(x1+1)(x2+2)(x1+3)(x2+4)=15
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NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 4 2022
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{3}{2}\\x_1x_2=-\dfrac{1}{2}\end{matrix}\right.\)
\(A=\dfrac{1}{x_1-3}+\dfrac{1}{x_2-3}=\dfrac{x_2-3+x_1-3}{\left(x_1-3\right)\left(x_2-3\right)}=\dfrac{x_1+x_2-6}{x_1x_2-3\left(x_1+x_2\right)+9}\)
\(=\dfrac{\dfrac{3}{2}-6}{-\dfrac{1}{2}-3.\dfrac{3}{2}+9}=...\) (em tự bấm máy)
\(B=x_1^2x_2-4-x_1x_2+x_1x_2^2=x_1x_2\left(x_1+x_2\right)-4-x_1x_2\)
\(=-\dfrac{1}{2}.\dfrac{3}{2}-4-\left(-\dfrac{1}{2}\right)=...\)
\(C=1-\left(x_1^2+x_2^2\right)=1-\left(x_1+x_2\right)^2+2x_1x_2=1-\left(\dfrac{3}{2}\right)^2+2.\left(-\dfrac{1}{2}\right)=...\)
\(D=x_1^3x_2^3+x_1^3+x_2^3=\left(x_1x_2\right)^3+\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\)
\(=\left(-\dfrac{1}{2}\right)^3+\left(\dfrac{3}{2}\right)^3-3.\left(-\dfrac{1}{2}\right).\dfrac{3}{2}=...\)
13 tháng 11 2021
\(1,\Leftrightarrow x^2+10x+25=x^2-4x-21\\ \Leftrightarrow14x=-46\\ \Leftrightarrow x=-\dfrac{23}{7}\\ 2,\Leftrightarrow x^3+8=15+x^3+2x\\ \Leftrightarrow2x=-7\Leftrightarrow x=-\dfrac{7}{2}\\ 3,\Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x=-3\\ 4,\Leftrightarrow x^3-9x^2+27x-27=0\\ \Leftrightarrow\left(x-3\right)^3=0\\ \Leftrightarrow x-3=0\Leftrightarrow x=3\\ 5,\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\\ \Leftrightarrow-12x=24\Leftrightarrow x=-2\\ 6,\Leftrightarrow x^2-3x+5x-15=0\\ \Leftrightarrow\left(x-3\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
1 tháng 2 2022
a: \(\left\{{}\begin{matrix}x_1+x_2=8\\x_1x_2=6\end{matrix}\right.\)
\(D=x_1^4-x_2^4=\left(x_1+x_2\right)\left(x_1-x_2\right)\left(x_1^2+x_2^2\right)\)
\(=8\cdot\left[\left(x_1+x_2\right)^2-2x_1x_2\right]\cdot\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}\)
\(=8\cdot\left[8^2-2\cdot6\right]\cdot\sqrt{8^2-4\cdot6}\)
\(=8\cdot52\cdot2\sqrt{10}=832\sqrt{10}\)
b: \(E=\left(x_1^2+x_2^2\right)^2-2x_1^2\cdot x_2^2\)
\(=52^2-2\cdot\left(x_1\cdot x_2\right)^2=52^2-2\cdot6^2=2632\)
c: \(F=\dfrac{3x_2^2+3x_1^2}{\left(x_1\cdot x_2\right)^2}=\dfrac{3\cdot52}{6^2}=\dfrac{13}{3}\)
HH
1
24 tháng 6 2022
\(x_1^2-x_2^2=\left(x_1-x_2\right)\left(x_1+x_2\right)\)
\(=\pm\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}\cdot\left(x_1+x_2\right)\)
\(x_1^3-x_2^3\)
\(=\left(x_1-x_2\right)^3+3x_1x_2\left(x_1-x_2\right)\)
\(=\pm\left[\left(x_1+x_2\right)^2-4x_1x_2\right]^3+3\cdot x_1x_2\cdot\pm\left(\left(x_1+x_2\right)^2-4x_1x_2\right)\)