ta có: a + b=-2 ; a^2 + b^2 = 52

=> (a+b)^2 = 4 => a^2 + 2ab + b^2 = 4

=> 52 + 2ab= 4

=> 48= -2ab

=> ab= -24

a^3 + b^3 = (a+b)( a^2-ab+ b^2)

=> a^3 + b^3 = -2.(52+24)= -2. 76= -152

Ta có \(\left\{{}\begin{matrix}a+b=\dfrac{-2+\sqrt{3}}{3}+\dfrac{-2-\sqrt{3}}{3}=-\dfrac{4}{3}\\ab=\dfrac{\left(-2+\sqrt{3}\right)\left(-2-\sqrt{3}\right)}{9}=\dfrac{1}{9}\end{matrix}\right.\)

\(\left(a+b\right)^2=a^2+b^2+2ab=16\\ \Leftrightarrow a^2+b^2=\dfrac{16}{9}-2\cdot\dfrac{1}{9}=\dfrac{14}{9}\left(1\right)\\ \left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)=-\dfrac{64}{27}\\ \Leftrightarrow a^3+b^3+\dfrac{1}{3}\cdot\left(-\dfrac{4}{3}\right)=-\dfrac{64}{27}\\ \Leftrightarrow a^3+b^3=-\dfrac{64}{27}+\dfrac{4}{9}=-\dfrac{52}{27}\left(2\right)\\ \left(1\right)\left(2\right)\Leftrightarrow\left(a^2+b^2\right)\left(a^3+b^3\right)=a^5+b^5+a^2b^2\left(a+b\right)=\dfrac{14}{9}\cdot\left(-\dfrac{52}{27}\right)=-\dfrac{728}{243}\\ \Leftrightarrow a^5+b^5+\dfrac{1}{81}\cdot\left(-\dfrac{4}{3}\right)=-\dfrac{728}{243}\\ \Leftrightarrow a^5+b^5=-\dfrac{728}{243}+\dfrac{4}{243}=-\dfrac{724}{243}\left(3\right)\)

\(\left(1\right)\left(3\right)\Rightarrow\left(a^2+b^2\right)\left(a^5+b^5\right)=a^7+b^7+a^2b^2\left(a^3+b^3\right)=\dfrac{14}{9}\cdot\left(-\dfrac{724}{243}\right)=-\dfrac{10136}{2187}\\ \Leftrightarrow a^7+b^7+\dfrac{1}{81}\cdot\left(-\dfrac{52}{27}\right)=-\dfrac{10136}{2187}\\ \Leftrightarrow a^7+b^7=-\dfrac{10136}{2187}-\dfrac{52}{2187}=-\dfrac{10188}{2187}=\dfrac{1132}{243}\)

thay b=2/3, ta có: A=1/3x(2/3-2/5):7/5=4/63

thay A= 3/7, ta có: 3/7=1/3x(b-2/5):7/5

1/3x(b-2/5):7/5=3/7

1/3x(b-2/5)       =3/7x7/5

1/3x(b-2/5)       = 3/5

        b-2/5       = 3/5:1/3

        b-2/5       =    9/5

           b         =   9/5+2/5

          b          = 11/5

(a-b)^2 = a^2-2ab+b^2

3^2 =7 - 2ab

9= 7 -2ab

-2ab=7-9

-2ab= -2

ab= 1

Có a^3-b^3= (a-b)(a^2+ab+b^2)

a^3-b^3= 3. (7+1)

a^3-b^3= 24

Ta co : (a-b)²=a²-2ab+b² 

(a-b)²=a²+b²-2ab

Ma : a²+b² va a-b=3

\(\Rightarrow\)3²=7-2ab

7-3²=-2ab

-2=-2ab

\(\Leftrightarrow ab=1\) 

Ta lai co : a³-b³

=(a-b)(a²+ab+b²)

=(a-b)(a²+b²+ab)

=3.(7+1)

=24

a) Ta có: \(a^3+b^3\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)\)

Thay \(ab=40\) và \(a+b=-6\) vào biểu thức ta có

\(\left(-6\right)^3-3\cdot7\cdot\left(-6\right)=-90\)

b) Ta có: \(a^3-b^3\)

\(=\left(a-b\right)^3+3ab\left(a-b\right)\)

Thay \(ab=40\) và \(a-b=3\) vào biểu thức ta có:

\(3^3+3\cdot40\cdot3=387\)

a: a^3+b^3=(a+b)^3-3ab(a+b)

=(-6)^3-3*7*(-6)

=-90

b: a^3-b^3=(a-b)^3+3ab(a-b)

=3^3+3*40*3

=387

Ta có: \(a+b\sqrt{3}=\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)

\(\Leftrightarrow a+b\sqrt{3}=2-\sqrt{3}-2-\sqrt{3}\)

\(\Leftrightarrow a+b\sqrt{3}=-2\sqrt{3}\)

\(\Leftrightarrow a=0;b=-2\)

T=a+b=0+(-2)=-2

\(S=\sqrt{\left(\sqrt{3}\right)^2-2\cdot2\sqrt{3}+2^2}-\sqrt{\left(\sqrt{3}\right)^2+2\cdot2\cdot\sqrt{3}+2^2}\)

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

\(S=\left|\sqrt{3}-2\right|-\left|\sqrt{3}+2\right|=-\sqrt{3}+2-\sqrt{3}-2=0+\left(-2\right)\sqrt{3}\)

\(a=0,b=-2\)

\(T=0+-2=-2\)