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Theo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=6;x_1x_2=\frac{c}{a}=-4\)
\(\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1x_2\)
\(=6^2-4\cdot\left(-4\right)=36+16=52\)
=>\(x_1-x_2=\pm2\sqrt{13}\)
\(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\)
\(=6^2-2\cdot\left(-4\right)=36+8=44\)
TH1: \(x_1-x_2=2\sqrt{13}\)
\(x_1^4-x_2^4\)
\(=\left(x_1^2+x_2^2\right)\cdot\left(x_1^2-x_2^2\right)\)
\(=\left(x_1^2+x_2^2\right)\cdot\left(x_1+x_2\right)\left(x_1-x_2\right)\)
\(=44\cdot6\cdot2\sqrt{13}=528\sqrt{13}\)
TH2: \(x_1-x_2=-2\sqrt{13}\)
\(x_1^4-x_2^4\)
\(=\left(x_1^2+x_2^2\right)\cdot\left(x_1^2-x_2^2\right)\)
\(=\left(x_1^2+x_2^2\right)\cdot\left(x_1+x_2\right)\left(x_1-x_2\right)\)
\(=44\cdot6\cdot\left(-2\sqrt{13}\right)=-528\sqrt{13}\)
Theo định lí Viet \(\left\{{}\begin{matrix}x_1+x_2=5\\x_1x_2=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y_1+y_2=x_1^4+x_2^4=\left[\left(x_1+x_2\right)^2-2x_1x_2\right]^2-2x_1^2x_2^2=727\\y_1y_2=x_1^4x_2^4=1\end{matrix}\right.\)
Phương trình cần tìm có dạng \(ax^2+bx+c=0\left(1\right)\)
\(\Rightarrow\left\{{}\begin{matrix}-\dfrac{b}{a}=727\\\dfrac{c}{a}=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}b=-727a\\c=a\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow ax^2-727ax+a=0\)
\(\Leftrightarrow x^2-727x+1=0\)
\(y=\frac{1}{x^2+\sqrt{x}}\frac{\left(\int^4_2^5^2_5_8\vec{\log_1\Rightarrow\beta}\right)}{100462}\)
a: \(D=\frac{10}{100}+\frac{10}{150}+\frac{10}{210}+\cdots+\frac{10}{1200}\)
\(=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\cdots+\frac{1}{120}\)
\(=\frac{2}{20}+\frac{2}{30}+\cdots+\frac{2}{240}=2\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\cdots+\frac{1}{15\cdot16}\right)\)
\(=2\left(\frac14-\frac15+\frac15-\frac16+\cdots+\frac{1}{15}-\frac{1}{16}\right)=2\left(\frac14-\frac{1}{16}\right)=2\cdot\frac{3}{16}=\frac38\)
b: \(E=1\cdot2+2\cdot3+\cdots+99\cdot100\)
\(=1\left(1+1\right)+2\left(2+1\right)+\cdots+99\left(99+1\right)\)
\(=\left(1^2+2^2+\cdots+99^2\right)+\left(1+2+\cdots+99\right)\)
\(=\frac{99\left(99+1\right)\left(2\cdot99+1\right)}{6}+\frac{99\left(99+1\right)}{2}=\frac{99\cdot100\cdot199}{6}+99\cdot\frac{100}{2}\)
\(=33\cdot50\cdot199+99\cdot50\)
\(=33\cdot50\cdot\left(199+3\right)=33\cdot50\cdot202=33\cdot101\cdot100=333300\)
c: \(F=1^2+2^2+\cdots+98^2\)
\(=\frac{98\left(98+1\right)\left(2\cdot98+1\right)}{6}=\frac{98\cdot99\cdot197}{6}=49\cdot33\cdot197=318549\)
TL
10-2.3+4-2=6
HT
10-2.3+4-2 = 10-6+2 = 4=2=6
kết quả = 6 nha
chúc bé học tốt