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\(a;43^2+43.17=43\left(43+17\right)=43.60⋮60\left(đpcm\right)\)
\(b;27^5-3^{11}=3^{15}-3^{11}=3^{11}\left(3^4-1\right)=3^{11}.80⋮80\left(đpcm\right)\)
=x^2-3xy+3y
nho k de ung ho mik nhe
\(\frac{x^4-y^4}{y^3-x^3}\)
\(=\frac{\left(x^2+y^2\right)\left(x^2-y^2\right)}{\left(y-x\right)\left(y^2+xy+x^2\right)}\)
\(=\frac{\left(x^2+y^2\right)\left(x+y\right)\left(x-y\right)}{\left(y-x\right)\left(y^2+xy+x^2\right)}\)
\(=-\frac{\left(x^2+y^2\right)\left(x+y\right)\left(x-y\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=-\frac{\left(x^2+y^2\right)\left(x+y\right)}{x^2+xy+y^2}\)
\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(P=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(P=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(P=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(P=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(P=\frac{1}{2}\left(5^{32}+1\right)\)
\(x^2-2x-3=0\)
\(\Leftrightarrow x^2-x+3x-3=0\)
\(\Leftrightarrow x\left(x-1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=1\end{cases}}\)
\(\left(2x^2-2x+3\right)\left(2x^2+2x-3\right)\)
\(=\left(2x^2\right)^2-\left(2x-3\right)^2\)
\(=4x^4-\left(4x^2-12x+9\right)\)
\(=4x^4-4x^2+12x-9\)
\(\left(3x^2-x-1\right)\left(3x^2+x-1\right)\)
\(=\left(3x^2-1\right)^2-x^2\)
\(=9x^4-6x^2+1-x^2\)
\(=9x^4-7x^2+1\)
a)ta co: 125x^3+y^6=(5x)^3+(y^2)^3=(5x+y^2)(5x-5xy^2+y^2) b)ta co 5xy^2-10xyz+5xz^2=5x(y^2-2yz+z^2)=5x(y-z)^2 (may cau sau gan giong ban tu lam nha)
b) \(5xy^2-10xyz+5xz^2\)
\(=5xy^2-5xyz-5xyz+5xz^2\)
\(=5xy\left(y-z\right)-5xz\left(y-z\right)\)
\(=\left(y-z\right)\left(5xy-5xz\right)\)
\(=5x\left(y-z\right)\left(y-z\right)\)
\(=5x\left(y-z\right)^2\)