1. Cho pt \(3x^2+4x+1=0\)
có nghiệm x1,x2, không giải pt, hãy tính giá trị biểu thức \(C=\dfrac{x_1}{x_2-1}+\dfrac{x_2}{x_1-1}\)
2. . Cho pt \(3x^2-5x-1=0\)
có nghiệm x1,x2, không giải pt, hãy tính giá trị biểu thức \(D=\dfrac{x_1-x_2}{x_1}+\dfrac{x_2-1}{x_2}\)
3. . Cho pt \(3x^2-7x-1=0\)
có nghiệm x1,x2, không giải pt, hãy tính giá trị biểu thức \(B=\dfrac{2x^2_2}{x_1+x_2}+2x_1\)
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 \(B=\dfrac{104}{21}\)
3.Theo hệ thức Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{7}{3}\\x_1x_2=-\dfrac{1}{3}\end{matrix}\right.\)
\(B=\dfrac{2x_2^2}{x_1+x_2}+2x_1=\dfrac{2x_2^2+2x_1\left(x_1+x_2\right)}{x_1+x_2}\)
\(=\dfrac{2x_2^2+2x_1^2+2x_1x_2}{x_1+x_2}=\dfrac{2\left(x_1^2+x_2^2+x_1x_2\right)}{x_1+x_2}\)
\(=\dfrac{2\left[\left(x_1+x_2\right)^2-2x_1x_2+x_1x_2\right]}{x_1+x_2}\)
\(=\dfrac{2\left[\left(\dfrac{7}{3}\right)^2-\left(-\dfrac{1}{3}\right)\right]}{\dfrac{7}{3}}=\dfrac{\dfrac{104}{9}}{\dfrac{7}{3}}=\dfrac{104}{21}\)
