奥数课笔记

简介

暂时记录奥数课上的笔记，由于过于垃圾，不放在主页了。同学：beautiful chicken。

题目

\[a^2+a=210 \]

\[\begin{aligned} 解:x&=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ &=\frac{-1\pm \sqrt{841}}{2} \\ &=\frac{-1\pm 29}{2} \end{aligned} \]

\[x_1=14,x_2=-15 \]

\[\frac{x-1}{90}+\frac{x}{91}+\frac{x+1}{92}=3 \]

\[\begin{aligned} 解：\frac{x-91+90}{90}+\frac{x-91+91}{91}+\frac{x-91+92}{92}&=3 \\ \frac{x-91}{90}+1+\frac{x-91}{91}+1+\frac{x-91}{92}+1&=3 \end{aligned} \]

\[\because 1+1+1=3 \]

\[\begin{aligned} \therefore \frac{x-91}{90}+\frac{x-91}{91}+\frac{x-91}{92}&=0 \\ x-91&=0 \\ x&=0 \end{aligned} \]

\[4x^2-6x+n~为完全平方式的倍数，求~n \]

\[\begin{aligned} &\ \ \ \ \ 4x^2-6x+n \\ &=(2x)^2-1.5\times 2\times 2x+1.5^2-1.5^2+n \\ &=[(2x)^2-1.5\times 4x+1.5^2]+n-1.5^2 \\ &=(2x+1.5)^2+n-1.5^2 \end{aligned} \]

\[\because (2x+1.5)^2+n-1.5^2~为完全平方式的倍数 \\ \]

\[\begin{aligned} \therefore n-1.5^2&=0 \\ n&=\frac{3}{2}\times \frac{3}{2} \\ n&=\frac{9}{4} \end{aligned} \]

\[\left\{ \begin{aligned} &5x+y=12 \\ &4x-2y=5 \end{aligned} \right. \]

\[\begin{aligned} 解：10x+2y&=24 \\ 4x-2y+10x+2y&=24+5 \\ 14x&=29 \\ x&=\frac{29}{14} \\ \end{aligned} \]

\[\begin{aligned} \frac{29}{14}\times 4&=\frac{145}{14} \\ y&=\frac{23}{14} \end{aligned} \]

\[\left\{ \begin{aligned} &x+y+z=8 \\ &3x-2y+z=5 \\ &2x-y-z=1 \end{aligned} \right. \]

\[\begin{aligned} 解：6x+6y+6z&=48 \\ 6x-4y+2z&=10 \\ 6x-3y-3z&=3 \end{aligned} \]

\[\begin{aligned} 4y+6z+4y-2z&=38 \\ 10y+4z&=38 \end{aligned} \]

\[\begin{aligned} -4y+2z+3y+3z&=7 \\ -y+5z&=7 \\ -10y+50z&=70 \end{aligned} \]

\[\begin{aligned} 10y+4z-10y+50z&=108 \\ 54z&=108 \\ z&=2 \end{aligned} \]

\[\begin{aligned} 5\times 2&=10 \\ 7-10&=-3 \\ -y&=-3 \\ y&=3 \end{aligned} \]

\[\begin{aligned} 8-(2+3)&=3 \\ x&=3 \end{aligned} \]

\[\left[ \begin{array}{ccc|c} 3 & 2 & 5 & 1 \\ 1 & 3 & 2 & 2 \\ 5 & 8 & 7 & 3 \end{array} \right] \]

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\[\left[ \begin{array}{ccc|c} 1 & 3 & 2 & 2 \\ 0 & 7 & 1 & 5 \\ 0 & 7 & 3 & 7 \end{array} \right] \]

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\[\left[ \begin{array}{ccc|c} 7 & 0 & 11 & -1 \\ 0 & 7 & 1 & 5 \\ 0 & 7 & 3 & 7 \end{array} \right] \]

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\[\left[ \begin{array}{ccc|c} 7 & 0 & 11 & -1 \\ 0 & 7 & 1 & 5 \\ 0 & 0 & 2 & 2 \end{array} \right] \]

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\[\left[ \begin{array}{ccc|c} 14 & 0 & 0 & -24 \\ 0 & 7 & 1 & 5 \\ 0 & 0 & 2 & 2 \end{array} \right] \]

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\[\left[ \begin{array}{ccc|c} 14 & 0 & 0 & -24 \\ 0 & 14 & 0 & 8 \\ 0 & 0 & 2 & 2 \end{array} \right] \]

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\[\left[ \begin{array}{ccc|c} 7 & 0 & 0 & -12 \\ 0 & 7 & 0 & 4 \\ 0 & 0 & 1 & 1 \end{array} \right] \]

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\[\left[ \begin{array}{ccc|c} 1 & 0 & 0 & -\dfrac{12}{7} \\ 0 & 1 & 0 & \dfrac{4}{7} \\ 0 & 0 & 1 & 1 \end{array} \right] \]

\[\begin{aligned} ax^2+bx+c&=0 \\ 解：x^2+\frac{bx}{a}+\frac{c}{a}&=0 \\ x^2+\frac{bx}{a}&=-\frac{c}{a} \\ x^2+\frac{b}{a}x&=-\frac{c}{a} \\ x^2+2\cdot\frac{b}{2a}x&=-\frac{c}{a} \\ x^2+2\cdot\frac{b}{2a}x+(\frac{b}{2a})^2&=-\frac{c}{a}+(\frac{b}{2a})^2 \\ (x+\frac{b}{2a})^2&=-\frac{c}{a}+(\frac{b}{2a})^2 \\ (x+\frac{b}{2a})^2&=-\frac{c}{a}+\frac{b^2}{4a^2} \\ (x+\frac{b}{2a})^2&=-\frac{c\cdot 4a}{4a^2}+\frac{b^2}{4a^2} \\ (x+\frac{b}{2a})^2&=\frac{b^2-4ac}{4a^2} \\ x&=\pm \sqrt{\frac{b^2-4ac}{4a^2}}-\frac{b}{2a}\\ x&=\frac{\pm\sqrt{b^2-4ac}}{2a}-\frac{b}{2a} \\ x&=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \end{aligned} \]