a/

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

\(\Leftrightarrow x^2+y^2-2.35=4\Leftrightarrow x^2+y^2=74\)

\(\Rightarrow\left(x^2+y^2\right)^2=x^4+y^4+2x^2y^2=74^2\)

\(\Rightarrow x^4+y^4=74^2-2.\left(-35\right)^2\)

b/

\(\left(x^4+y^4\right)\left(x+y\right)=x^5+x^4y+xy^4+y^5\)

\(\Leftrightarrow x^5+y^5=\left(x^4+y^4\right)\left(x+y\right)-xy\left(x^3+y^3\right)\)(1)

Ta có

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

Thay các giá trị đã tính được vào (1) Bạn tự tính nốt nhé

\(A=2^{2+4+6+...+100}=2^{\dfrac{50\left(2+100\right)}{2}}=2^{2250}\)

20=1.20=2.10=4.5

=> b={5;10;20} => a={4;2;1}

\(=3^{25}.15⋮15\)

\(=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}=3^{28}-3^{27}-3^{26}=\)

\(=3^{26}\left(3^2-3-1\right)=3^{26}.5⋮5\)

\(3^{200}=\left(3^2\right)^{100}=9^{100}>8^{100}=\left(2^3\right)^{100}=2^{300}\)

\(\overline{6a14b}⋮20\) Khi đồng thời chia hết cho 4 và 5 

\(\overline{6a14b}⋮4\) => b chẵn 

\(\overline{6a14b}⋮5\Rightarrow b=\left\{0;5\right\}\) do b chẵn => b=0

\(\Rightarrow\overline{6a14b}=\overline{6a140}⋮3\Rightarrow6+a+1+4=11+a⋮3\)

\(\Rightarrow a=\left\{1;4;7\right\}\)

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

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

\(=\left(x+y\right)^2-4xy=15^2-4.40=65\)

\(VT=\cos^2a-2.\dfrac{1}{2}\left[\cos\left(a+b\right)+\cos\left(a-b\right)\right].\cos\left(a+b\right)+\cos^2\left(a+b\right)=\)

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

\(=\cos^2a-\dfrac{1}{2}\left(\cos2a+\cos2b\right)=\)

\(=\dfrac{2\cos^2a-\cos^2a+\sin^2a-1+2\sin^2b}{2}=\)

\(=\dfrac{\left(\cos^2a+\sin^2a\right)-1+2\sin^2b}{2}=\sin^2b=VP\)

\(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}=\)

\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{100-99}{99.100}=\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}< 1\)