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

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

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

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

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

\(\)

1. a.) \(\cot a=\dfrac{1}{\tan a}=\dfrac{1}{\sqrt{3}}\)

\(\tan\sqrt{3}=60\Rightarrow a=60^o\)

\(\sin60=\dfrac{\sqrt{3}}{2}\)

\(\cos60=\dfrac{1}{2}\)

b.) \(\cos^2a=1-\left(\dfrac{15}{17}\right)^2=\dfrac{64}{289}\Rightarrow\cos a=\dfrac{8}{17}\)

\(\tan a=\dfrac{\sin a}{\cos a}=\dfrac{\dfrac{15}{17}}{\dfrac{8}{17}}=\dfrac{15}{17}.\dfrac{17}{8}=\dfrac{15}{8}\)

2. \(\left(\sin a+\cos a\right)^2+\left(\sin a-\cos a\right)^2+2\)

\(=\sin^2a+2.\sin a.\cos a+\cos^2a+\sin^2a\cdot2.\sin a.\cos a+\cos^2a+2\)

\(=2\sin^2a+2\cos^2a+2\)

\(=2\left(\sin^2a+\cos^2a\right)+2\)

\(=2.1+2=4\)

=> biểu thức trên ko phụ thuộc vào a

( 12,32 - 1,765 ) x 3,9

= 10,555 x 3,9

= 41,1645

\(\dfrac{x}{2}=\dfrac{y}{3}\Leftrightarrow3x=2y\Rightarrow x=\dfrac{2y}{3}\) thay vào xy=10 ta có:

\(\dfrac{2y}{3}.y=10\Leftrightarrow2y^2=30\Leftrightarrow y^2=15\Leftrightarrow y=\sqrt{15}\)

\(\Rightarrow x=\dfrac{10}{\sqrt{15}}\)

Thay a+c=2b vào 2bd=c(b+d) ta có:

(a+c)d=cd+cb

<=> ad+cd=cd+cb

<=> ad=cb

<=> \(\dfrac{a}{b}=\dfrac{c}{d}\)

(a+b+c)³ - a³-b³-c³

= a³ +b³ +c³ + 3(a+b)(a+c)(b+c) - a³ - b³ - c³

= 3(a+b)(a+c)(b+c)

(a + b + c)³ - a³ - b³ - c³

= (a+b)³ + 3(a+b)c² +3(a+b)²c+c³- a³ - b³ - c³

= a³+3a²b+3ab²+b³+3(a²+2ab+b²)c+3(a+b)c²+c³- a³ - b³ - c³

= 3a²b+3ab³+3a²c+6abc+3b²c+3c²a+3c²b

= (3c²b+6abc+3a²b)+3ac²+(3ab²+3b²c)+3a²c

= 3b(c²+2ac+a²)+3ac²+3b²(a+c)+3a²c

= 3b(c+a)²+3ac(a+c)+3b²(a+c)

= 3(a+c)[b(a+c)+ac(a+c)+b²]

\(\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\)

\(=x^3+1-x^3+1\)

\(=2\)