrowđpcm\)

a, \(x^2-2x=0\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

Vậy...

b, \(\left(3x-1\right)^2-16=0\)

\(\Leftrightarrow\left(3x-5\right)\left(3x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3x-5=0\\3x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)

Vậy...

a, \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)

\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)

\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)

Do \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\)

\(\Rightarrow x+1=0\Rightarrow x=-1\)

Vậy x = -1

b, \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

Vì \(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\)

\(\Rightarrow x+2004=0\Rightarrow x=-2004\)

Vậy...

a, \(6x^2-xy-y^2\)

\(=6x^2-3xy+2xy-y^2\)

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

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

b, \(8x^2-23x-3\)

\(=8x^2-24x+x-3\)

\(=8x\left(x-3\right)+\left(x-3\right)=\left(8x+1\right)\left(x-3\right)\)

c, \(10x^2-11x-6\)

\(=10x^2-15x+4x-6\)

\(=5x\left(2x-3\right)+2\left(2x-3\right)\)

\(=\left(5x+2\right)\left(2x-3\right)\)

d, \(x^3-6x^2+11x-6\)

\(=x^3-3x^2-3x^2+9x+2x-6\)

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

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

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

\(=\left(x-1\right)\left(x-2\right)\left(x-3\right)\)

a, Ta có: \(4\equiv1\left(mod3\right)\)

\(\Rightarrow4^{2018}\equiv1\left(mod3\right)\)

\(\Rightarrow4^{2018}-1⋮3\)

b, Ta có: \(5\equiv1\left(mod4\right)\)

\(\Rightarrow5^{2019}\equiv1\left(mod4\right)\)

\(\Rightarrow5^{2019}-1⋮4\)

c, \(4\equiv-1\left(mod5\right)\)

\(\Rightarrow4^{2019}\equiv-1\left(mod5\right)\)

\(\Rightarrow4^{2019}+1⋮5\)

d, \(5\equiv-1\left(mod6\right)\)

\(\Rightarrow5^{2017}\equiv-1\left(mod6\right)\)

\(\Rightarrow5^{2017}+1⋮6\)

\(x^3-6x^2-x+30=0\)

\(\Leftrightarrow x^3-5x^2-x^2+5x-6x+30=0\)

\(\Leftrightarrow x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)=0\)

\(\Leftrightarrow\left(x^2-x-6\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left(x^2-3x+2x-6\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[x\left(x-3\right)+2\left(x-3\right)\right]\left(x-5\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-3\right)\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\x=5\end{matrix}\right.\)

Vậy...