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

Bài 2: 

\(a^2+b^2=\left(a+b\right)^2-2ab=5^2-2\cdot\left(-2\right)=9\)

\(\dfrac{1}{a^3}+\dfrac{1}{b^3}=\dfrac{a^3+b^3}{a^3b^3}=\dfrac{\left(a+b\right)^3-3ab\left(a+b\right)}{\left(ab\right)^3}\)

\(=\dfrac{5^3-3\cdot5\cdot\left(-2\right)}{\left(-2\right)^3}=\dfrac{125+30}{8}=\dfrac{155}{8}\)

\(a-b=-\sqrt{\left(a+b\right)^2-4ab}=-\sqrt{5^2-4\cdot\left(-2\right)}=-\sqrt{33}\)

Ta có công thức tổng quát như sau:

\(A=n^k+n^{k+1}+n^{k+2}+...+n^{k+x}\Rightarrow A=\dfrac{n^{k+x+1}-n^k}{n-1}\)

Áp dụng ta có:

\(A=1+4+4^2+...+4^6=\dfrac{4^7-1}{3}\) 

\(\Rightarrow B-3A=4^7-3\cdot\dfrac{4^7-1}{3}=1\)

______

\(A=2^0+2^1+...+2^{2008}=2^{2009}-1\)

\(\Rightarrow B-A=2^{2009}-2^{2009}+1=1\)

_____

\(A=1+3+3^2+....+3^{2006}=\dfrac{3^{2007}-1}{2}\)

\(\Rightarrow B-2A=3^{2007}-2\cdot\dfrac{3^{2007}-1}{2}=1\)

\(\begin{array}{l}A + B + C\\ = (3{x^4} - 2{x^3} - x + 1) + ( - 2{x^3} + 4{x^2} + 5x) + ( - 3{x^4} + 2{x^2} + 5)\\ = 3{x^4} - 2{x^3} - x + 1 - 2{x^3} + 4{x^2} + 5x - 3{x^4} + 2{x^2} + 5\\ = (3{x^4} - 3{x^4}) + ( - 2{x^3} - 2{x^3}) + (4{x^2} + 2{x^2}) + ( - x + 5x) + (1 + 5)\\ = 0 + ( - 4{x^3}) + 6{x^2} + 4x + 6\\ =  - 4{x^3} + 6{x^2} + 4x + 6\\A - B + C\\ = (3{x^4} - 2{x^3} - x + 1) - ( - 2{x^3} + 4{x^2} + 5x) + ( - 3{x^4} + 2{x^2} + 5)\\ = 3{x^4} - 2{x^3} - x + 1 + 2{x^3} - 4{x^2} - 5x - 3{x^4} + 2{x^2} + 5\\ = (3{x^4} - 3{x^4}) + ( - 2{x^3} + 2{x^3}) + ( - 4{x^2} + 2{x^2}) + ( - x - 5x) + (1 + 5)\\ = 0 + 0 + ( - 2{x^2}) - 6x + 6\\ =  - 2{x^2} - 6x + 6\\A - B - C\\ = (3{x^4} - 2{x^3} - x + 1) - ( - 2{x^3} + 4{x^2} + 5x) - ( - 3{x^4} + 2{x^2} + 5)\\ = 3{x^4} - 2{x^3} - x + 1 + 2{x^3} - 4{x^2} - 5x + 3{x^4} - 2{x^2} - 5\\ = (3{x^4} + 3{x^4}) + ( - 2{x^3} + 2{x^3}) + ( - 4{x^2} - 2{x^2}) + ( - x - 5x) + (1 - 5)\\ = 6{x^4} + 0 + ( - 6{x^2}) - 6x + ( - 4)\\ = 6{x^4} - 6{x^2} - 6x - 4\end{array}\)

2:

a: Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{24}{9}=\dfrac{8}{3}\)

=>x=16/3; y=8; z=32/3

A=3x+2y-6z

=3*16/3+2*8-6*32/3

=16+16-64

=-32

b: Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y+z}{5-6+7}=\dfrac{6\sqrt{2}}{6}=\sqrt{2}\)

=>x=5căn 2; y=6căn 2; y=7căn 2

B=xy-yz

=y(x-z)

=6căn 2(5căn 2-7căn 2)

=-6căn 2*2căn 2

=-24

bài 1 a)áp dụng dãy tỉ số bằng nhau ta có:\(\dfrac{a+b+c}{3+4+5}\)=\(\dfrac{24}{12}\)=2

a=2.3=6 ; b=2.4=8 ;c=2.5=10

M=ab+bc+ac=6.8+8.10+6.10=48+80+60=188

"nhưng bài còn lại làm tương tự"

Bài 1

a) 3 2/5 - 1/2

= 17/5 - 1/2

= 34/10 - 5/10

= 29/10

b) 4/5 + 1/5 × 3/4

= 4/5 + 3/20

= 16/20 + 3/20

= 19/20

c) 3 1/2 × 1 1/7

= 7/2 × 8/7

= 4

d) 4 1/6 : 2 1/3

= 25/6 : 7/3

= 25/14

Bài 2

a) 3 × 1/2 + 1/4 × 1/3

= 3/2 + 1/12

= 18/12 + 1/12

= 19/12

b) 1 4/5 - 2/3 : 2 1/3

= 9/5 - 2/3 : 7/3

= 9/5 - 2/7

= 63/35 - 10/35

= 53/35

Bài 1

a) 3 2/5 - 1/2

= 17/5 - 1/2

= 34/10 - 5/10

= 29/10

b) 4/5 + 1/5 × 3/4

= 4/5 + 3/20

= 16/20 + 3/20

= 19/20

c) 3 1/2 × 1 1/7

= 7/2 × 8/7

= 4

d) 4 1/6 : 2 1/3

= 25/6 : 7/3

= 25/14

bủh

\(\begin{array}{l}A + B = (6{x^4} - 4{x^3} + x - \dfrac{1}{3}) + ( - 3{x^4} - 2{x^3} - 5{x^2} + x + \dfrac{2}{3})\\ = 6{x^4} - 4{x^3} + x - \dfrac{1}{3} - 3{x^4} - 2{x^3} - 5{x^2} + x + \dfrac{2}{3}\\ = (6{x^4} - 3{x^4}) + ( - 4{x^3} - 2{x^3}) - 5{x^2} + (x + x) + ( - \dfrac{1}{3} + \dfrac{2}{3})\\ = 3{x^4} - 6{x^3} - 5{x^2} + 2x + \dfrac{1}{3}\\A - B = (6{x^4} - 4{x^3} + x - \dfrac{1}{3}) - ( - 3{x^4} - 2{x^3} - 5{x^2} + x + \dfrac{2}{3})\\ = 6{x^4} - 4{x^3} + x - \dfrac{1}{3} + 3{x^4} + 2{x^3} + 5{x^2} - x - \dfrac{2}{3}\\ = (6{x^4} + 3{x^4}) + ( - 4{x^3} + 2{x^3}) + 5{x^2} + (x - x) + ( - \dfrac{1}{3} - \dfrac{2}{3})\\ = 9{x^4} - 2{x^3} + 5{x^2} - 1\end{array}\)\(\begin{array}{l}A + B = (6{x^4} - 4{x^3} + x - \dfrac{1}{3}) + ( - 3{x^4} - 2{x^3} - 5{x^2} + x + \dfrac{2}{3})\\ = 6{x^4} - 4{x^3} + x - \dfrac{1}{3} - 3{x^4} - 2{x^3} - 5{x^2} + x + \dfrac{2}{3}\\ = (6{x^4} - 3{x^4}) + ( - 4{x^3} - 2{x^3}) - 5{x^2} + (x + x) + ( - \dfrac{1}{3} + \dfrac{2}{3})\\ = 3{x^4} - 6{x^3} - 5{x^2} + 2x + \dfrac{1}{3}\\A - B = (6{x^4} - 4{x^3} + x - \dfrac{1}{3}) - ( - 3{x^4} - 2{x^3} - 5{x^2} + x + \dfrac{2}{3})\\ = 6{x^4} - 4{x^3} + x - \dfrac{1}{3} + 3{x^4} + 2{x^3} + 5{x^2} - x - \dfrac{2}{3}\\ = (6{x^4} + 3{x^4}) + ( - 4{x^3} + 2{x^3}) + 5{x^2} + (x - x) + ( - \dfrac{1}{3} - \dfrac{2}{3})\\ = 9{x^4} - 2{x^3} + 5{x^2} - 1\end{array}\)

\(A=a^2+b^2=\left(a+b\right)^2-2ab=3^2-2.\left(-5\right)=19\)

\(B=a^4+b^4=\left(a^2+b^2\right)^2-2\left(ab\right)^2=19^2-2.\left(-5\right)^2=311\)