Đặt \(CD=x,BC=y\left(x,y>0\right)\)

Ta có \(AB=\sqrt{BD^2-AD^2}=12\)

Ta có hệ phương trình: \(\hept{\begin{cases}\frac{x}{y}=\frac{AD}{AB}=\frac{4}{12}=\frac{1}{3}\\12^2+\left(4+x\right)^2=y^2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=3x\\144+\left(4+x\right)^2=\left(3x\right)^2\end{cases}}\Leftrightarrow\hept{\begin{cases}y=3x\\x=5\left(h\right)x=-4\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=5\\y=15\end{cases}}\)(Vì \(x,y>0\))

Vậy \(S_{ABC}=\frac{AB.\left(AD+CD\right)}{2}=\frac{12.\left(4+5\right)}{2}=54.\)

Đặt \(\sqrt{x}=a\ge0\)

\(\Rightarrow A=\frac{a^2+a+1}{a^2+2a+1}\)

\(\Leftrightarrow\left(A-1\right)a^2+\left(2A-1\right)a+A-1=0\)

Để PT theo nghiệm a có nghiệm thì 

\(\Delta=\left(2A-1\right)^2-4\left(A-1\right)\left(A-1\right)\ge0\)

\(\Leftrightarrow4A-3\ge0\)

\(\Leftrightarrow A\ge\frac{3}{4}\)

Ta lại có: \(A=\frac{a^2+a+1}{a^2+2a+1}=1-\frac{a}{a^2+2a+1}\le1\)

Vậy ...

Ta có:

\(x=\frac{1}{2}.\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}=\frac{\sqrt{2}-1}{2}\)

\(\Rightarrow x\left(x+1\right)=\frac{\sqrt{2}-1}{2}.\frac{\sqrt{2}+1}{2}=\frac{1}{4}\)

Thế vô bài toán ta được

\(A=\left(4x^5+4x^4-5x^3+5x-2\right)^{2016}+2017\)

\(=\left(4x^4\left(x+1\right)-5x^3+5x-2\right)^{2016}+2017\)

\(=\left(-4x^3+5x-2\right)^{2016}+2017\)

\(=\left(\left(-4x^3-4x^2\right)+\left(4x^2+4x\right)+x-2\right)^{2016}+2017\)

\(=\left(-x+1+x-2\right)^{2016}+2017\)

\(=\left(-1\right)^{2016}+2017=2018\)

bạn làm rõ hơn được k ạ? mik k hiểu lắm

Ta có:

\(\frac{n\left(n+2\right)}{\left(n+1\right)^2}=1-\frac{1}{\left(n+1\right)^2}>1-\frac{1}{n\left(n+2\right)}=1+\frac{1}{2}.\left(\frac{1}{n+2}-\frac{1}{n}\right)\)

Thế vô bài toán ta được

\(B=\frac{2.4}{3^2}+\frac{4.6}{5^2}+...+\frac{200.202}{201^2}\)

\(>1+1+...+1+\frac{1}{2}.\left(\frac{1}{4}-\frac{1}{2}+\frac{1}{6}-\frac{1}{4}+...+\frac{1}{202}-\frac{1}{200}\right)\)

\(=100+\frac{1}{2}.\left(\frac{1}{202}-\frac{1}{2}\right)=\frac{10075}{101}>99,75\)

Ta có đánh giá sau:\(\frac{n\left(n+2\right)}{\left(n+1\right)^2}=1-\frac{1}{\left(n+1\right)^2}\)

\(>1-\frac{1}{x\left(x+2\right)}=1-\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)\)

Suy ra \(B=\frac{2\cdot4}{3^2}+\frac{4\cdot6}{5^2}+\frac{6\cdot8}{7^2}+...+\frac{200\cdot202}{201^2}\)

\(>1-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}\right)+1-\frac{1}{2}\left(\frac{1}{4}-\frac{1}{6}\right)+...+1-\frac{1}{2}\left(\frac{1}{200}-\frac{1}{202}\right)\)

\(=100-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{200}-\frac{1}{202}\right)\)

\(=100-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{202}\right)\)\(=100-\frac{1}{2}\cdot\frac{50}{101}\)

\(>100-\frac{1}{2}\cdot\frac{50}{100}=100-0,25=99,75\)

Tức là \(B>99,75\) 

bấm máy tính

được thế đã tốt =))) phải giải hẳn ra cơ

2 cái đầu quy đồng r tính.... cái sau rút gọn cho căn 2