a, sinC = \(\frac{AB}{BC}\); tanC = \(\frac{AB}{AC}\)

cosC = \(\frac{AC}{BC}\); cotC = \(\frac{AC}{AB}\)

b, Xét tam giác ABC vuông tại A, đường cao AH

tanB = \(\frac{AC}{AB}=\sqrt{2}\Rightarrow AC=\sqrt{2}AB\)

* Áp dụng hệ thức : \(\frac{1}{AH^2}=\frac{1}{AB^2}+\frac{1}{AC^2}\Rightarrow\frac{1}{12}=\frac{1}{AB^2}+\frac{1}{2AB^2}\Rightarrow AB\approx4,24\)cm 

\(\Rightarrow AC\approx4,24\sqrt{2}\)cm

Theo định lí Pytago tam giác ABC vuông tại A

\(BC=\sqrt{AB^2+AC^2}\approx\sqrt{4,24^2+\left(4,24\sqrt{2}\right)^2}\approx7,34\)cm 

\(Q=\frac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{\sqrt{x}-2}{1-\sqrt{x}}\)ĐK : \(x\ge0;x\ne1\)

\(=\frac{3x+3\sqrt{x}-3-\left(1-x\right)+x-4}{\left(\sqrt{x}+2\right)\left(1-\sqrt{x}\right)}=\frac{5x+3\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(1-\sqrt{x}\right)}=\frac{-8-5\sqrt{x}}{\sqrt{x}+2}\)

ta có : 

ta có :

\(2x-2+2\sqrt{2x-1}=14\)

\(\Leftrightarrow2x-1+2\sqrt{2x-1}+1=16\)

\(\Leftrightarrow\left(\sqrt{2x-1}+1\right)^2=16\Leftrightarrow\sqrt{2x-1}+1=4\)

\(\Leftrightarrow\sqrt{2x-1}=3\Leftrightarrow2x-1=9\Leftrightarrow x=5\)

ta có : 

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

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

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

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

a. ta có  : \(NP^2=MN^2+MP^2\Rightarrow\Delta MNP\) vuông tại M

b. ta có : \(MH=\frac{MN.MP}{NP}=\frac{12.5}{13}=\frac{60}{13}cm\)

\(PH=\frac{MP^2}{PN}=\frac{12^2}{13}=\frac{144}{13}cm\)

c. \(sin\widehat{N}=\frac{MP}{PN}=\frac{12}{13}\Rightarrow\widehat{N}\simeq67^0\)\(\Rightarrow\text{\widehat{P}=}90^0-\widehat{N}=23^0\)

\(\widehat{P}=90^0-\widehat{N}=23^0\)

Ta có: 

\(y=3x\left(k^2-k+1\right)=-2\)

\(\Rightarrow3\left(-3\right)\left(k^2-k+1\right)=-2\)

\(\Rightarrow-9\left(k^2-k+1\right)=-2\)

\(\Rightarrow k^2-k+1=\frac{2}{9}\)

\(\Rightarrow\left(k-\frac{1}{2}\right)^2=\frac{2}{9}-\frac{3}{4}=\frac{35}{36}\) (Vô nghiệm)