( a + 2 )³ - a( a - 3 )²

= a³ + 6a² + 12a + 8 - a( a² - 6a + 9 )

= a³ + 6a² + 12a + 8 - a³ + 6a² - 9a

= 12a² + 3a + 8

cách của symbolab:

\(\left(a+2\right)^3-a\left(a-3\right)^2\)

\(=a^3+6a^2+12a+8-a\left(a-3\right)^2\)

\(=a^3+6a^2+12a+8-a\left(a^2-6a+9\right)\)

\(=a^3+6a^2+12a+8-a^3+6a^2-9a\)

\(=12a^2+3a+8\)

\(\dfrac{2.3.5.\left(-7\right).12}{3.\left(-5\right).7.14}=\dfrac{2.3.5.\left(-7\right).12}{3.5.\left(-7\right).2.7}=\dfrac{12}{7}\)

\(\dfrac{2.3.5.\left(-7\right).12}{3.\left(-5\right).7.14}=\dfrac{1.1.1.\left(-1\right).12}{1.\left(-1\right).1.7}=\dfrac{-12}{-7}=\dfrac{12}{7}\)

Ta có : \(\frac{a}{b}=\frac{3}{4}\)(1)

 \(\frac{a+15}{b}=\frac{7}{6}\)

\(\Rightarrow\frac{a}{b}+\frac{15}{b}=\frac{7}{6}\)(2)

Thay (1) vào (2) ta có :

\(\frac{3}{4}+\frac{15}{b}=\frac{7}{6}\)

\(\Rightarrow\frac{15}{b}=\frac{7}{6}-\frac{3}{4}\)

\(\Rightarrow\frac{15}{b}=\frac{5}{12}\)

\(\Rightarrow5b=12.15\)

\(\Rightarrow b=12.15:5\)

\(\Rightarrow b=36\)

Thay b\(\Rightarrow a=27\) vào (1) ta có : 

\(\frac{a}{b}=\frac{a}{36}=\frac{3}{4}\)

\(\Rightarrow4a=36.3\)

\(\Rightarrow a=36.3:4\)

\(\Rightarrow a=27\)

\(\Rightarrow\frac{a}{b}=\frac{27}{36}\)

Cảm ơn bạn nhiều nha Xyz

a) \(A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)

\(A=\dfrac{x-5+2x+10-2x-10}{\left(x+5\right)\left(x-5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}=\dfrac{1}{x+5}\)

b) \(A=-3\Rightarrow\dfrac{1}{x+5}=-3\)

\(\Leftrightarrow x+5=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{1}{3}-5=\dfrac{-16}{3}\)

\(9x^2-42x+49=\left(3x-7\right)^2=\left(3.\dfrac{-16}{3}-7\right)^2=\left(-23\right)^2=529\) \(\left(x=\dfrac{-16}{3}\right)\)

chắc chết quá 

Bằng ???

a,\(\left(\sqrt{3}-\sqrt{2}\right)+\sqrt{2}=\sqrt{3}\) (vi \(\sqrt{3}>\sqrt{2}\) )

b,\(3\sqrt{5}-\left(\sqrt{5}-1\right)\) =\(3\sqrt{5}-\sqrt{5}+1=2\sqrt{5}+1\)  

c,\(\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)

Bạn ỏi, bài này mk làm đc rồi nhé ^^. Bạn có cần trợ giúp hông ??? Rất sẵn lòng :)