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

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

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

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

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

a) Ta có: \(BC=\sqrt{AB^2+AC^2}=\sqrt{16^2+12^2}=20\left(cm\right)\)

Ta có: \(AB.AC=AH.BC\Rightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{12.16}{20}=\dfrac{48}{5}\left(cm\right)\)

Ta có: \(AB^2=BH.BC\Rightarrow BH=\dfrac{AB^2}{BC}=\dfrac{16^2}{20}=\dfrac{64}{5}\left(cm\right)\)

Ta có: \(sinB=\dfrac{AC}{BC}=\dfrac{12}{20}=\dfrac{3}{5}\Rightarrow\angle B\approx37\)

b) tam giác AHE vuông tại H có HN là đường cao \(\Rightarrow AN.AE=AH^2\)

tam giác ABC vuông tại A có AH là đường cao \(\Rightarrow AH^2=HB.HC\)

\(\Rightarrow AN.AE=HB.HC\)

c) tam giác AHB vuông tại H có HM là đường cao \(\Rightarrow AH^2=AM.AB\)

\(\Rightarrow AN.AE=AM.AB\Rightarrow\dfrac{AM}{AE}=\dfrac{AN}{AB}\)

Xét \(\Delta AMN\) và \(\Delta AEB:\) Ta có: \(\left\{{}\begin{matrix}\angle EABchung\\\dfrac{AM}{AE}=\dfrac{AN}{AB}\end{matrix}\right.\)

\(\Rightarrow\Delta AMN\sim\Delta AEB\left(c-g-c\right)\Rightarrow\dfrac{AE}{AM}=\dfrac{BE}{MN}\)

mà \(BE=3MN\Rightarrow\dfrac{BE}{MN}=3\Rightarrow\dfrac{AE}{AM}=3\Rightarrow AE=3AM\)

giả sử tam giác ABC vuông tại A có \(tanC=2\)

\(\Rightarrow\dfrac{AB}{AC}=2\Rightarrow AB=2AC\Rightarrow BC=\sqrt{AC^2+AB^2}\)

\(=\sqrt{AC^2+4AC^2}=\sqrt{5}AC\)

\(\Rightarrow sinC=\dfrac{AB}{BC}=\dfrac{2AC}{\sqrt{5}AC}=\dfrac{2}{\sqrt{5}}\Rightarrow sin\alpha=\dfrac{2}{\sqrt{5}}\)

\(cosC=\dfrac{AC}{BC}=\dfrac{AC}{\sqrt{5}AC}=\dfrac{1}{\sqrt{5}}\Rightarrow cos\alpha=\dfrac{1}{\sqrt{5}}\)

\(cotC=\dfrac{AC}{AB}=\dfrac{AC}{2AC}=\dfrac{1}{2}\Rightarrow cot\alpha=\dfrac{1}{2}\)

a) \(\sqrt{6,25.4,9.640}=\sqrt{\dfrac{25}{4}.4,9.10.64}=\sqrt{\dfrac{25}{4}.64.49}=\sqrt{25.16.49}\)

\(=\sqrt{\left(5.4.7\right)^2}=\left|5.4.7\right|=140\)

b) \(\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}\)

\(=\sqrt{5-\sqrt{21}}.\sqrt{5+\sqrt{21}}.\sqrt{5+\sqrt{21}}.\sqrt{2}.\left(\sqrt{7}-\sqrt{3}\right)\)

\(=\sqrt{25-21}.\sqrt{10+2\sqrt{21}}\left(\sqrt{7}-\sqrt{3}\right)\)

\(=2.\sqrt{\left(\sqrt{7}\right)^2+2.\sqrt{7}.\sqrt{3}+\left(\sqrt{3}\right)^2}\left(\sqrt{7}-\sqrt{3}\right)\)

\(=2.\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}\left(\sqrt{7}-\sqrt{3}\right)=2\left|\sqrt{7}+\sqrt{3}\right|\left(\sqrt{7}-\sqrt{3}\right)\)

\(=2\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)=2.4=8\)

c) \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}=\sqrt{\dfrac{8+2\sqrt{7}}{2}}-\sqrt{\dfrac{8-2\sqrt{7}}{2}}-\sqrt{2}\)

\(=\sqrt{\dfrac{\left(\sqrt{7}\right)^2+2.\sqrt{7}.1+1^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}\right)^2-2.\sqrt{7}.1+1^2}{2}}-\sqrt{2}\)

\(=\sqrt{\dfrac{\left(\sqrt{7}+1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}-1\right)^2}{2}}-\sqrt{2}=\dfrac{\left|\sqrt{7}+1\right|}{\sqrt{2}}-\dfrac{\left|\sqrt{7}-1\right|}{\sqrt{2}}-\sqrt{2}\)

\(=\dfrac{\sqrt{7}+1}{\sqrt{2}}-\dfrac{\sqrt{7}-1}{\sqrt{2}}-\sqrt{2}=\dfrac{2}{\sqrt{2}}-\sqrt{2}=0\)

d) \(\sqrt{14+\sqrt{40}+\sqrt{56}+\sqrt{140}}=\sqrt{14+\sqrt{4.10}+\sqrt{4.14}+\sqrt{4.35}}\)

\(=\sqrt{14+2\sqrt{10}+2\sqrt{14}+2\sqrt{35}}\)

\(=\sqrt{\left(\sqrt{5}\right)^2+\left(\sqrt{2}\right)^2+\left(\sqrt{7}\right)^2+2.\sqrt{2}.\sqrt{7}+2.\sqrt{2}.\sqrt{5}+2.\sqrt{5}.\sqrt{7}}\)

\(=\sqrt{\left(\sqrt{5}+\sqrt{7}+\sqrt{2}\right)^2}=\sqrt{5}+\sqrt{7}+\sqrt{2}\)

Vì \(sin\alpha=\dfrac{\sqrt{3}}{2}\Rightarrow\alpha=60\)

\(\Rightarrow cos\alpha=cos60=\dfrac{1}{2}\)

\(tan\alpha=tan60=\sqrt{3}\)

\(cot\alpha=cot60=\dfrac{1}{\sqrt{3}}\)

\(sinC=\dfrac{AB}{BC}=\dfrac{3}{5}\)

\(cosC=\dfrac{AC}{BC}=\dfrac{4}{5}\)

\(tanC=\dfrac{AB}{AC}=\dfrac{3}{4}\)

\(cotC=\dfrac{AC}{AB}=\dfrac{4}{3}\)

trừ là khi hắn cùng dấu,thí dụ như là \(\left\{{}\begin{matrix}ax+by=c\\ax+my=n\end{matrix}\right.\)

còn cộng là khi hắn trái dấu như là \(\left\{{}\begin{matrix}ax+by=c\\-ax+my=n\end{matrix}\right.\)

Gọi các điểm như hình vẽ:

\(cos\alpha=\dfrac{AC}{BC}=\dfrac{250}{320}=\dfrac{25}{32}\Rightarrow\alpha\approx38\)

Ta có: \(\angle B=90-45=45\Rightarrow\Delta ABC\) vuông cân tại A

\(\Rightarrow b=c=10\left(cm\right)\Rightarrow a=\sqrt{b^2+c^2}=\sqrt{10^2+10^2}=10\sqrt{2}\left(cm\right)\)

\(\left\{{}\begin{matrix}5x+3y=1\\-x+2y=-8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}5x+3y=1\left(1\right)\\-5x+10y=-40\left(2\right)\end{matrix}\right.\)

Lấy \(\left(1\right)+\left(2\right)\Rightarrow13y=-39\Rightarrow y=-3-x=-8-2.\left(-3\right)=-2\)

\(\Rightarrow x=2\)