\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)

\(=\sqrt{1}=1\)

\(\frac{2}{x^2-4x+3}+\frac{2}{x^2-8x+15}+\frac{2}{x^2-12x+35}=-\frac{1}{2}\)(x khác 1;3;5;7)

<=>\(\frac{2}{x^2-3x-x+3}+\frac{2}{x^2-5x-3x+15}+\frac{2}{x^2-5x-7x+35}=-\frac{1}{2}\)

<=>\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{2}{\left(x-3\right)\left(x-5\right)}+\frac{2}{\left(x-5\right)\left(x-7\right)}=-\frac{1}{2}\)

<=>\(\frac{1}{x-3}-\frac{1}{x-1}+\frac{1}{x-5}-\frac{1}{x-3}+\frac{1}{x-7}-\frac{1}{x-5}=-\frac{1}{2}\)

<=>\(\frac{1}{x-7}-\frac{1}{x-1}=-\frac{1}{2}\)

<=>\(2x-2-2x+14=-x^2+8x-7\)

<=>\(x^2-8x+19=0\)

<=>(x-4)²+3=0(vô lí)

Vậy PT vô nghiệm

\(=3.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)=3.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(=3.\left(1-\frac{1}{100}\right)=\frac{297}{100}\)

\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)

\(=1-\frac{1}{n}< 1\)

=>điều cần chứng minh

\(A=\frac{100^{2016}+8}{9}=\frac{100....0\left(4032\text{ số }0\right)+8}{9}\)

\(=\frac{999....9\left(4032\text{ số }9\right)+9}{9}=111...1\left(4032\text{ số }1\right)+1=111...12\left(4031\text{ số }1\right)\)

Vậy A là số tự nhiên

\(A=\frac{2016^{2016}+1}{2016^{2017}+1}\Rightarrow2016A=\frac{2016^{2017}+2016}{2016^{2017}+1}=1+\frac{2015}{2016^{2017}+1}\)

\(B=\frac{2016^{2017}-3}{2016^{2018}-3}\Rightarrow2016B=\frac{2016^{2018}-6048}{2016^{2018}-3}=1+\frac{-6045}{2016^{2018}-3}\)

Vì \(\frac{2015}{2016^{2017}+1}>0;\frac{-6045}{2016^{2018}-3}< 0\)

Nên: A>B

2n²+3n+2 = 2n²+2n+n+1+1=2n.(n+1)+(n+1)+1

Để (2n²+3n+2) chia hết cho n+1 thì: 1 chia hết cho n+1

=>n+1 thuộc Ư(1)={1;-1}

=>n=0;-2

Vậy n=0,-2 thì (2n²+3n+2) chia hết cho n+1

hình tự vẽ

a) Vì góc xOy và góc yOz là 2 góc kề bù nên : góc xOy+góc yOz = 180^o

=>góc yOz=180^o-góc xOy=180^o-60^o=120^o

b)Vì Ot là tia p/g của góc xOy nên: \(\widehat{xOt}=\widehat{yOt}=\frac{\widehat{xOy}}{2}=\frac{60^o}{2}=30^o\)

\(P=\frac{2!}{3!}+\frac{2!}{4!}+\frac{2!}{5!}+...+\frac{2!}{n!}=2!.\left(\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+...+\frac{1}{n!}\right)\)

\(< 2!.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{\left(n-1\right).n}\right)=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{n}\right)=1-\frac{1}{n}< 1\)

=>điều phải chứng minh

301 đúng thì like hk đúng thì thôi