= 13/5 x 1/4 x 3/2

= 13 x 1 x 3/ 5 x 4 x 2

= 39/40

=13/5x1/4x3/2

=13x1x3/5x4x2

=39/40

= \(\dfrac{13}{5}\)x\(\dfrac{1}{4}\)x\(\dfrac{3}{2}\)
=  \(\dfrac{39}{40}\)

1. Theo hệ thức Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{4}{3}\\x_1.x_2=\dfrac{1}{3}\end{matrix}\right.\)

\(C=\dfrac{x_1}{x_2-1}+\dfrac{x_2}{x_1-1}=\dfrac{x_1\left(x_1-1\right)+x_2\left(x_2-1\right)}{\left(x_1-1\right)\left(x_2-1\right)}\)

   \(=\dfrac{x_1^2-x_1+x_2^2-x_2}{x_1x_2-x_1-x_2+1}=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2-\left(x_1+x_2\right)}{x_1x_2-\left(x_1+x_2\right)+1}\)

  \(=\dfrac{\left(-\dfrac{4}{3}\right)^2-2.\dfrac{1}{3}-\left(-\dfrac{4}{3}\right)}{\dfrac{1}{3}-\left(-\dfrac{4}{3}\right)+1}=\dfrac{\dfrac{22}{9}}{\dfrac{8}{3}}=\dfrac{11}{12}\)

\(1,3x^2+4x+1=0\)

Do pt có 2 nghiệm \(x_1,x_2\) nên theo đ/l Vi-ét ta có :

\(\left\{{}\begin{matrix}S=x_1+x_2=\dfrac{-b}{a}=-\dfrac{4}{3}\\P=x_1x_2=\dfrac{c}{a}=\dfrac{1}{3}\end{matrix}\right.\)

Ta có :

\(C=\dfrac{x_1}{x_2-1}+\dfrac{x_2}{x_1-1}\)

\(=\dfrac{x_1\left(x_1-1\right)+x_2\left(x_2-1\right)}{\left(x_2-1\right)\left(x_1-1\right)}\)

\(=\dfrac{x_1^2-x_1+x_2^2-x_2}{x_1x_2-x_2-x_1+1}\)

\(=\dfrac{\left(x_1^2+x_2^2\right)-\left(x_1+x_2\right)}{x_1x_2-\left(x_1+x_2\right)+1}\)

\(=\dfrac{S^2-2P-S}{P-S+1}\)

\(=\dfrac{\left(-\dfrac{4}{3}\right)^2-2.\dfrac{1}{3}-\left(-\dfrac{4}{3}\right)}{\dfrac{1}{3}-\left(-\dfrac{4}{3}\right)+1}\)

\(=\dfrac{11}{12}\)

Vậy \(C=\dfrac{11}{12}\)

\(3,3x^2-7x-1=0\)

Do pt có 2 nghiệm \(x_1,x_2\) nên theo đ/l Vi-ét ta có :

\(\left\{{}\begin{matrix}S=x_1+x_2=-\dfrac{b}{a}=\dfrac{7}{3}\\P=x_1x_2=\dfrac{c}{a}=-\dfrac{1}{3}\end{matrix}\right.\)

Ta có :

\(B=\dfrac{2x_2^2}{x_1+x_2}+2x_1\)

\(=\dfrac{2x_2^2+2x_1\left(x_1+x_2\right)}{x_1+x_2}\)

\(=\dfrac{2x_2^2+2x_1^2+2x_1x_2}{x_1+x_2}\)

\(=\dfrac{2\left(x_1^2+x_2^2\right)+2x_1x_2}{x_1+x_2}\)

\(=\dfrac{2\left(S^2-2P\right)+2P}{S}\)

\(=\dfrac{2\left(\dfrac{7}{3}^2-2\left(-\dfrac{1}{3}\right)\right)+2\left(-\dfrac{1}{3}\right)}{\dfrac{7}{3}}\)

\(=\dfrac{104}{21}\)

Vậy \(B=\dfrac{104}{21}\)

a: A=x1+x2=-5/2

b: \(=\dfrac{x_1+x_2}{x_1x_2}=\dfrac{-5}{2}:\left(-1\right)=\dfrac{5}{2}\)

c: \(=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\)

\(=\left(-\dfrac{5}{2}\right)^3-3\cdot\dfrac{-5}{2}\cdot\left(-1\right)\)

\(=-\dfrac{125}{8}-\dfrac{15}{2}=\dfrac{-185}{8}\)

e: \(E=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}\)

\(=\sqrt{\left(-\dfrac{5}{2}\right)^2-4\cdot\left(-1\right)}=\sqrt{\dfrac{25}{4}+4}=\dfrac{\sqrt{41}}{2}\)

Lời giải:
Áp dụng hệ thức Viet:

$x_1+x_2=\frac{-4}{3}; x_1x_2=\frac{1}{3}$

Khi đó:
\(B=\frac{x_1}{x_2-1}+\frac{x_2}{x_1-1}=\frac{x_1(x_1-1)+x_2(x_2-1)}{(x_1-1)(x_2-1)}\)

\(=\frac{x_1^2+x_2^2-(x_1+x_2)}{x_1x_2-(x_1+x_2)+1}=\frac{(x_1+x_2)^2-2x_1x_2-(x_1+x_2)}{x_1x_2-(x_1+x_2)+1}\) 

\(=\frac{(\frac{-4}{3})^2-2.\frac{1}{3}-\frac{-4}{3}}{\frac{1}{3}-\frac{-4}{3}+1}=\frac{11}{12}\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{5}{3}\\x_1x_2=-2\end{matrix}\right.\)

\(\dfrac{x_1}{x_2-1}+\dfrac{x_2}{x_1-1}=\dfrac{x_1\left(x_1-1\right)+x_2\left(x_2-1\right)}{\left(x_1-1\right)\left(x_2-1\right)}\)

\(=\dfrac{x_1^2+x_2^2-\left(x_1+x_2\right)}{x_1x_2-\left(x_1+x_2\right)+1}=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2-\left(x_1+x_2\right)}{x_1x_2-\left(x_1+x_2\right)+1}\)

\(=\dfrac{\left(-\dfrac{5}{3}\right)^2-2.\left(-2\right)-\left(-\dfrac{5}{3}\right)}{-2-\left(-\dfrac{5}{3}\right)+1}=...\)

\(...=\dfrac{3}{2}x\dfrac{4}{3}x\dfrac{5}{4}x\dfrac{6}{5}x\dfrac{7}{6}....x\dfrac{1000}{999}\)

\(=\dfrac{1}{2}x\dfrac{1000}{1}=500\)

=3/2x4/3x5/4x....x1000/999

=1/2x1000=500

mình chưa chắc là đúng đâu nhé