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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 \(C=\dfrac{11}{12}\)
Theo viet: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-b}{a}=\dfrac{1}{1}=1\\x_1x_2=\dfrac{c}{a}=-\dfrac{3}{1}=-3\end{matrix}\right.\)
a
\(A=x_1^2+x_2^2=x_1^2+2x_1x_2+x_2^2-2x_1x_2\)
\(=\left(x_1+x_2\right)^2-2x_1x_2=1^2-2.\left(-3\right)=1+6=7\)
b
\(B=x_1^2x_2+x_1x_2^2=x_1x_2\left(x_1+x_2\right)=\left(-3\right).1=-3\)
c
\(C=\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_2}{x_1x_2}+\dfrac{x_1}{x_1x_2}=\dfrac{x_1+x_2}{x_1x_2}=\dfrac{1}{-3}=-\dfrac{1}{3}\)
d
\(D=\dfrac{x_2}{x_1}+\dfrac{x_1}{x_2}=\dfrac{x_2^2}{x_1x_2}+\dfrac{x_1^2}{x_1x_2}=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1x_2}=\dfrac{1^2-2.\left(-3\right)}{-3}=\dfrac{1+6}{-3}=\dfrac{7}{-3}=-\dfrac{3}{7}\)
Lời giải:
Áp dụng hệ thức Viet:
$x_1+x_2=\frac{-4}{3}; x_1x_2=\frac{1}{3}$
Khi đó:
\(B=\frac{x_1}{x_2-1}+\frac{x_2}{x_1-1}=\frac{x_1(x_1-1)+x_2(x_2-1)}{(x_1-1)(x_2-1)}\)
\(=\frac{x_1^2+x_2^2-(x_1+x_2)}{x_1x_2-(x_1+x_2)+1}=\frac{(x_1+x_2)^2-2x_1x_2-(x_1+x_2)}{x_1x_2-(x_1+x_2)+1}\)
\(=\frac{(\frac{-4}{3})^2-2.\frac{1}{3}-\frac{-4}{3}}{\frac{1}{3}-\frac{-4}{3}+1}=\frac{11}{12}\)
a. Em tự giải
b.
\(\Delta=4-3\left(m+5\right)>0\Rightarrow m< -\dfrac{11}{3}\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{4}{3}\\x_1x_2=\dfrac{m+5}{3}\end{matrix}\right.\)
Để biểu thức đề bài xác định \(\Rightarrow x_1x_2\ne0\Rightarrow m\ne-5\)
\(\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{4}{7}\) \(\Leftrightarrow\dfrac{x_1+x_2}{x_1x_2}=\dfrac{4}{7}\)
\(\Leftrightarrow\dfrac{4}{m+5}=\dfrac{4}{7}\)
\(\Rightarrow m+5=7\)
\(\Rightarrow m=2\) (ktm)
Vậy ko tồn tại m thỏa mãn yêu cầu đề bài
Có cả điều kiện delta lúc đầu nữa em, \(m< -\dfrac{11}{3}\) mà \(m=2>-\dfrac{11}{3}\) nên không thỏa mãn
,có \(ac< 0\)=>pt đã cho luôn có 2 nghiệm phân biệt
vi ét \(=>\left\{{}\begin{matrix}x1+x2=2\\x1x2=-1\end{matrix}\right.\)
a,\(A=\left(x1+x2\right)^2-2x1x2=.....\) thay số tính
b,\(B=\left(x1+x2\right)^3-3x1x2\left(x1+x2\right)=.......\)
c,\(C=x1^{2^2}+x2^{2^2}=\left(x1^2+x2^2\right)^2-2\left(x1x2\right)^2=\left[\left(x1+x2\right)^2-2x1x2\right]^2-2\left(x1x2\right)^2=....\)
\(D=x1x2\left(x1+x2\right)=.....\)
\(x1,x2\ne0=>E=\dfrac{\left(x1+x2\right)^3-3x1x2\left(x1+x2\right)}{x1x2}=...\)
\(F=\sqrt{\left(x1-x2\right)^2}=\sqrt{\left(x1+x2\right)^2-4x1x2}=....\)
\(x1,x2\ne-1=>G=\dfrac{\left(x1+x2\right)^2-2x1x2+x1x2}{x1x2+x1+X2+1}=...\)
\(x1,x2\ne0=>H=\left(\dfrac{x1x2+2}{x2}\right)\left(\dfrac{x1x2+2}{x1}\right)=\dfrac{\left(x1x2+2\right)^2}{x1x2}\)
\(=\dfrac{\left(x1x2\right)^2+4x1x2+4}{x1x2}=..\)
Câu 1: B
Câu 2: C
Câu 3: C
Câu 4: D
Câu 5: C
Câu 6: B
Câu 7: B
Câu 8: C
Câu 9: C
Câu 10: C