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![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(C=\dfrac{11}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta nhận thấy tổng các hệ số của pt bậc 2 đã cho là \(1-a+a-1=0\) nên pt này có 1 nghiệm là 1, nghiệm kia là \(a-1\), nhưng do không được giải pt nên ta sẽ làm theo cách sau:
Ta thấy pt này luôn có 2 nghiệm phân biệt. Theo hệ thức Viète:
\(\left\{{}\begin{matrix}x_1+x_2=a\\x_1x_2=a-1\end{matrix}\right.\)
Vậy, \(M=\dfrac{3\left(x_1^2+x_2^2\right)-3}{x_1x_2\left(x_1+x_2\right)}\)
\(M=\dfrac{3\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-3}{a\left(a-1\right)}\)
\(M=\dfrac{3\left(a^2-2\left(a-1\right)\right)-3}{a\left(a-1\right)}\)
\(M=\dfrac{3\left[\left(a-1\right)^2-1\right]}{a\left(a-1\right)}\)
\(M=\dfrac{3a\left(a+2\right)}{a\left(a-1\right)}\)
\(M=\dfrac{3a+6}{a-1}\)
b) Ta có \(P=\left(x_1+x_2\right)^2-2x_1x_2=a^2-2\left(a-1\right)=\left(a-1\right)^2\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow a=1\). Vậy để P đạt GTNN thì \(a=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\Delta=\left(-10\right)^2-4.3.2=100-24=76>0\)
Suy ra pt luôn có 2 nghiệm phân biệt
Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{10}{3}\\x_1x_2=\dfrac{2}{3}\end{matrix}\right.\)
\(A=\dfrac{x_1-1}{x_2}+\dfrac{x_2-1}{x_1}-x_1^2x_2^2\\ =\dfrac{x_1\left(x_1-1\right)+x_2\left(x_2-1\right)}{x_1x_2}-\left(x_1x_2\right)^2\\ =\dfrac{x_1^2-x_1+x_2^2-x_2}{\dfrac{2}{3}}-\left(\dfrac{2}{3}\right)^2\\ =\dfrac{\left(x_1+x_2\right)^2-2x_1x_2-\left(x_1+x_2\right)}{\dfrac{2}{3}}-\dfrac{4}{9}\)
\(=\dfrac{\left(\dfrac{10}{3}\right)^2-2.\dfrac{2}{3}-\dfrac{10}{3}}{\dfrac{2}{3}}-\dfrac{4}{9}\\ =\dfrac{83}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=4\\x_1x_2=-5\end{matrix}\right.\)
\(D=5-\dfrac{x_2}{x_1}-\dfrac{x_1}{x_2}+3=8-\dfrac{x_1^2+x_2^2}{x_1x_2}=8-\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1x_2}=8-\dfrac{\left(-4\right)^2-10}{5}=...\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{\left(x_1+x_2\right)^2+3x_1x_2}{4x_1x_2\left(x_1+x_2\right)}=\dfrac{9+3}{4\cdot1\left(-3\right)}=\dfrac{12}{-12}=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Áp dụng hệ thức Viet:
$x_1+x_2=\frac{-4}{3}; x_1x_2=\frac{1}{3}$
Khi đó:
\(B=\frac{x_1}{x_2-1}+\frac{x_2}{x_1-1}=\frac{x_1(x_1-1)+x_2(x_2-1)}{(x_1-1)(x_2-1)}\)
\(=\frac{x_1^2+x_2^2-(x_1+x_2)}{x_1x_2-(x_1+x_2)+1}=\frac{(x_1+x_2)^2-2x_1x_2-(x_1+x_2)}{x_1x_2-(x_1+x_2)+1}\)
\(=\frac{(\frac{-4}{3})^2-2.\frac{1}{3}-\frac{-4}{3}}{\frac{1}{3}-\frac{-4}{3}+1}=\frac{11}{12}\)
Ptr có: `\Delta'=(-2)^2-3.(-2)=10 > 0`
`=>` Ptr có `2` `n_o` pb
`=>` Áp dụng Vi-ét có: `{(x_1+x_2=[-b]/a=4/3),(x_1.x_2=c/a=[-2]/3):}`
Có: `A=[3x_1 ^2-2]/[x_1]+[3x_2 ^2-2]/[x_2]`
`A=[x_2(3x_1 ^2-2)+x_1(3x_2 ^2-2)]/[x_1.x_2]`
`A=[3x_1 ^2.x_2-2x_2+3x_1.x_2 ^2-2x_1]/[x_1.x_2]`
`A=[3x_1.x_2(x_1+x_2)-2(x_1+x_2)]/[x_1.x_2]`
`A=[3 . [-2]/3 . 4/3-2 . 4/3]/[[-2]/3]=8`