Tổng quát: 

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

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

Suy ra: \(2\left(\sqrt{n+1}-\sqrt{n}\right)< \frac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)

\(A=1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}< 1+2\left(\sqrt{2}-\sqrt{1}+...+\sqrt{100}-\sqrt{99}\right)\)

\(=1+2\left(\sqrt{100}-\sqrt{1}\right)=19\)

\(A=1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\right)\)

\(=2\left(\sqrt{100}-\sqrt{1}\right)=18\)

Do đó ta có đpcm. 

Dựng hình bình hành \(ABEC\).

Khi đó \(E\in DC\).

Vì \(BD\perp AC\)mà \(AC//BE\)nên \(BE\perp BD\).

Kẻ \(BH\perp DE\). 

Xét tam giác \(BED\)vuông tại \(B\)đường cao \(BH\): 

\(\frac{1}{BH^2}=\frac{1}{BD^2}+\frac{1}{BE^2}\Leftrightarrow\frac{1}{4^2}=\frac{1}{5^2}+\frac{1}{BE^2}\Leftrightarrow BE=\frac{20}{3}\left(cm\right)\)

\(S_{ABCD}=\frac{1}{2}.AC.BD=\frac{1}{2}.BD.BE=\frac{1}{2}.5.\frac{20}{3}=\frac{50}{3}\left(cm^2\right)\)

Có ai biết đổi tên cho mình hông?

\(\frac{3+\sqrt{3}}{3}+\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}=1+\frac{\sqrt{3}}{3}+\frac{\sqrt{3}.\sqrt{2}-\sqrt{3}}{1-\sqrt{2}}=1+\frac{\sqrt{3}}{3}+\frac{\sqrt{3}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}\)

\(=1+\frac{\sqrt{3}}{3}-\sqrt{3}\)

\(\frac{\sqrt{3}\left(\sqrt{3}+1\right)}{3}-\frac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}\)

\(\frac{\sqrt{3}+1}{\sqrt{3}}-\sqrt{3}\)

\(\frac{\sqrt{3}-2}{\sqrt{3}}\)

\(=\frac{3-2\sqrt{3}}{3}\)

\(\sqrt{193-132\sqrt{2}}+\sqrt{193+132\sqrt{2}}=\sqrt{121-2.11.6\sqrt{2}+72}+\sqrt{121+2.11.6\sqrt{2}+72}\)

\(=\sqrt{11^2-2.11.6\sqrt{2}+\left(6\sqrt{2}\right)^2}+\sqrt{11^2+2.11.6\sqrt{2}+\left(6\sqrt{2}\right)^2}\)

\(=\sqrt{\left(11-6\sqrt{2}\right)^2}+\sqrt{\left(11+6\sqrt{2}\right)^2}=\left|11-6\sqrt{2}\right|+\left|11+6\sqrt{2}\right|\)

\(=11-6\sqrt{2}+11+6\sqrt{2}=22\)

p/s : cách khác 

Đặt \(Nghia=\sqrt{193-132\sqrt{2}}+\sqrt{193+132\sqrt{2}}\)

\(\Rightarrow Nghia^2=193-132\sqrt{2}+193+132\sqrt{2}+2\sqrt{\left(193-132\sqrt{2}\right)\left(193+132\sqrt{2}\right)}\)

\(=386+2\sqrt{2401}=386+2.49=484\)

\(\Rightarrow Nghia=\sqrt{484}=22\)

\(\sqrt{28-10\sqrt{3}}+\sqrt{28+10\sqrt{3}}=\sqrt{25-2.5.\sqrt{3}+3}+\sqrt{25+2.5.\sqrt{3}+3}\)

\(=\sqrt{5^2-2.5.\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{5^2+2.5.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(5-\sqrt{3}\right)^2}+\sqrt{\left(5+\sqrt{3}\right)^2}\)

\(=\left|5-\sqrt{3}\right|+\left|5+\sqrt{3}\right|=5-\sqrt{3}+5+\sqrt{3}=10\)

\(\frac{1+sin^2x}{1-sin^2x}=\frac{cos^2x+sin^2x+sin^2x}{cos^2x+sin^2x-sin^2x}=\frac{cos^2x+2sin^2x}{cos^2x}=1+2\left(\frac{sinx}{cosx}\right)^2=1+2tan^2x\)

a) \(cos^4x-sin^4x=\left(cos^2x+sin^2x\right)\left(cos^2x-sin^2x\right)=cos^2x-sin^2x\)

b) \(\frac{1}{1+tanx}+\frac{1}{1+cotx}=\frac{1}{1+tanx}+\frac{tanxcotx}{tanxcotx+cotx}=\frac{1}{1+tanx}+\frac{tanx}{tanx+1}\)

\(=\frac{1+tanx}{1+tanx}=1\)

c) Ta có: \(1+tan^2x=1+\frac{sin^2x}{cos^2x}=\frac{cos^2x+sin^2x}{cos^2x}=\frac{1}{cos^2x}\)

\(\Rightarrow\frac{1}{1+tan^2x}=cos^2x\)

Tương tự \(\frac{1}{1+tan^2y}=cos^2y\)

\(\Rightarrow cos^2x-cos^2y=\frac{1}{1+tan^2x}-\frac{1}{1+tan^2y}\)

\(cos^2x-cos^2y=\left(1-sin^2x\right)-\left(1-sin^2y\right)=sin^2y-sin^2x\)

d) \(\frac{1+sin^2x}{1-sin^2x}=\frac{cos^2x+sin^2x+sin^2x}{cos^2x+sin^2x-sin^2x}=\frac{cos^2x+2sin^2x}{cos^2x}=1+2\left(\frac{sinx}{cosx}\right)^2=1+2tan^2x\)

a, \(P=\left(\frac{x\sqrt{x}}{\sqrt{x}+1}+\frac{x^2}{x\sqrt{x}+1}\right)\left(2-\frac{1}{\sqrt{x}}\right)\)ĐK : \(x\ge0;\sqrt{x}+1>0\)

\(=\left(\frac{x\sqrt{x}\left(x-\sqrt{x}+1\right)+x^2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\left(\frac{2\sqrt{x}-1}{\sqrt{x}}\right)\)

\(=\left(\frac{x^2\sqrt{x}-x^2+x\sqrt{x}+x^2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\left(\frac{2\sqrt{x}-1}{\sqrt{x}}\right)\)

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

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

b, \(P=0\Rightarrow\frac{x\left(x+1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=0\Leftrightarrow x\left(x+1\right)\left(2\sqrt{x}-1\right)=0\)

\(\Leftrightarrow x=0;x=-1;x=\frac{1}{4}\)Kết hợp với đk vậy \(x=0;x=\frac{1}{4}\)

Trên mạng nó thế

\(x^3-7x^2+11x-4+2\sqrt{\left(x-1\right)^3}=0\) (ĐK: \(x\ge1\)) 

\(\Leftrightarrow\left(x-1\right)^3-4\left(x-1\right)^2+2\sqrt{\left(x-1\right)^3}+1=0\)

Đặt \(t=\sqrt{x-1}\ge0\)

Phương trình ban đầu tương đương với: 

\(t^6-4t^4+2t^3+1=0\)

\(\Leftrightarrow\left(t^3+1\right)^2-\left(2t^2\right)^2=0\)

\(\Leftrightarrow\left(t^3-2t^2+1\right)\left(t^3+2t^2+1\right)=0\)

\(\Leftrightarrow t^3-2t^2+1=0\)(vì \(t^3+2t^2+1>0\)) 

\(\Leftrightarrow\left(t-1\right)\left(t^2-t-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t=1\\t=\frac{1+\sqrt{5}}{2}\end{cases}}\)(vì \(t\ge0\)) 

\(\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{5+\sqrt{5}}{2}\end{cases}}\)(thỏa mãn)