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

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)

\(ĐKXĐ:x\ne1;x\ge0\)

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

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

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

\(P=\frac{-3\sqrt{x}}{x\sqrt{x}-1}\)

Xét tam giác \(BGA\)vuông tại \(G\): 

\(BA^2=BG^2+GA^2=\frac{4}{9}\left(BE^2+AM^2\right)\Leftrightarrow BE^2+\frac{BC^2}{4}=\frac{27}{2}\)(1)

Xét tam giác \(ABE\)vuông tại \(A\): 

\(BE^2=AB^2+AE^2=6+\frac{1}{4}AC^2\)(2)

Từ (1) và (2) suy ra \(BC^2+AC^2=30\)

mà \(BC^2=AC^2+6\)

suy ra \(BC^2=18\Rightarrow BC=3\sqrt{2}\left(cm\right)\).

\(AB^2=BH.BC=HB.\left(HB+HC\right)=HB^2+15HB\)

\(\Leftrightarrow HB^2+15HB=16\Leftrightarrow HB=1\left(cm\right)\)

\(AB^2=BH.BC=\frac{1}{5}BC.BC\)

\(\Rightarrow BC=\sqrt{5AB^2}=10\left(cm\right)\)