a⁵+a+1=a⁵+a⁴+a³+a²+a+1-a⁴-a³-a²

=a³.(a²+a+1)+(a²+a+1)-a².(a²+a+1)

=(a²+a+1)(a³-a²+1)

Ta có : \(a^5+a+1=\left(a^5-a^2\right)+\left(a^2+a+1\right)\)

\(=a^2\left(a^3-1\right)+\left(a^2+a+1\right)\)

\(=a^2\left(a-1\right)\left(a^2+a+1\right)+\left(a^2+a+1\right)\)

\(=\left(a^2+a+1\right)\left(a^3-a^2+1\right)\)

a) \(x^2+4y^2+4xy\)

\(=x^2+2.x.2y+\left(2y\right)^2\)

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

b) \(\left(x+y\right)^2-\left(x-y\right)^2\)

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

\(=2y.2x\)

\(=4xy\)

c) \(\left(3x+1\right)^2-\left(x+1\right)^2\)

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

a) \(x^6-y^6=\left(x^2\right)^3-\left(y^2\right)^3\)

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

a) \(2x\left(3x+5\right)=3x\left(10+2x\right)+15\)

\(\Leftrightarrow6x^2+10x=30x+6x^2+15\)

\(\Leftrightarrow6x^2+10x-30x-6x^2=15\)

\(\Leftrightarrow-20x=15\)

\(\Leftrightarrow x=-0.75\)

b) \(3x\left(x+5\right)-x\left(3x-10\right)+7=0\)

\(\Leftrightarrow3x^2+15x-3x^2+10x+7=0\)

\(\Leftrightarrow25x+7=0\)

\(\Leftrightarrow25x=-7\)

\(\Leftrightarrow x=-0.28\)

Ta có : \(\left(x-3\right)^2\ge0\Leftrightarrow-\left(x-3\right)^2\le0\Leftrightarrow-\left(x-3\right)^2-17\le-17\)

Vậy Max = -17 <=> x = 3

\(x^2-2\)

\(=x^2-\left(\sqrt{2}\right)^2\)

\(=\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\)

\(y^2-13\)

\(=y^2-\left(\sqrt{13}\right)^2\)

\(=\left(y-\sqrt{13}\right)\left(y+\sqrt{13}\right)\)

a) Ta xét: Tam giác ADE có: AD = AE

=> Tam giác ADE cân tại A

\(\Rightarrow\widehat{AED}=\widehat{ACB}\)

=> DE//BC

Ta xét: Tứ giác DECB có: DE//BC

\(\Rightarrow\widehat{ABC}=\widehat{ACB}\)

=> BDEC là hình thang cân

b) \(\widehat{ABC}=\frac{1}{2}\left(180^o-50^o\right)=65^o\)

\(\widehat{ACB}=\widehat{ABC}=65^o\)

\(\widehat{DEC}=180^o-65^o=115^o\)

\(\widehat{EDB}=\widehat{EDC}=115^o\)

Áp dụng bất đẳng thức \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\) với \(x=a^2+2bc;y=b^2+2ac;z=c^2+2ab\)

Ta có : \(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge\frac{9}{a^2+b^2+c^2+2\left(ab+bc+ac\right)}=\frac{9}{\left(a+b+c\right)^2}\)

\(\Rightarrow\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge9\)( Vì a + b + c = 1)