cos_a+sin_a=\(\dfrac{5}{4}\)

( cos_a+sin_a)²=\(\dfrac{5}{4}\)

1+2.sin_a.cos_a=\(\dfrac{25}{16}\)

sin_a.cos_a=\(\dfrac{9}{32}\)

28)  Ta có :  cot_B=\(\dfrac{BH}{AH}\)  và cot_C=\(\dfrac{HC}{AH}\)

=> cot_B+ cot_C=\(\dfrac{BH+HC}{AH}\) =\(\dfrac{BC}{AH}\) =\(\dfrac{2.BC.AH}{2.AH^2}\)=\(\dfrac{4h^2}{2h^2}\)=2

26) a) Vẽ AH vuông góc với BC 

=> Góc BAH = 30⁰ 

=> Tam giác ABH có góc B = 60⁰ , góc BAH = 90⁰ , góc BAH = 30⁰ và AB=16      

=> BH = 8  =>  AH = \(8\sqrt{3}\) 

Tam giác AHC vuông tại H suy ra HC = 2 

=> BC = BH + HC = 8 + 2 = 10 

b) S_ABC=10.\(\dfrac{10.8\sqrt{3}}{2}=40\sqrt{3}\)

a) P=\(\left(\dfrac{x-1}{\sqrt{x}}\right):\left(\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}\right)+\sqrt{x}\left(1-\sqrt{x}\right)}{\sqrt{x}\left(x+\sqrt{x}\right)}\right)\)

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

    P= \(\dfrac{x+\sqrt{x}}{\sqrt{x}}\)

a)S_MNPQ=   (MQ+NP).MN:2= (32+40).17:2= 612 cm²

b)  Kẻ QH vuông góc với NP => HP= 8 cm

Tam giác HQP vuông tại H => QP = \(\sqrt{353}\)

Sin_P=\(\dfrac{17}{\sqrt{353}}\) => Góc P= 64.79887635⁰∼\(65^{^{^0}}\)

Q=\(\dfrac{\left(3+2\sqrt{3}\right)\left(\sqrt{2}+1\right)+\sqrt{3}\left(2+\sqrt{2}\right)}{\sqrt{3}\left(\sqrt{2}+1\right)}-\left(\sqrt{2}+\sqrt{3}\right)\)

Q=\(\dfrac{3+4\sqrt{3}+3\sqrt{6}+3\sqrt{2}}{\sqrt{3}\left(\sqrt{2}+1\right)}\)-\(\left(\sqrt{2}+\sqrt{3}\right)\)

Q=\(\dfrac{3+4\sqrt{3}+3\sqrt{6}+3\sqrt{2}-2\sqrt{3}-3\sqrt{2}-\sqrt{6}-3}{\sqrt{3}\left(\sqrt{2}+1\right)}\)

Q=\(\dfrac{2\sqrt{3}-2\sqrt{6}}{\sqrt{3}\left(\sqrt{2}+1\right)}\)=\(\dfrac{\sqrt{4}-\sqrt{8}}{\sqrt{2}+1}\)

2)   Ta có  :\(\left(-3\right)^6=3^6=27^2\) 

                   \(\sqrt{\left(-2\right)^8}=\sqrt{2^8}=16\)

\(-4\sqrt{\left(-3\right)^6}+\sqrt{\sqrt{\left(-2\right)^8}}=-108+4=-104\)    

1) Ta có :   \(\left(-3\right)^6=3^6\)  = 27²

                  \(\left(-2\right)^4=2^4\)  = 4²

\(2\sqrt{\left(-3\right)^6}+3\sqrt{\left(-2\right)^4}\) =\(2\sqrt{27^2}+3\sqrt{4^2}=54+12=66\)

x>3 và x≥4 có giống nhau không