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Tự xử lí delta nha
Theo vi-et: \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1.x_2=-\left(m-1\right)\left(m-3\right)\end{matrix}\right.\)
Theo đề: \(\frac{1}{4}.\left(2m\right)^2-\left(m-1\right)\left(m-3\right)-2.2m+3=0\)
<=> \(m^2-m^2+4m-3-4m+3=0\) (TM)
Vậy vs mọi m thỏa delta thì ...
Tự xử lí delta nha
Ta có: \(\frac{1}{x_1}+\frac{1}{x_2}=2\) <=> \(\frac{x_1+x_2}{x_1.x_2}=2\) <=> 2.x1.x2 = x1 + x2
Theo vi-et: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1.x_2=1-m\end{matrix}\right.\)
Theo đề: 2.(1 - m) = 2(m + 1)
<=> 2 - 2m = 2m + 2
<=> 4m = 0
<=> m = 0 (đối chiếu ĐK)
Vậy ...
Ta có: \(x^2-5x+3=0\)
Áp dụng định lí viet ta có: \(\hept{\begin{cases}x_1+x_2=5\\x_1x_2=3\end{cases}}\)
a) \(A=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=5^2-2.3=19\)
b) \(B=x_1^3+x_2^3=\left(x_1+x_2\right)^3-3\left(x_1+x_2\right)x_1x_2=5^3-3.5.3=80\)
c) \(C=\left|x_1-x_2\right|\)>0
=> \(C^2=x_1^2+x_2^2-2x_1x_2=19-2.3=13\)
=> C = căn 13
d) \(D=x_2+\frac{1}{x_1}+x_1+\frac{1}{x_2}=\left(x_1+x_2\right)+\frac{x_1+x_2}{x_1x_2}=5+\frac{5}{3}=5\frac{5}{3}\)
e) \(E=\frac{1}{x_1+3}+\frac{1}{x_2+3}=\frac{\left(x_1+x_2\right)+6}{x_1x_2+3\left(x_1+x_2\right)+9}=\frac{5+6}{3+3.5+9}=\frac{11}{27}\)
g) \(G=\frac{x_1-3}{x_1^2}+\frac{x_2-3}{x_2^2}=\left(\frac{1}{x_1}+\frac{1}{x_2}\right)-3\left(\frac{1}{x_1^2}+\frac{1}{x_2^2}\right)\)
\(=\frac{x_1+x_2}{x_1x_2}-3\frac{x_1^2+x_2^2}{x_1^2.x_2^2}=\frac{5}{3}-3.\frac{19}{3^2}=-\frac{14}{3}\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=\sqrt{5}\\x_1x_2=1\end{matrix}\right.\)
\(A=\left(x_1+x_2\right)^2-5x_1x_2=\left(\sqrt{5}\right)^2-5.1=0\)
\(B=\frac{1}{\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)}=\frac{1}{\left(\sqrt{5}\right)^3-3.1.\sqrt{5}}=\frac{1}{2\sqrt{5}}\)
\(C=\frac{x_1+x_2}{x_1x_2}=\sqrt{5}\)
\(D=\frac{x_1^2+x_2^2}{\left(x_1x_2\right)^2}=\frac{\left(x_1+x_2\right)^2-2x_1x_2}{\left(x_1x_2\right)^2}=\frac{5-2}{1^2}=3\)
\(E=\sqrt{x_1x_2}\left(\sqrt{x_1}+\sqrt{x_2}\right)\Rightarrow E^2=x_1x_2\left(x_1+x_2+2\sqrt{x_1x_2}\right)\)
\(\Rightarrow E^2=1\left(\sqrt{5}+2.1\right)\Rightarrow E=\sqrt{2+\sqrt{5}}\)
\(F=\frac{3\left(x_1+x_2\right)+5x_1x_2}{x_1x_2\left(x_1^2+x_2^2\right)}=\frac{3\left(x_1+x_2\right)-5x_1x_2}{x_1x_2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]}=\frac{3\sqrt{5}-5}{3}\)