Xét ta giác ABC vuông tại A, đường cao AH

* Áp dụng hệ thức :

 \(AB^2=BH.BC=BH.\left(CH+BH\right)\Rightarrow25=BH\left(\frac{144}{13}+BH\right)\Rightarrow BH=\frac{25}{13}\)cm 

\(\Rightarrow BC=HB+HC=\frac{144}{13}+\frac{25}{13}=\frac{196}{13}\)

* Áp dụng hệ thức : \(AC^2=HC.BC=\frac{144}{13}.\frac{169}{13}=144\Rightarrow AC=12\)cm 

sin 2a = 2 sin a cos a

cos 2a = cos²a - sin²a = 2cos²a - 1 = 1 - 2sin²a

\(tan2a=\frac{2tana}{1-tan^2a}\)

Bạn áp dụng công thức này để tính cos ( 2x ) 

Xét vế trái (VT) ta có 

VT = \(\left(ac+bd\right)^2+\left(ad-bc\right)^2\)

\(=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\)

\(=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)

\(=\left(a^2c^2+a^2d^2\right)+\left(b^2d^2+b^2c^2\right)\)

\(=a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)\)

\(=\left(a^2+b^2\right)\left(c^2+d^2\right)\)= VP

Vậy ...

hc tốt

chịu thôi

Bài 1 : Với \(x>0;x\ne1\)

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

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

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

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

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

b, Ta có : \(\frac{2\sqrt{x}+2}{\sqrt{x}-1}=\frac{2\left(\sqrt{x}-1\right)+5}{\sqrt{x}-1}=2+\frac{5}{\sqrt{x}-1}\)

\(\Rightarrow\sqrt{x}-1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)