giải pt \(\frac{120}{x}+\frac{120}{x+10}=7\)
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Áp dụng hệ thức Vi-ét,ta có : \(\hept{\begin{cases}x_1+x_2=-2\left(m-1\right)\\x_1x_2=4m\end{cases}}\)
Ta có : \(4x_1^2\left(1+x_2\right)+4x_2\left(1+x_1\right)+x_1^2x_2^2=36\)
\(\Rightarrow4\left(x_1^2+x_2^2\right)+4x_1x_2\left(x_1+x_2\right)+x_1^2x_2^2=36\)
\(\Rightarrow4\left[\left(x_1+x_2\right)^2-2x_1x_2\right]+4x_1x_2\left(x_1+x_2\right)+x_1^2x_2^2=36\)
thay vào rồi tìm m thôi
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) \(G=\frac{x_1-3}{x_1^2}+\frac{x_2-3}{x_2^2}=\left(\frac{1}{x_1}+\frac{1}{x_2}\right)-3\left(\frac{1}{x_1^2}+\frac{1}{x_2^2}\right)\)
\(=\frac{x_1+x_2}{x_1x_2}-3\frac{x_1^2+x_2^2}{x_1^2.x_2^2}=\frac{5}{3}-3.\frac{19}{3^2}=-\frac{14}{3}\)
\(\sqrt{32}+10\sqrt{7}+\sqrt{32}-10\sqrt{7}\)
\(=\left(\sqrt{32}+\sqrt{32}\right)+\left(10\sqrt{7}-10\sqrt{7}\right)\)
\(=\sqrt{16\times2}\times2\)
\(=\sqrt{\left(4\right)^2\times2}\times2\)
\(=4\sqrt{2}\times2\)
\(=8\sqrt{2}\)
\(\frac{120}{x}+\frac{120}{x+10}=7\)
\(\Leftrightarrow\frac{120\left(x+10\right)}{x\left(x+10\right)}+\frac{120x}{x\left(x+10\right)}=\frac{7x\left(x+10\right)}{x\left(x+10\right)}\)
\(\Leftrightarrow120\left(x+10\right)+120x=7x\left(x+10\right)\)
\(\Leftrightarrow120x+1200+120x=7x^2+70x\)
\(\Leftrightarrow240x+1200=7x^2+70x\)
\(\Leftrightarrow240x+1200-7x^2-70x=0\)
\(\Leftrightarrow170x+1200-7x^2=0\)
\(\Leftrightarrow-7x^2+170x+1200=0\)
\(\Leftrightarrow7x^2+40x-210x-1200=0\)
\(\Leftrightarrow x\left(7x+40\right)-30\left(7x+40\right)=0\)
\(\Leftrightarrow\left(x-30\right)\left(7x+40\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=30\\7x=-40\end{cases}\Leftrightarrow\orbr{\begin{cases}x=30\\x=-\frac{40}{7}\end{cases}}}\)